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NCERT Exemplar Class 9 Maths Solutions Chapter 14 Statistics and Probability

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NCERT Exemplar Class 9 Maths Solutions Chapter 14 Statistics and Probability

Edited By Safeer PP | Updated on Aug 31, 2022 03:40 PM IST

NCERT exemplar Class 9 Maths solutions chapter 14 provides the basics of statistics and probability. The concepts used here will be useful in higher classes too. These NCERT exemplar Class 9 Maths chapter 14 solutions are designed in a manner to provide the students with a correct and exhaustive approach towards the problems while practicing the questions of NCERT Book for Class 9 Maths.

The NCERT exemplar Class 9 Maths chapter 14 solutions explores all the concepts of statistics and probability recommended by the CBSE for Class 9. NCERT exemplar Class 9 Maths solutions chapter 14 are prepared by the highly experienced subject experts at Careers 360.

Also read - NCERT Solutions for Class 9 Maths

Question:1

The class mark of the class 90-120 is :

(A) \; 90

(B) \; 105

(C) \; 115

(D) \; 120

Answer: B
Classmark is the mid-value or the central value of a class
It is calculated as follows:
\frac{\text {upper limit + lower limit}}{2}
\Rightarrow Class\ mark=\frac{90+120}{2}=\frac{210}{2}=105
The class mark of class 90-120 is 105 i.e. option B

Question:2

\text{The range of the data} :\ 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 \ is
(A) \; 10
(B) \; 15
(C) \; 18
(D) \; 26

Answer : D
The range is the difference between the largest (maximum value) and the smallest number (minimum value) of a data
25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20
Here largest number=32
Smallest number =6
Range =32-6=26
Hence, 26 is the range of the data.
Therefore option (D) is correct.

Question:3

In a frequency distribution, the mid-value of a class is 10 and the width of the class is 6. The lower limit of the class is :
(A) \; 6
(B) \; 7
(C) \; 8
(D) \; 12

Answer : B
Let x and y be the upper and lower class limit in the frequency distribution.
\text{The mid-value of given class}=\frac{x+y}{2}
Here, the mid-value of the class is given as 10.
\frac{x+y}{2}=10
x+y=20\; \; \; \; \; \; \; \; \; \; \; \; ....(1)
Also, given that width of the class is 6
x-y=6\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(2)
Now, adding equation (1) and (2),
x+y+x-y=20+6
x=\frac{26}{2}=13
Putting x=13 in equation (1)
x+y=20
13+y=20
y=20-13=7
Hence, the lower limit of the class is 7
Therefore option (B) is correct.

Question:4

The width of each of five continuous classes in a frequency distribution is 5 and the lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
(A) \; 15
(B) \; 25
(C) \; 35
(D) \; 40

Answer: C
Here, lower class-limit of the lowest class is 10 and the width of the class is 5
So, starting from 10 the first class will be:
10 - 15
Now we have been given that this is the lowest class. So next 4 continuous classes will be:
\\15 - 20\\ 20 - 25\\ 25 - 30\\ 30 - 35
As the total number of classes is 5, so 35 is the upper limit of the highest class.
Therefore option (C) is correct.

Question:5

Let m be the mid-point and l be the upper-class limit of a class in a continuous frequency distribution. The lower class limit of the class is :
(A)\; 2m+l
(B)\; 2m-l
(C)\; m-l
(D)\; m-2l

Answer : B
Given, l is the upper limit of class
Let n be the lower class limit
Also given, m is the mid-point of this class.
Mid value of a class is calculated as follows:
\frac{\text {upper limit+lower limit}}{2}
Then, mid-point
m=\frac{n+l}{2}
2m=n+l
Therefore, n=2m-l
Therefore option (B) is correct.

Question:6

The class marks of a frequency distribution are given as follows: 15, 20, 25, ..... The class corresponding to the class mark 20 is :
(A)\; 12.5-17.5
(B)\; 17.5-22.5
(C)\; 18.5-21.5
(D)\; 19.5-20.5

Answer : B
The class marks of a frequency distribution are given as: 15, 20, 25, ...
Class size for the given frequency distribution is: 20-15=5
To find the class corresponding to the class mark 20:
We will use the formula
\text{Upper limit}=\text {class mark}+\frac{\text {class size}}{2}
=20+\frac{5}{2}=\frac{40+5}{2}=\frac{45}{2}
=22.5
\text{Lower limit}=\text {class mark}-\frac{\text {class size}}{2}
=20-\frac{5}{2}=\frac{35}{2}=17.5
Hence, 17.5-22.5 is the corresponding class to the classmark 20
Therefore option (B) is correct.

Question:7

In the class intervals 10-20, 20-30, the number 20 is included in :
(A) 10-20
(B) 20-30
(C) both the intervals
(D) none of these intervals

Answer : B
Here, in these class intervals 10-20,20-30, the number 20 is included in 20-30,
Because in 10-20 there are only (10, 11....19)
And in 20-30 there are only (20, 21,.....29)
Hence we don’t count the upper limit in every class interval.
The number is always included in the lower limit of the class interval.
Therefore option (B) is correct.

Question:8

A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class intervals is constructed for the following data :
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.
The frequency of the class 310-330 is:
\\(A)\; 4\\(B)\; 5\\(C)\; 6\\(D)\; 7

Answer : C
Firstly we will arrange the given data into intervals of equal sizes as mentioned i.e., 210-230, 230-250 and so on.
The given data can be arranged as:
\\210-230: 220, 210, 215\\ 230-250: 242, 240, 236\\ 250-270: 258, 268, 258
\\270-290: 272, 278\\ 290-310: 290, 304, 300, 292\\ 310-330: 310, 310, 320, 319, 318, 316\\ 330-350: 342, 330\\ 350-370: 368, 354
370-390: No element
390-410: 402, 406, 406,

So, we can see that 6 is the frequency of 310-330..
Therefore option (C) is correct.

Question:9

A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data :
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20,15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44.
The number of classes in the distribution will be :
\\(A) 9 \\ (B) 10 \\ (C) 11 \\ (D) 12

Answer: B
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20,15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44.
The minimum number in the data is 14 and the maximum number is 112.
Classes of equal sizes using 63-72 are to be constructed.
So, the intervals are as following:
13-22, 23-32, 33-42, 43-52, 53-62, 63-72, 73-82, 83-92, 93-102, 103-112
The number of class interval is 10.
Therefore option (B) is correct.

Question:10

To draw a histogram to represent the following frequency distribution :

Class interval
5-10
10-15
15-25
25-45
45-75
Frequency
6
12
10
8
15
the adjusted frequency for the class 25-45 is :
\\(A) 6 \\ (B) 5\\ (C) 3 \\ (D) 2
Answer : D
In the given data, we have to find the width to adjust the frequency.
We have, width = 10 - 5 = 5
Width of 25-45 = 45 -25 = 20
So interval 25-45 includes 4 classes with class interval 5
\text {AF}=\frac{\text {given frequency}\times \text {lower limit}}{\text {Class width}}
=\frac{8 \times 5}{20}=\frac{8}{4}=2
Therefore option (D) is correct.

Question:11

The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is :

\\(A) 28\\ (B) 30\\ (C) 35\\ (D) 38

Answer : D
Let the numbers be
a+b+c+d+e
\Rightarrow Mean =\frac{a+b+c+d+e}{5}=30
\Rightarrow a+b+c+d+e=30 \times 5=150\; \; \; \; \; \; \; \; \; \; \; \; ....(i)
Let the excluded number be a
New mean =28
\Rightarrow \frac{b+c+d+e}{4}=28
\Rightarrow b+c+d+e=28\times 4=112
Putting the value of b+c+d+e in equation (i)
a+112=150
\Rightarrow a=150-112
\Rightarrow a=38\; \; \; \; \; \; \; \; \; \; \rightarrow\text{excluded number}
Therefore option (D) is correct.

Question:12

If the mean of the observations : x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
(A)\ 10\frac{1}{3}
(B) \ 10\frac{2}{3}
(C)\ 11\frac{1}{3}
(D)\ 11\frac{2}{3}

Answer:

In the given data the mean of the observations x, x + 3, x + 5, x + 7, x + 10 is given as 9
\Rightarrow Mean=\frac{x+x+3+x+5+x+7+x+10}{5}=9
\Rightarrow 5x+25=45
\Rightarrow 5x=45-25
\Rightarrow 5x=20
\Rightarrow x=\frac{20}{5}=4
\therefore Last three observation are
x + 5, x + 7, x + 10
= 4 + 5, 4 + 7, 4 + 10 = 9, 11, 14
\therefore So, the mean of the last three observations
\Rightarrow \frac{9+11+14}{3}=\frac{34}{3}=11\frac{1}{3}
Therefore option (C) is correct.

Question:13

\text{If x represents the mean of n observations }\sum_{i=l}^{n}\left ( x_{i}-\bar{x} \right )x_{1},x_{2}......x_{n},\text{then value of} \sum_{i=l}^{n}\left ( x_{i}-\bar{x} \right )\text{ is}
(A) -1
(B) 0
(C) 1
(D) n-1

Answer : B
Here we are given,
\sum_{i=l}^{n}\left ( x_{i}-\bar{x} \right )
\Rightarrow \bar{x}\text{ is the mean of all observations. }
\text {mean}=\frac{\text {sum of observations}}{\text {Number of observations}}
\bar{x}=\frac{\left ( x_{1}+x_{2}+x_{3}+......+x_{n} \right )}{n}
\Rightarrow n\bar{x}=\left ( x_{1}+x_{2}+.....+x_{n} \right )
\Rightarrow n\bar{x}=\sum_{i=1}^{n}x_{i}
\Rightarrow \sum_{i=1}^{n}x_{i}-n\bar{x}=0
\Rightarrow \sum_{i=1}^{n} \left ( x_{1}-\bar{x} \right )=0
Here the value of
\sum_{i=1}^{n} \left ( x-\bar{x} \right )=0
Therefore option (B) is correct.

Question:14

If each observation of the data is increased by 5, then their mean
(A) remains the same
(B) becomes 5 times the original mean
(C) is decreased by 5
(D) is increased by 5

Answer : D
Let x_{1},x_{2}......x_{n} be the n observations.
\Rightarrow old mean
\bar{x}_{old}=\frac{\sum_{i=1}^{n}x_{i}}{n} \; \; \; \; \; \; \; \; \; ..(i)
Now adding 5 to each observation, the new mean becomes.
\bar{x}_{new}=\frac{\left ( x_{1}+5 \right )+\left ( x_{2}+5 \right )+....+\left ( x_{n}+5 \right )}{n}
\bar{x}_{new}=\frac{\left ( x_{1}+x_{2}+.....+x_{n} \right )+\left ( 5+5+5..... \right ) \text {n times}}{n}
\bar{x}_{new}=\frac{\left ( x_{1}+x_{2}+.....+x_{n} \right )+5n}{n}
\bar{x}_{new}=\frac{\sum_{i=1}^{n}x_{i}}{n}+5=\bar{x}_{old}+5 (from eq. (i))
\bar{x}_{new}=\bar{x}_{old}+5
Hence, the new mean is increased by 5.
Therefore option (D) is correct.

Question:15

Let \bar{x} be the mean of x_1,\ x_2........., x_n and \bar{y} the mean of y_{1}, y_{2}, ... , y_{n}. If \bar{z} is the mean of x_1,\ x_2........., x_n, y_{1}, y_{2}, ... , y_{n} then \bar{z} is equal to
(A) \bar{x}+\bar{y}
(B) \frac{\bar{x}+\bar{y}}{2}
(C) \frac{\bar{x}+\bar{y}}{n}
(D) \frac{\bar{x}+\bar{y}}{2n}

Answer : B
\bar{x} is the mean of x_{1},x_{2},.....,x_{n} then
\bar{x}=\frac{x_{1}+x_{2}+.....+x_{n}}{n}
\bar{y} is the mean of y_{1},y_{2},......y_{n} then
\bar{y}=\frac{y_{1}+y_{2}+......y_{n}}{n}
\bar{z} is the mean of x_{1},x_{2},.......x_{n},y_{1},y_{2}............,y_{n}
\bar{z} = \frac{x_{1}+x_{2}+........+x_{n}+y_{1}+y_{2}+.....y_{n}}{2n}
\bar{z} = \frac{1}{2}\left ( \frac{x_{1}+x_{2}+....+x_{n}+y_{1}+y_{2}+.......y_{n}}{n} \right )
\bar{z} = \frac{1}{2}\left ( \frac{x_{1}+x_{2}+....+x_{n}}{n} +\frac{y_{1}+y_{2}+.......y_{n}}{n}\right )
\bar{z} = \frac{\bar{x}+\bar{y}}{2}
Hence, the value of \bar{z} is \frac{\bar{x}+\bar{y}}{2}
Therefore option (B) is correct.

Question:16

If \bar{x} is \ the \ mean \ ofx_{1},x_{2},........,x_{n}, \text{then for} a\neq 0\text{,the mean of }ax_{1},ax_{2},........ax_{n},\frac{x_{1}}{a},\frac{x_{2}}{a},........,\frac{x_{n}}{a} is
(A) \left ( a+\frac{1}{a} \right )\bar{x}
(B) \left ( a+\frac{1}{a} \right )\frac{\bar{x}}{2}
(C)\left ( a+\frac{1}{a} \right )\frac{\bar{x}}{n}
(D)\frac{\left ( a+\frac{1}{a} \right )\bar{x}}{2n}

Answer : B
\frac{x_{1}+x_{2}...+x_{n}}{n}=\bar{x} \; \; \; \; \; \; \; \; ....(i)
Multiply both sides by a
\frac{ax_{1}+ax_{2}...+ax_{n}}{n}=a\bar{x} \; \; \; \; \; \; \; \; ....(i)
ax_{1}+ax_{2}......ax_{n}=na\bar{x}\; \; \; \; \; \; \; \; \; \; .....(ii)
Dividing (i) by a, we get
\frac{\frac{x_{1}}{a}+\frac{x_{2}}{a}.......+\frac{x_{n}}{a}}{n}=\frac{\bar{x}}{a}
\frac{x_{1}}{a}+\frac{x_{2}}{a}.......+\frac{x_{n}}{a}=\frac{n\bar{x}}{a}\; \; \; \; \; \; \; .....(iii)
The mean of
ax_{1}+ax_{2}.........+ax_{n}+ \frac{x_{1}}{a},\frac{x_{2}}{a}.......\frac{x_{n}}{a} \text{is :}
\frac{ax_{1}+ax_{2}+ax_{n}+\frac{x_{1}}{a}+\frac{x_{2}}{a}+.......+\frac{x_{n}}{a}}{2n}
from (ii) and (iii) we get
\frac{ax_{1}+ax_{2}+ax_{n}+\frac{x_{1}}{a}+\frac{x_{2}}{a}+.......+\frac{x_{n}}{a}}{2n}=\frac{\left ( na\bar{x}+n\frac{\bar{x}}{a} \right )}{2n}
=\frac{\bar{x}n}{2n}\left ( a+\frac{1}{a} \right )
=\frac{\bar{x}}{2}\left ( a+\frac{1}{a} \right )
Therefore option (B) is correct.

Question:17

If \bar{x}_{1},\bar{x}_{2},\bar{x}_{3},........\bar{x}_{n}, are the means of n groups with n_{1},n_{2},......n_{n} number of observations respectively, then the mean \bar{x} of all the groups taken together is given by :
(A) \sum_{i=1}^{n}n_{i}\bar{x}_{i}
(B) \frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{n^{2}}
(C) \frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{\sum_{i=1}^{n}n_{i}}
(D) \frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{2n}

Answer : C
\text {Mean is defined as}\; \bar{x}=\frac{\text {sum of observations}}{\text {Number of observations}}
\Rightarrow Sum of observations = (mean) (Number of observations).
Given : \bar{x}_{1},\bar{x}_{2},\bar{x}_{3},........\bar{x}_{n} are the means of n groups with n_{1},n_{2},......n_{n} number of observations respectively.
\sum x_{1}=\bar{x}_{1}n_{1}
\sum x_{2}=\bar{x}_{2}n_{2}
\sum x_{n}=\bar{x}_{n}n_{n}
Now all the groups are taken together
Sum of all the observations:
\sum x_{1}+\sum x_{2}+.......+\sum x_{n}=\bar{x}_{1}n_{1}+\bar{x}_{2}n_{2}+....+\bar{x}_{n}n_{n}
Mean of all observations:
\bar{x}=\frac{\bar{x}_{1}n_{1}+\bar{x}_{2}n_{2}+.....+\bar{x}_{n}n_{n}}{n_{1}+n_{2}+.......+n_{n}}=\frac{\sum_{i=1}^{n}n_{i}\bar{x}_{i}}{\sum_{i=1}^{n}n_{i}}
Therefore option (C) is correct.

Question:18

The mean of 100 observations is 50. If one of the observations which was 50 is replaced by 150, the resulting mean will be :
(A) 50.5
(B) 51
(C) 51.5
(D) 52

Answer : B
Given that mean of 100 observations is 50.
\Rightarrow \text {mean}=\frac{\text {sum of observation}}{\text {No.of observation}}
Here number of obervations 100
Let the sum of 50 observations be x
\Rightarrow 50=\frac{x}{100}
\Rightarrow x=50\times100=5000
Given that one observation 50 is replaced by 150
New \ sum=x-50+150=5000-50+150=5100
New \ mean=\frac{5100}{100}=51
Hence the mean is 51
Therefore option (B) is correct.

Question:19

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is :
(A) 46.5
(B) 49.5
(C) 53.5
(D) 56.5

Answer : D
Given that there are 50 numbers.
Let the numbers be
x,y,z.........\left ( 50\; nos. \right )
\text {Mean is defined as }=\frac{\text {sum of observations}}{\text {Number of observations}}
\text {Original mean }=\frac{(x+y+z+......)}{50}
Each number is subtracted from 53
New numbers =53-x,53-y,53-z......
Sum of new numbers =53-x+53-y+53-z+......
The new mean is given as -3.5
-3.5=\frac{53-x+53-y+53-z+.....}{50}
-3.5=\frac{\left ( 53 \right )\left ( 50 \right )-\left ( x+y+z+..... \right )}{50}
-3.5=\frac{(53)(50)}{50}-\frac{\left ( x+y+z+.... \right )}{50}
-3.5=53-\text {original mean}
Original mean =53+3.5=56.5
Therefore option (D) is correct.

Question:20

The mean of 25 observations is 36. Out of these observations if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is :
\\(A) 23 \\ (B) 36\\ (C) 38 \\ (D) 40

Answer : B
\text {Mean}=\frac{\text {Sum of observations}}{\text {No.of observation}}
The mean of 25 observations is 36.
36=\frac{\text {sum of 25 observations}}{25}
\Rightarrow Sum of 25 observations =900.
For first 13 observations, mean is given as 32
32=\frac{\text {sum of 13 observations}}{13}
\Rightarrow Sum of last 13 observations
=40 \times 13 = 520
Now, 13th observation = sum of 13 observation+sum of last 13 observation-sum of 25 observations.
\Rightarrow 13th observation =416+520-900=36
The value of 13th observation is 36
Therefore option (B) is correct.

Question:21

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
\\(A) 45\\ (B) 49.5\\ (C) 54\\ (D) 56

Answer : C
Given terms : 78, 56, 22, 34, 45, 54, 39, 68, 54, 84
To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
In ascending order:
22, 34, 39, 45, 54, 54, 56, 68, 78, 84
Number of terms
=10\left ( \text {even} \right )
\text {Median}=\text {average of }\left ( \frac{n}{2} \right )^{th}\; \text {and}\; \left ( \frac{n}{2}+1 \right )^{th} \text {term}
\left ( \frac{n}{2} \right )^{th}\text {term}=\left ( \frac{10}{2} \right )^{th}=5^{th}\text {term}=54
\text {Median}=\frac{\text {term}+\text {6th term}}{2}=\frac{108}{2}=54
Therefore option (C) is correct.

Question:22

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively :
(A) upper limits of the classes
(B) lower limits of the classes
(C) class marks of the classes
(D) upper limits of preceding classes

Answer : C
For drawing a frequency polygon of a continuous frequency distribution, we plot the frequency of the classes on the ordinates and the class marks of the classes on the abscissae.
Class mark is the mid value or the central value of a class
It is calculated as follows:
\frac{\text {upper limit+lower limit}}{2}
For example
Class interval
Class marks (x-coordinate)
Frequency (y-coordinate)
10-20
15
7
20-30
25
8
30-40
35
9
Therefore option (C) is correct.

Question:23

Median of the following numbers : 4, 4, 5, 7, 6, 7, 7, 12, 3 is
\\(A) 4\\ (B) 5\\ (C) 6\\ (D) 7

Answer : C
Given numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3
To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
Firstly, we will arrange the numbers in ascending order –
3, 4, 4, 5, 6, 7, 7, 7, 12
There are 9 number of observations.
For odd number of observations, middle term
=\left ( \frac{n+1}{2} \right )^{th}\text {term}=\left ( \frac{9+1}{2} \right )^{th}\text {term}=\left ( \frac{10}{2} \right )^{th}\text {term}=5^{th}\text {term}
5th term is 6, hence it is the required median.
Therefore option (C) is correct.

Question:24

Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is
(A) 14
(B) 15
(C) 16
(D) 17

Answer: B
Mode is the value that occurs the most number of times in a given set of values
Given data, 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15
Arranging in ascending order
14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20
Here, 15 occurs 5 times, which is the most number of times in the given set of values.
Mode =15
Therefore option (B) is correct.

Question:25

In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is :
\\(A) 0.5\\ (B) 0.6\\ (C) 0.7\\ (D) 0.8

Answer : D
We will use the formula of probability:-
\text {P}=\frac{\text {favourable outcome}}{\text {Total possible outcome}}
Total possible outcomes =642
Favourable outcomes =514
Required probability
=\frac{514}{642}=0.800
So if a person is selected at random, the probability that the person has a high school certificate is 0.8
Therefore option (D) is correct.

Question:26

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is:
\\(A) 0.25\\ (B) 0.50\\ (C) 0.75\\ (D) 0.80

Answer : C
In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips.
If a child is selected at random, the probability that he/she likes to eat potato chips is
=\frac{91}{364}
If a child is selected at random, the probability that he/she does not like to eat potato chips is
=1-\frac{91}{364}
=\frac{364-91}{361}=\frac{273}{361}=0.756
In the options given, the closest to 0.756 is 0.75
Therefore option (C) is correct.

Question:27

In a medical examination of students of a class, the following blood groups are recorded:

Blood Group
A
AB
B
O
Number of students
10
13
12
5
A student is selected at random from the class. The probability that he/she has blood group B, is:
(A)\ \frac{1}{4}
(B)\ \frac{13}{40}
(C)\ \frac{3}{10}
(D)\ \frac{1}{8}

Answer:

Answer : B

\text {P}=\frac{\text {favourable outcomes}}{\text {Total possible outcome}}
Number of students with blood group B =12
Total number of students =10+13+12+5
Required probability
=\frac{12}{10+13+12+5}=\frac{12}{40}=\frac{3}{10}
Therefore option (B) is correct.

Question:28

Two coins are tossed 1000 times and the outcomes are recorded as below :

Number of heads
2
1
0
Frequency
200
550
250
Based on this information, the probability for at most one head is:
(A)\ \frac{1}{5}
(B)\ \frac{1}{4}
(C)\ \frac{4}{5}
(D)\ \frac{3}{4}
Answer : C
Total number of outcomes =200+550+250=1000
At most one head = 0head+ 1 head
Number of favourable outcomes =250+550=800
Probability will be
=\frac{800}{1000}=\frac{4}{5}
Therefore option (C) is correct.

Question:29

80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below :

Life time (in hours)
300
500
700
900
1100
Frequency
10
12
23
25
10
One bulb is selected at random from the lot. The probability that its life is 1150 hours, is
(A)\ \frac{1}{80}
(B)\ \frac{7}{16}
(c)\ 0
(D)\ 1
Answer : C
\text {P}=\frac{\text {favourable outcome}}{\text {Total outcomes}}
Bulb’s life is given as 1150 hours.
From the table, we can see that no bulb has life as1150 hours.
So favourable outcomes =0
Required probability =\frac{0}{10+12+23+25+10}=0
Therefore option (C) is correct.

Question:30

80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below :

Life time (in hours)
300
500
700
900
1100
Frequency
10
12
23
25
10
The probability that bulbs selected randomly from the lot has life less than 900 hours is :
(A)\ \frac{11}{40}
(B)\ \frac{5}{16}
(C)\ \frac{7}{16}
(D) \ \frac{9}{16}

Answer: D
\text {P}=\frac{\text {favourable outcome}}{\text {Total outcomes}}
Bulb’s life is given as less than 900 hours = Bulbs with life 300, 500 and 700 hours
From the table, we can see that bulbs with life less than 900 hours =10+12+23
\text {P}=\frac{\text {favourable outcome}}{\text {Total favourable outcomes}}=\frac{45}{10+12+23+25+10}
=\frac{45}{80}=\frac{9}{16}
Therefore option (D) is correct.

Question:1

The frequency distribution :

Marks
0-20
20-40
40-60
60-100
Number of Students
10
15
20
25
has been represented graphically as follows :

Do you think this representation is correct? Why?

Answer:

Answer : yes
Given :
Marks
0-20
20-40
40-60
60-100
Number of Students
10
15
20
25

Class interval (x-axis)
Frequency (y-axis)
0-20
10
20-40
15
40-60
20
60-100
25
The above will be plotted as:

Hence, the given graph is correct.

Question:2

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded:
46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44
Which ‘average’ will be a good representative of the above data and why?

Answer: Median

Given data is : 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44
The mode is best used with categorical data.
Mean is best used with continuous and symmetrical data.
Median is best used with skewed distributions as we can find the middle of the data set.
In ascending order, we have:
11, 40, 41, 44, 46, 48, 52, 53, 54, 62, 96, 98.
In this case,
Median will be a good representative of data, because each value occurs once, hence we can draw a line right at the middle of the data.
Hence the correct answer is median.

Question:3

A child says that the median of 3, 14, 18, 20, 5 is 18. What doesn’t the child understand about finding the median?

Answer: The child has not arranged the data in ascending order before finding the median
To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
Firstly, we will arrange the given data in ascending order as,
3, 5, 14, 18, 20.
Total number of terms = 5( odd )
\text{Middle term }=( \frac{n+1}{2})^{th}\ \text{term}
(where n is the number of terms)
\text{Middle term}=( \frac{5+1}{2} )^{th}\text { term = 3rd term =14}
So the median is not 18
The child has not arranged the data in ascending order before finding the median.

Question:4

A football player scored the following number of goals in the 10 matches: 1, 3, 2, 5, 8, 6, 1, 4, 7, 9
Since the number of matches is 10 (an even number), therefore, the median

=\frac{5th\; \text {observation}+6th \; \text {observation}}{2}
=\frac{8+6}{2}=7
Is it the correct answer and why?

Answer:

Answer : [No, as the data is not arranged in ascending order before finding the middle terms]
Given terms : 1, 3, 2, 5, 8, 6, 1, 4, 7, 9
To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
Firstly, we will arrange the data in ascending order as:
1, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Here the total number of observations, n=10 which is even.
In case of even number of terms:
\text{Median = average of }( \frac{n}{2} )^{th} \text{ and} \left ( \frac{n}{2}+1 \right )^{th}\text{ term}
\text{Median} =\frac{\left ( \frac{n}{2} \right )^{th}observation+\left [ \frac{n}{2}+1 \right ]^{th}observation}{2}
\Rightarrow \frac{\left ( \frac{10}{2} \right )^{th}observation+\left [ \frac{10}{2}+1 \right ]^{th}observation}{2}
\Rightarrow \frac{5^{th}\; \text {observation}+6^{th}\text {observation}}{2}
\Rightarrow \frac{4+5}{2}=\frac{9}{2}=4.5
Hence, the median is 4.5

A football player scored the following number of goals in the 10 matches: 1, 3, 2, 5, 8, 6, 1, 4, 7, 9 Since the number of matches is 10 (an even number), therefore, the median Is it the correct answer and why?


Question:5

Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement.

Answer: No
In a histogram, the area of each rectangle is proportional to the frequency of its class.
The frequency of a class may change for any class interval. But for a specific class, the class size will remain the same.
We know that the frequency is represented on the y-axis and class size on the x-axis.
So we can say that for all the rectangles of a histogram, width (class size) remains the same & length (frequency) changes.
Hence, the correct statement is that in a histogram, the area of each rectangle is proportional to the frequency of its class.

Question:6

The class marks of a continuous distribution are :
1.04, 1.14, 1.24, 1.34, 1.44, 1.54\text { and}\ 1.64
Is it correct to say that the last interval will be 1.55 - 1.73? Justify your answer.

Answer: As we know that class is the mid-point of a class interval.
\Rightarrow So, Difference between two consecutive class marks is always equal to the class size.
Class-marks
Difference
1.14-1.04
0.1
1.24-1.14
0.1
1.34-1.24
0.1
1.44-1.34
0.1
1.54-1.44
0.1
1.55-1.73
0.8
So, we can see that the last class size is 0.8, which is not equal to the class size found previously.
Hence it cannot be the correct class interval for the given class marks.

Question:7

30 children were asked about the number of hours they watched TV programs last week. The results are recorded as under :

Number of hours
0-5
5-10
10-15
15-20
Frequency
8
16
4
2
Can we say that the number of children who watched TV for 10 or more hours a week is 22 ? Justify your answer.
Answer : No
Total number of students, who watched TV for 10 or more hours are:
From the table:
C.I. = frequency
10-15= children
10-20 children
Total = 6 children
Hence, the total number of students who watched TV for 10 or more hours are 6 and not 22.

Question:8

Can the experimental probability of an event be a negative number? If not, why?

Answer: No
Probability of a number is determined by: Favourable Events / Total Events
Now, the number of trials in which the event can happen cannot be negative. So the total number of trials is always positive. Hence, the experimental probability of an event is always positive.

Question:9

Can the experimental probability of an event be greater than 1? Justify your answer.

Answer : No
We know that,
\text {Probability}=\frac{\text {No. of favourable outcomes}}{\text {Total number of outcomes}}
As we know the value of favourable outcomes can never be greater than the total outcomes.
So the probability can never be greater than 1. The probability of an event can be between only 0 and 1.
Hence, the experimental probability of an event can never be greater than 1

Question:10

As the number of tosses of a coin increase, the ratio of the number of heads to the total number of tosses will be 0.5. Is it correct? If not, write the correct one.

Answer : No
As the number of tosses of a coin increase, we may have a head or a tail as an outcome. Probability of 0.5 is achieved when the total number of heads = total number of tails. So obviously we can say that it is not necessary that the favourable outcomes of tails will be equal to favourable outcomes of heads.
So this ratio may or may not be 0.5.
\\\text{ For example, if we toss for three times and one head is obtained, so ratio}=\frac{1}{3}\neq \frac{1}{2}
\text{If we toss for three times and two heads are obtained, so ratio}=\frac{2}{3}\neq \frac{1}{2}
\text{If we toss for four times and no head is obtained, so ratio}=\frac{0}{4}\neq \frac{1}{2}
\text{If we toss for four times and 2 heads are obtained, so ratio }=\frac{2}{4}= \frac{1}{2}
So we are not sure what the ratio will be.
Hence the given statement is not true.

Question:1

The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B, A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.

Answer:

Given data :
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B,
A, A, O, A, AB, B, A, O, B, A, B, A
This can be arranged as:
A, A, A, A, A, A, A, A, A, A, A, A
B, B, B, B, B, B, B, B
AB, AB, AB, AB
O, O, O, O, O, O
Frequency Table can be represented as:
Blood group
Frequency distribution table
A
12
B
8
AB
4
O
6
Total
30
Hence, we can be observed clearly that the most common blood group is (A) and least common is (AB).

Question:2

The value of p upto 35 decimal places is given below :3.14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after the decimal point.

Answer:

We have to make a frequency distribution of the digits 0 to 9 after the decimal point.
The digits are

1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8
Hence the frequency distribution table can be represented as:

Digit
Frequency
0
1
1
2
2
5
3
6
4
3
5
4
6
3
7
2
8
5
9
4
Total
35

Question:3

The scores (out of 100) obtained by 33 students in a mathematics test are as follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84, 66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency distribution.

Answer:

Given data :
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84, 66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
The data can be arranged in ascending order as:
48, 48, 48, 58, 58, 58, 64, 64, 64, 64, 66, 66, 66, 66, 66, 66, 66, 69, 69, 69, 69, 69,
69, 71, 71, 71, 73, 73, 81, 83, 83, 84, 84
Frequency distribution table:
Scores (X)
No.of students (f)
48
3
58
3
64
4
66
7
69
6
71
3
73
2
81
1
83
2
84
2

N=33



Question:4

Prepare a continuous grouped frequency distribution from the following data:

Mid-Point
Frequency
5
4
15
8
25
13
35
12
45
6
Also find the size of class intervals.

Answer:

Answer: Class size is 10
Classmark is the mid-value or the central value of a class
It is calculated as follows :
\frac{\text {upper limit}+\text {lower limit}}{2}
Here midpoints are given as 5, 15, 25, 35, 45
So corresponding classes will be: 0-10, 10-20, 20-30, 30-40, 40-50
The given frequencies are :
0-10: 4, 10-20: 8, 20-30: 13, 30-40: 12, 40-50: 6
Continuous grouped frequency distribution table:

Hence, the class size is 10

Question:5

Convert the given frequency distribution into a continuous grouped frequency distribution:

Class interval
Frequency
150-153
7
154-157
7
158-161
15
162-165
10
166-169
5
170-173
6
In which intervals would 153.5 and 157.5 be included?

Answer:

Given classes are :
150 - 153, 154 - 157, 158-161 and so on.
Consider any one class: 150-153
The lower limit of 154 -157 = 154
Upper limit of 150 - 153 = 153
Difference = 154 - 153 = 1
Half of the difference = 0.5
The new class interval formed = (150 - 0.5) - (153 + 0.5) = 149.5 - 153.5.
Similarly for other classes as well. So we get:
Class interval
Frequency
149.5-153-5
7
153.5-157.5
7
157.5-161.5
15
161.5-165.5
10
165.5-169.5
5
169.5-173.5
6
Hence 153.5 belongs to the class interval 153.5 -157.5 and 157.5 belongs to the class interval 157.5 - 161.5

Question:6

The expenditure of a family on different heads in a month is given below:

Head
Food
Education
Clothing
House Rent
Others
Savings
Expenditure (in Rs)
4000
2500
1000
3500
2500
1500
Draw a bar graph to represent the data above.

Answer:

Here, we will draw a bar graph to represent the expenditure of a family for different heads in a month as follows:

Question:7

Expenditure on Education of a country during a five year period (2002-2006), in crores of rupees, is given below:

Elementary education
240
Secondary Education
120
University Education
190
Teacher's Training
20
Social Education
10
Other Educational Programmes
115
Cultural programmes
25
Technical Education
125
Represent the information above by a bar graph.

Answer:

Expenditure on Education of a country during a five year period (2002-2006), in crores of rupees, is given below:
Elementry education
240
Secondary Education
120
University Education
190
Teacher's Training
20
Social Education
10
Other Educational Programmes
115
Cultural programmes
25
Technical Education
125
Bar graph to represent the same:

Question:8

The following table gives the frequencies of most commonly used letters a, e, i, o, r, t, u from a page of a book:

Letters
a
e
i
o
r
t
u
Frequency
75
125
80
70
80
95
75
Represent the information above by a bar graph.

Answer:

We will draw the graph with letters (a, e, i, o, r, t, u) as follows:

Question:9

If the mean of the following data is 20.2, find the value of p:

x
10
15
20
25
30
f
6
8
p
10
6
Answer : \left [ p=20 \right ]
The frequency distribution table can be prepared as follows-
x_{i}
f_{i}
x_{i}f_{i}
10
6
60
15
8
120
20
p
20 p
25
10
250
30
6
180

\sum f_{i}=30+p
\sum f_{i}x_{i}=610+20p
Mean =\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{610+20p}{30+p}=20.2
610+20p=606+20.2p
610-606=20.2p-20p
4=0.2 p
p=\frac{4}{0.2}=\frac{40}{2}=20
Hence, the value of p is 20

Question:10

Obtain the meaning of the following distribution:

Frequency
Variable
4
4
8
6
14
8
11
10
3
12

Answer :
\left [ 8.05 \right ]
The frequency distribution table can be prepared as follows-
x_{i}
f_{i}
x_{i}f_{i}
4
4
16
6
8
48
8
14
112
10
11
110
12
3
36

\sum f_{i}=40
\sum x_{i}f_{i}=322
Mean
\bar{x}=\frac{\sum x_{i}f_{i}}{\sum f_{i}}=\frac{322}{40}=\frac{161}{20}=8.05
Hence, the mean is 8.05

Question:11

A class consists of 50 students out of which 30 are girls. The mean of marks scored by girls in a test is 73 (out of 100) and that of boys is 71. Determine the mean score of the whole class.

Answer: 72.2
A class consists of 50 students out of which 30 are girls.
The mean marks of 30 girls =73
Total score of 30 girls
= 73 \times 30 = 2190.
Now, number of boys are
50 - 30 = 20.
The mean marks of 20 boys =71
Total score of boys
= 71 \times 20 = 1420
\RightarrowHere, the total score of whole class = Total score of all girls + Total score of all boys
= 2190 + 1420 = 3610.
\text{Mean of all students}=\frac{\text{Total marks of all students} }{\text {Total number of students}}
=\frac{3610}{50}=72.2

Hence, the mean of whole class is 72.2.

Question:12

Mean of 50 observations was found to be 80.4. But later on, it was discovered that 96 was misread as 69 at one place. Find the correct mean.

Answer :
\left [ 80.94 \right ]
\text{Mean}=\frac{\text {sum of all observations}}{\text {No. of observations}}
Mean of 50 observations was found to be 80.4 (Incorrect mean)
The incorrect sum of all the numbers = Incorrect mean \times Total numbers
=80.4 \times 50=4020
It was discovered that 96 was misread as 69 at one place
Hence the correct sum of all the numbers
= 4020 - 69 + 96 = 4047
So correct mean
\frac{4047}{50}=80.94
Hence, the correct mean is 80.94

Question:13

Ten observations 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 are written in an ascending order. The median of the data is 24. Find the value of x.

Answer : 20
Here, the observations are :
6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43
To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
The terms are already given in ascending order so we have to find the middle term.
Number of terms, n=10(even)
\text{So, Median = average of}\left ( \frac{n}{2} \right )^{th} \text{ and}\left ( \frac{n}{2} +1\right )^{th}\text{term}
\Rightarrow Median =24 (given)
\text{Median}=\frac{\left ( \frac{n}{2} \right )^{th}\text {observation}+\left ( \frac{n}{2}+1 \right )^{th}\text {observation}}{2}
24=\frac{\left ( \frac{10}{2} \right )^{th}\text {obs.}+\left ( \frac{10}{2}+1 \right )\text {obs.}}{2}
24=\frac{5^{th}\text {obs.}+6^{th}\text {obs.}}{2}
24=\frac{x+1+2x-13}{2}
\Rightarrow 2 \times 24=x+1+2x-13
\Rightarrow 48=3x-12
\Rightarrow 48+12=3x
\Rightarrow 60=3x
\Rightarrow x=20
Hence, the correct answer is 20.

Question:14

The points scored by a basketball team in a series of matches are as follows:
17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28
Find the median and mode for the data.

Answer:

Answer : [median = 12 and mode = 10]
Solution. To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
Here total elements, n = 16 (even)
The terms as arranged in ascending order:
2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48
Number of observation = 16 (even number)
Now using the formula of the median in case number of terms is even
\Rightarrow \text{Median}=\frac{\left ( \frac{n}{2} \right )^{th}\text {obs}+\left ( \frac{n}{2}+1 \right )^{th}\text {obs}}{2}
=\frac{\left ( \frac{16}{2} \right )^{th}\text {obs}+\left ( \frac{16}{2}+1 \right )^{th}\text {obs}}{2}
median=\frac{\left ( 8^{th}\text {obs.}+9^{th}obs. \right )}{2}
\Rightarrow \frac{10+14}{2}=\frac{24}{2}
Median =12
The mode is the value that occurs the most number of times in a given set of values
Now mode is 10 because it is the most repeating number.
Hence, median = 12 and mode = 10.

Question:15

In Fig. 14.2, there is a histogram depicting the daily wages of workers in a factory. Construct the frequency distribution table.

Answer:

Based on the given bar graph :
Class-Interval
Frequency
150-200
50
200-250
30
250-300
35
300-350
20
350-400
10
Total workers
145


Question:16

A company selected 4000 households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information so obtained is listed in the following table:

Monthly income (in Rs)
Number of Televisions/household

0
1
2
Above 2
<10000
20
80
10
0
10000-14999
10
240
60
0
15000-19999
0
380
120
30
20000-24999
0
520
370
80
25000 and above
0
1100
760
220
Find the probability:
(i) of a household earning Rs 10000 – Rs 14999 per year and having exactly one television.
(ii) of a household earning Rs 25000 and more per year and owning 2 televisions.
(iii) of a household not having any television.

Answer:

Here, total events =4000
(i) A household earning Rs 10000 – Rs 14999 per year and having exactly one television.
Favourable outcomes =240
\text{Required probability}=\frac{240}{4000}=\frac{3}{50}
(ii) A household earning Rs 25000 and more per year and owning 2 televisions.
Favourable outcomes =760
\text{Required probability}=\frac{760}{4000}=0.19
(iii) A household not having any television.
Favourable outcomes =20+10
\text{Required probability}=\frac{30}{4000}=\frac{3}{400}

Question:17

Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:

Sum
Frequency
2
14
3
30
4
42
5
55
6
72
7
75
8
70
9
53
10
46
11
28
12
15
If the dice are thrown once more, what is the probability of getting a sum
(i) 3? (ii) more than 10?
(iii) less than or equal to 5? (iv) between 8 and 12?

Answer:

\text{Probability }=\frac{\text {Favourable outcomes}}{\text {Total number of events}}
Here, total events = 14 + 30 + 42 + 55 + 72 + 75 + 70 + 53 + 46 + 28 +15 = 500
(i) probability of getting a sum = 3
Favourable events = 30
\text{P[of getting sum 3]}=\frac{30}{500}=0.06
(ii) probability of getting a sum more than 10
\text{Favourable events}= 28 + 15 = 43
\text{P[of getting sum 10]}=\frac{43}{500}=0.086
(iii) probability of getting a sum less than or equal to 5
Favourable events = 14 + 30 + 42 + 55
\text{p[sum less than or equal to 5]}=\frac{141}{500}=\frac{28.2}{100}=2.282
(iv) probability of getting a sum between 8 and 12
Favourable events = 53 + 46 + 28 = 127
\text{p[sum between 8 and 12]}=\frac{127}{500}=0.254

Question:18

Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:

Number of defective bulbs
0
1
2
3
4
5
6
more than 6
Frequency
400
180
48
41
18
8
3
2
One carton was selected at random. What is the probability that it has
(i) no defective bulb?
(ii) defective bulbs from 2 to 6?
(iii) defective bulbs less than 4?

Answer:

\text{Probability is defined as }=\frac{\text {Favourable outcomes}}{\text {Total number of events}}
Here, total events = total cartons =700
(i) no defective bulb
Favourable outcomes =400
\text{p (cartoon has no defective bulb) =}\frac{400}{700}=\frac{4}{7}
(ii) defective bulbs from 2 to 6 = 2 or 3 or 4 or 5 or 6 defective bulbs
Favourable outcomes = 48 + 41 + 18 + 8 + 3 = 118
\text{p(defective bulb from 2 to 6)}=\frac{118}{700}=\frac{59}{350}
(iii) defective bulbs less than 4 = defective bulbs equal to 0 or 1 or 2 or 3
Favourable outcomes = 400 + 180 + 48 + 41 = 669
\text{p(defective bulbs less than 4)}=\frac{669}{700}

Question:19

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:

Number of defective parts
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Days
50
32
22
18
12
12
10
10
10
8
6
6
2
2
Determine the probability that tomorrow’s output will have
(i) no defective part
(ii) at least one defective part
(iii) not more than 5 defective parts
(iv) more than 13 defective parts

Answer:

\text{Probability is defined as}=\frac{\text {Favourable outcomes}}{\text {Total number of events}}
Here, total events = total number of working days =200
(i) no defective part
Favourable outcomes =50 days =50
\text{p (no defective part)}=\frac{50}{200}=0.25
(ii) Probability that at least one defective part = 1 - the probability that no defective part
\text{p (no defective part)}=\frac{50}{200}=0.25
\text{p (at least one defective part)}=1-\frac{50}{200}=0.75
(iii) not more than 5 defective parts = 0 or 1 or 2 or 3 or 4 or 5 defective parts
p(not more than 5 defective parts) = p(no. defective part) + p(1 defective part) + p(2 defective part) + p(3 defective part) + p(4 defective part) + p(5 defective part)
=\frac{50}{200}+\frac{32}{200}+\frac{22}{200}+\frac{18}{200}+\frac{12}{200}+\frac{12}{200}
=\frac{50+32+22+18+12+12}{200}=\frac{146}{200}
=\frac{73}{200}=0.73
(iv) more than 13 defective parts = not possible
Favourable outcomes = 0
p (more than 13 defective parts) = 0

Question:20

A recent survey found that the ages of workers in a factory is distributed as follows:

Age (in years)
20-29
30-39
40-49
50-59
60 and above
Number of workers
38
27
86
46
3
If a person is selected at random, find the probability that the person is:
(i) 40 years or more
(ii) under 40 years
(iii) having age from 30 to 39 years
(iv) under 60 but over 39 years

Answer:

\text{Probability}=\frac{\text {Favourable outcomes}}{\text {Total number of events}}
Here, total events = total number of workers= 38 + 27 + 86 + 46 + 3 = 200
(i) p (person is 40 years or more) = p(person having age 40 to 49 years) + p (person having age 50 to 59 years) + p (having age 60 and above)
=\frac{86}{200}+\frac{46}{200}+\frac{3}{200}=\frac{135}{200}=0.675
(ii) p(person is under 40 years) = p(person having age 20 to 29 years) + p(person having age 30 to 39 years)
=\frac{38}{200}+\frac{27}{200}=\frac{65}{200}=0.325
Hence the different age group decided the work.
\text{(iii) p(having age from 30 to 39 years) }=\frac{27}{200}=0.135

(iv) p(under 60 but over 39 years) = p(person having age 40 to 49 years) + p (person having age 50 to 59 years)

=\frac{86}{200}+\frac{46}{200}=\frac{132}{200}=0.66

Question:1

The following are the marks (out of 100) of 60 students in mathematics.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30.
Construct a grouped frequency distribution table with width 10 of each class starting from 0 - 9.

Answer:

Step 1: arrange these number in ascending order.
\\4, 5, 9, 7, 13, 15, 16, 17, 17, 19,\\ 24, 25, 26, 27, 28, 30, 31, 34, 34, 34, 35, 35, 36, 36, 36,\\ 42, 44, 45, 47, 48, 51, 52, 52, 54, 55, 55, 56, 56,\\ 61, 62, 62, 63, 68, 70, 72, 72, 72, 74, 75, 75, 78,\\ 80, 81, 85, 86, 92, 95, 97.
Step 2: Make the frequency distribution table as follows:
The class interval of (0-9), (10-19) etc. are given. So we can write the class interval and the corresponding frequency.
Class interval
Frequency
0-9
4
10-19
7
20-29
5
30-39
10
40-49
5
50-59
8
60-69
5
70-79
8
80-89
5
90-99
3
Total
60

Question:2

The following are the marks (out of 100) of 60 students in mathematics.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30.
Construct a grouped frequency distribution table with width 10 of each class, in such a way that one of the classes is 10 - 20 (20 not included).

Answer:

Step 1: arrange these number in ascending order.
4, 5, 9, 7, 13, 15, 16, 17, 17, 19,
24, 25, 26, 27, 28, 30, 31, 34, 34, 34, 35, 35, 36, 36, 36,
42, 44, 45, 47, 48, 51, 52, 52, 54, 55, 55, 56, 56,
61, 62, 62, 63, 68, 70, 72, 72, 72, 74, 75, 75, 78,
80, 81, 85, 86, 92, 95, 97.
Step 2: Make the frequency distribution table as follows:
The class interval of (0-10), (10-20) etc. are given. So we can write the class interval and the corresponding frequency.
A value like 10 will always be counted in the interval where it is the lower limit, i.e., 10-20 and not in 0-10 (where it is the higher limit)
Class interval
Frequency
0-10
4
10-20
7
20-30
5
30-40
10
40-50
5
50-60
8
60-70
5
70-80
8
80-90
5
90-100
3
Total
60

Question:3

Draw a histogram of the following distribution :

Heights (in cm)
Number of students
150-153
7
153-156
8
156-159
14
159-162
10
162-165
6
165-168
5

Answer:

To draw the histogram, we have to plot the frequency, i.e., number of students on the y-axis and class interval, i.e., heights on the x-axis. We can construct the histogram as follows:

Question:4

Draw a histogram to represent the following grouped frequency distribution :

Ages (in years)
Number of teachers
20-24
10
25-29
28
30-34
32
35-39
48
40-44
50
45-49
35
50-54
12

Answer:

We can see that the intervals are 20-24, 25-29, 30-34....etc
For the interval 20-24: Upper limit is 24
For the interval 25-29: Lower limit is 25
Difference = 25 - 24 = 1
\frac{1}{2}\left ( \text {upper limit-lower limit} \right )=\frac{1}{2}\left ( 1 \right )=0.5
Now, we will subtract 0.5 from all lower limits & add 0.5 to all upper limits
Ages
No. of teachers
19.5-24.5
10
24.5-29.5
28
29.5-34.5
32
34.5-39.5
48
39.5-44.5
50
44.5-49.5
35
49.5-54.5
12
Now, we draw the diagram.

Question:5

The lengths of 62 leaves of a plant are measured in millimetres and the data is represented in the following table :

Length (in mm)
Number of leaves
118-126
8
127-135
10
136-144
12
145-153
17
154-162
7
163-171
5
172-180
3
Draw a histogram to represent the data above.

Answer:

We can see that the intervals are 118-126, 127-135, 136-144...etc
For the interval 118-126: Upper limit is 126
For the interval 127-135: Lower limit is 127
Difference = 127 - 126 = 1
\frac{1}{2}\left ( \text {upper limit - lower limit} \right )=\frac{1}{2}\left ( 1 \right )=0.5
Now, we will subtract 0.5 from all lower limits & add 0.5 to all upper limits
Length (in mm)
Number of leaves
117.5-126.5
8
126.5-135.5
10
135.5-144.5
12
144.5-153.5
17
153.5-162.5
7
162.5-171.5
5
171.5-180.5
3
Hence the histogram is :

Question:6

The marks obtained (out of 100) by a class of 80 students are given below :

Marks
Number of students
10-20
6
20-30
17
30-50
15
50-70
16
70-100
26
Construct a histogram to represent the data above.

Answer:

To draw the histogram, we have to plot the frequency, i.e., number of students on the y-axis and class interval, i.e., marks on the x-axis. We can construct the histogram as follows:

Question:7

Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way :

Class interval (km/h)
Frequency
30-40
3
40-50
6
50-60
25
60-70
65
70-80
50
80-90
28
90-100
14
Draw a histogram and frequency polygon representing the data above.

Answer:

To draw the histogram, we have to plot the frequency on the y-axis and class interval on the x-axis. We can construct the histogram as follows:

A frequency polygon is a graph constructed by using lines to join the midpoints of each interval.
Accordingly, the curve ABCDEFGHI is the frequency polygon.

Question:8

Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way :

Class interval (km/h)
Frequency
30-40
3
40-50
6
50-60
25
60-70
65
70-80
50
80-90
28
90-100
14
Draw the frequency polygon representing the above data without drawing the histogram.

Answer:

A frequency polygon is a graph constructed by using lines to join the midpoints of each interval.
Here we the midpoints as follows:
Class interval Mid-point Frequency
30 - 40 35 3
40 - 50 45 6
50 - 60 55 25
60 - 70 65 65
70 - 80 75 50
80 - 90 85 28
90 – 100 95 14
Hence we can construct the points using mid-point on x-axis and corresponding frequency on y-axis. So the frequency polygon will be as follows:

Question:9

Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them.

Section A
Section B
Marks
Frequency
Marks
Frequency
0-15
5
0-15
3
15-30
12
15-30
16
30-45
28
30-45
25
45-60
30
45-60
27
60-75
35
60-75
40
75-90
13
75-90
10
Represent the marks of the students of both the sections on the same graph by two frequency polygons. What do you observe?

Answer:

\text{Formula of to find the class marks }=\frac{\text {upper class limit}+\text {lower class limit}}{2}
Marks
Class marks
Frequency A
Frequency B
0-15
0+15=\frac{15}{2}=7.5
5
3
15-30
\frac{15+30}{2}=\frac{45}{2}=22.5
12
16
30-45
\frac{30+45}{2}=\frac{75}{2}=37.5
28
25
45-60
52.5
30
27
60-75
67.5
35
40
75-90
82.5
13
10
A frequency polygon is a graph constructed by using lines to join the midpoints of each interval.
Points where frequency is zero:
Difference = 22.5 - 7.5 = 15
First Point =7.5-15 = -7.5
Last point =82.5+7.5= 90
So we can construct the frequency polygon of A (orange) and B (blue) as follows:

Question:10

The mean of the following distribution is 50.

x
f
10
17
30
5a+3
50
32
70
7a-11
90
19
Find the value of a and hence the frequencies of 30 and 70.

Answer:

a = 5, f_{30} = 28, f_{70} = 24
The frequency distribution table is as follows:
x_{i}
f_{i}
f_{i}x_{i}
10
17
170
30
5a+3
150a+90
50
32
1600
70
7a-11
490a-770
90
19
1710
Total
60+12a
2800+640a
Mean =\frac{\sum f_{i}x_{i}}{\sum f_{i}}
\frac{50}{1}=\frac{2800+640a}{60+12a}
[Cross multiply]
50\left ( 60+12a \right )=2800+640\; a
3000+600a=2800+640a
3000 - 2800 = 640a - 600a
200 = 40a
\frac{200}{40}=a
a=5
\\f_{30} = 5a + 3 \\ f_{30} = 5(5) + 3 = 28\\ f_{70} = 7a - 11\\ f_{70} = 7(5) -11\\ f_{70} = 35 - 11 = 24

Question:10

The mean of the following distribution is 50.

x
f
10
17
30
5a+3
50
32
70
7a-11
90
19
Find the value of a and hence the frequencies of 30 and 70.

Answer:

a = 5, f_{30} = 28, f_{70} = 24
The frequency distribution table is as follows:
x_{i}
f_{i}
f_{i}x_{i}
10
17
170
30
5a+3
150a+90
50
32
1600
70
7a-11
490a-770
90
19
1710
Total
60+12a
2800+640a
Mean =\frac{\sum f_{i}x_{i}}{\sum f_{i}}
\frac{50}{1}=\frac{2800+640a}{60+12a}
[Cross multiply]
50\left ( 60+12a \right )=2800+640\; a
3000+600a=2800+640a
3000 - 2800 = 640a - 600a
200 = 40a
\frac{200}{40}=a
a=5
\\f_{30} = 5a + 3 \\ f_{30} = 5(5) + 3 = 28\\ f_{70} = 7a - 11\\ f_{70} = 7(5) -11\\ f_{70} = 35 - 11 = 24

Question:11

The mean marks (out of 100) of boys and girls in an examination are 70 and 73, respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.

Answer : [ 2:1 ]
Let the no. of boys and girls be x and y respectively.
\text{ Mean }=\frac{\text { sum of observations}}{\text {Number of observations}}
Now, mean marks of x boys in the examination =70
Sum of marks of x boys =70 x
Mean marks of y girls in the examination =73
Sum of marks of y girls =73y
Given, mean marks of all students (x + y) = 71
Sum of marks of all students (x + y) = 71(x + y)
Now, sum of marks of all students x + y =sum of marks of x boys and sum of marks of y girls
\\71(x + y) = 70x + 73 y\\ 71x + 71y = 70x + 73y\\ 71x - 70x = 73y- 71y\\ x = 2y
\frac{x}{y}=\frac{2}{1}
Hence, the ratio is x:y=2:1

Question:12

A total of 25 patients admitted to a hospital are tested for levels of blood sugar, (mg/dl) and the results obtained were as follows :
87 \; \; \; 71 \: \: \: 83 \; \; \; \;67 \; \; \; 85
77 \; \; \; 69 \; \; \; 76 \; \; \; 65 \; \; \; 85
\\85 \; \; \; 54 \; \; \; 70 \; \; \; 68 \; \; \; 80\\ 73 \; \; \; 78 \; \; \; 68 \; \; \; 85 \; \; \; 73\\ 81 \; \; \; 78 \; \; \; 81 \; \; \; 77 \; \; \; 75
Find mean, median and mode (mg/dl) of the above data.

Answer:

Answer : The mean is 75.88, mode is 85, median is 77
Arrange the data in ascending order
54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 78, 80,
81, 81, 83, 85, 85, 85, 85, 87
\text {Mean}=\frac{\text {sum of all obseravations}}{\text {Total number of observations}}
=\frac{1891}{25}=75.88
The mean is 75.88

Mode is the value that occurs the most number of times in a given set of values
So, 85 is the mode, and its frequency is 4 times.
To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
The data in ascending order is:
54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 78, 80,
81, 81, 83, 85, 85, 85, 85, 87
Number of terms, n=25
\left ( \frac{n+1}{2} \right )^{th}\text{ term is the median}
=\left ( \frac{25+1}{2} \right )^{th}\text{ term}=\left ( \frac{26}{2} \right )^{th}\text{ term}
13th term is 77
Hence, median is 77

Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 14:

The important topics covered in NCERT exemplar Class 9 Maths solutions chapter 14 are mentioned below:

  • Data Reading and their behavior are very important to establish any conclusion about any phenomena.
  • How the collection of data in tabular form is represented graphically in different manners helps to understand the variation of the data.
  • How the collection of data in tabular form is represented graphically in different manners helps to understand the central tendency of the data.
  • NCERT exemplar Class 9 Maths solutions chapter 14 explains the concept of probability which is the study of chances.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 14:

These Class 9 Maths NCERT exemplar chapter 14 solutions provide a basic knowledge of statistics and probability which has great importance in economics, politics, social science as well as in science. These solutions can be used as supporting material to better study Statistics and Probability-based practice problems. The Class 9 Maths NCERT exemplar solutions chapter 14 Statistics and Probability provides adequate practice and information on this chapter in a student-friendly process flow that can be further applied while solving other books such as Class 9 Mathematics by Pearson, NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths etcetera.

NCERT exemplar Class 9 Maths solutions chapter 14 pdf download is a free to use feature for students and can be used judiciously while studying NCERT exemplar Class 9 Maths chapter 14 in an offline environment.

Check the Solutions of Questions Given in the Book

Also, Read NCERT Solution Subject Wise

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Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. For any event, there are two possibilities: either it will occur or not. Does it mean the probability of occurrence is half?

No, the possibility of occurring or not occurring may have different methods. Therefore, we can say occurring or non-occurring are not equally likely. If two events are not equally likely their probability will not be the same.

2. What is the meaning of mean median and mode

Mean, median, and mode of any data represent the central tendency of the data.

Mode represents the most favourable data outcome.

Mean represent the average value of the data.

Median represents the middle value, if the data is arranged in decreasing or increasing order.

3. What do you mean by the sure event?

When the occurrence of any event cannot be denied, the event is called a sure event. For example, the sun will rise in the east can be seen as a sure event. The probability of a sure event is one.

4. Can the probability be more than one?

The probability of any event cannot be more than one. The maximum probability is 1 for sure events. The minimum probability is 0 for impossible events

5. What is a statistical probability?

On the basis of analysis of large data, we estimate the occurrence of any event this analysis is called statistical probability. Most of the recent scientific researches are based on statistical probability

6. What is the weightage of Statistics and Probability in the final exam?

The chapter on Statistics and Probability is a very important chapter from the exam’s perspective as it accounts for 10-12% marks of the final paper. NCERT exemplar Class 9 Maths solutions chapter 14 provides sufficient learning content to the students to develop a better understanding and score well in the exam.

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0.34\; J

Option 2)

0.16\; J

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2.45×10−3 kg

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2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J

Option 3)

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\; K\;

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2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

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0.02

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Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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