NCERT Exemplar Class 9 Maths Solutions Chapter 14 Statistics and Probability

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:
$\frac{\text{upper limit + lower limit}}{2}$
$\Rightarrow$ $Classmark=\frac{90+120}{2}=\frac{210}{2}=105$
The class mark of class 90-120 is 105 i.e. option B

Question:2

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

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:
$$\begin{gathered} 15-20 \\ 20-25 \\ 25-30 \\ 30-35 \end{gathered}$$
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:
$$\begin{gathered} (A)4 \\ (B)5 \\ (C)6 \\ (D)7 \end{gathered}$$

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:
$$\begin{gathered} 210-230:220,210,215 \\ 230-250:242,240,236 \\ 250-270:258,268,258 \end{gathered}$$
$$\begin{gathered} 270-290:272,278 \\ 290-310:290,304,300,292 \\ 310-330:310,310,320,319,318,316 \\ 330-350:342,330 \\ 350-370:368,354 \end{gathered}$$
$370-390$: 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,$$15,35,44,66,75,84,95,96,102,110,88,74,112,14,34,44.$
The number of classes in the distribution will be :
$$\begin{gathered} (A)9 \\ (B)10 \\ (C)11 \\ (D)12 \end{gathered}$$

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 :

the adjusted frequency for the class 25-45 is :
$$\begin{gathered} (A)6 \\ (B)5 \\ (C)3 \\ (D)2 \end{gathered}$$
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 :

$$\begin{gathered} (A)28 \\ (B)30 \\ (C)35 \\ (D)38 \end{gathered}$$

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)
$a+112=150$
$\Rightarrow a=150-112$
$\Rightarrow a=38\to \text{excluded number}$
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

$\text{If x represents the mean of n observations }\sum _{i=l}^n(x_i-\overline{x})x_1,x_2......x_n,\text{then value of}$ $\sum _{i=l}^n(x_i-\overline{x})\text{ is}$
(A) -1
(B) 0
(C) 1
(D) n-1

Answer : B
Here we are given,
$\sum _{i=l}^n(x_i-\overline{x})$
$\Rightarrow \overline{x}\text{ is the mean of all observations. }$
$\text{mean}=\frac{\text{sum of observations}}{\text{Number of observations}}$
$\overline{x}=\frac{(x_1+x_2+x_3+......+x_n)}{n}$
$\Rightarrow n\overline{x}=(x_1+x_2+.....+x_n)$
$\Rightarrow n\overline{x}=\sum _{i=1}^n{x}_i$
$\Rightarrow \sum _{i=1}^n{x}_i-n\overline{x}=0$
$\Rightarrow \sum _{i=1}^n(x_1-\overline{x})=0$
Here the value of
$\sum _{i=1}^n(x-\overline{x})=0$
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
${\overline{x}}_{old}=\frac{\sum _{i=1}^n{x}_i}{n}..(i)$
Now adding $5$ to each observation, the new mean becomes.
${\overline{x}}_{new}=\frac{(x_1+5)+(x_2+5)+....+(x_n+5)}{n}$
${\overline{x}}_{new}=\frac{(x_1+x_2+.....+x_n)+(5+5+5.....)\text{n times}}{n}$
${\overline{x}}_{new}=\frac{(x_1+x_2+.....+x_n)+5n}{n}$
${\overline{x}}_{new}=\frac{\sum _{i=1}^n{x}_i}{n}+5={\overline{x}}_{old}+5$ (from eq. (i))
${\overline{x}}_{new}={\overline{x}}_{old}+5$
Hence, the new mean is increased by 5.
Therefore option (D) is correct.

Question:15

Let $\overline{x}$ be the mean of $x_1,x_2.........,x_n$ and $\overline{y}$ the mean of $y_1,y_2,...,y_n$. If $\overline{z}$ is the mean of $x_1,x_2.........,x_n$, $y_1,y_2,...,y_n$ then $\overline{z}$ is equal to
$(A)$ $\overline{x}+\overline{y}$
$(B)$ $\frac{\overline{x}+\overline{y}}{2}$
$(C)$ $\frac{\overline{x}+\overline{y}}{n}$
$(D)$ $\frac{\overline{x}+\overline{y}}{2n}$

Answer : B
$\overline{x}$ is the mean of $x_1,x_2,.....,x_n$ then
$\overline{x}=\frac{x_1+x_2+.....+x_n}{n}$
$\overline{y}$ is the mean of $y_1,y_2,......y_n$ then
$\overline{y}=\frac{y_1+y_2+......y_n}{n}$
$\overline{z}$ is the mean of $x_1,x_2,.......x_n,y_1,y_2............,y_n$
$\overline{z}=\frac{x_1+x_2+........+x_n+y_1+y_2+.....y_n}{2n}$
$\overline{z}=\frac{1}{2}\left(\frac{x_1+x_2+....+x_n+y_1+y_2+.......y_n}{n}\right)$
$\overline{z}=\frac{1}{2}(\frac{x_1+x_2+....+x_n}{n}+\frac{y_1+y_2+.......y_n}{n})$
$\overline{z}=\frac{\overline{x}+\overline{y}}{2}$
Hence, the value of $\overline{z}$ is $\frac{\overline{x}+\overline{y}}{2}$
Therefore option (B) is correct.

Question:16

$If$ $\overline{x}$ $isthemeanof$$x_1,x_2,........,x_n$, $\text{then for}$ $a\ne 0$$\text{,the mean of }$$a{x}_1,a{x}_2,........a{x}_n,\frac{x_1}{a},\frac{x_2}{a},........,\frac{x_n}{a}$ $is$
$(A)$ $(a+\frac{1}{a})\overline{x}$
$(B)$ $(a+\frac{1}{a})\frac{\overline{x}}{2}$
$(C)$$(a+\frac{1}{a})\frac{\overline{x}}{n}$
$(D)$$\frac{(a+\frac{1}{a})\overline{x}}{2n}$

Answer : B
$\frac{x_1+x_2...+x_n}{n}=\overline{x}....(i)$
Multiply both sides by a
$\frac{a{x}_1+a{x}_2...+a{x}_n}{n}=a\overline{x}....(i)$
$a{x}_1+a{x}_2......a{x}_n=na\overline{x}.....(ii)$
Dividing (i) by a, we get
$\frac{\frac{x_1}{a}+\frac{x_2}{a}.......+\frac{x_n}{a}}{n}=\frac{\overline{x}}{a}$
$\frac{x_1}{a}+\frac{x_2}{a}.......+\frac{x_n}{a}=\frac{n\overline{x}}{a}.....(iii)$
The mean of
$a{x}_1+a{x}_2.........+a{x}_n+$ $\frac{x_1}{a},\frac{x_2}{a}.......\frac{x_n}{a}$ $\text{is :}$
$\frac{a{x}_1+a{x}_2+a{x}_n+\frac{x_1}{a}+\frac{x_2}{a}+.......+\frac{x_n}{a}}{2n}$
from (ii) and (iii) we get
$\frac{a{x}_1+a{x}_2+a{x}_n+\frac{x_1}{a}+\frac{x_2}{a}+.......+\frac{x_n}{a}}{2n}=\frac{(na\overline{x}+n\frac{\overline{x}}{a})}{2n}$
$=\frac{\overline{x}n}{2n}(a+\frac{1}{a})$
$=\frac{\overline{x}}{2}(a+\frac{1}{a})$
Therefore option (B) is correct.

Question:17

If ${\overline{x}}_1,{\overline{x}}_2,{\overline{x}}_3,........{\overline{x}}_n,$ are the means of n groups with $n_1,n_2,......n_n$ number of observations respectively, then the mean $\overline{x}$ of all the groups taken together is given by :
$(A)$ $\sum _{i=1}^n{n}_i{\overline{x}}_i$
$(B)$ $\frac{\sum _{i=1}^n{n}_i{\overline{x}}_i}{n^2}$
$(C)$ $\frac{\sum _{i=1}^n{n}_i{\overline{x}}_i}{\sum _{i=1}^n{n}_i}$
$(D)$ $\frac{\sum _{i=1}^n{n}_i{\overline{x}}_i}{2n}$

Answer : C
$\text{Mean is defined as}\overline{x}=\frac{\text{sum of observations}}{\text{Number of observations}}$
$\Rightarrow$ Sum of observations = (mean) (Number of observations).
Given : ${\overline{x}}_1,{\overline{x}}_2,{\overline{x}}_3,........{\overline{x}}_n$ are the means of n groups with $n_1,n_2,......n_n$ number of observations respectively.
$\sum x_1={\overline{x}}_1{n}_1$
$\sum x_2={\overline{x}}_2{n}_2$
$\sum x_n={\overline{x}}_n{n}_n$
Now all the groups are taken together
Sum of all the observations:
$\sum x_1+\sum x_2+.......+\sum x_n={\overline{x}}_1{n}_1+{\overline{x}}_2{n}_2+....+{\overline{x}}_n{n}_n$
Mean of all observations:
$\overline{x}=\frac{{\overline{x}}_1{n}_1+{\overline{x}}_2{n}_2+.....+{\overline{x}}_n{n}_n}{n_1+n_2+.......+n_n}=$$\frac{\sum _{i=1}^n{n}_i{\overline{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
$\Rightarrow 50=\frac{x}{100}$
$\Rightarrow x=50\times 100=5000$
Given that one observation 50 is replaced by 150
$Newsum=x-50+150=5000-50+150=5100$
$Newmean=\frac{5100}{100}=51$
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
$x,y,z.........(50nos.)$
$\text{Mean is defined as }=\frac{\text{sum of observations}}{\text{Number of observations}}$
$\text{Original mean }=\frac{(x+y+z+......)}{50}$
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
$-3.5=\frac{53-x+53-y+53-z+.....}{50}$
$-3.5=\frac{\left(53\right)\left(50\right)-(x+y+z+.....)}{50}$
$-3.5=\frac{(53)(50)}{50}-\frac{(x+y+z+....)}{50}$
$-3.5=53-\text{original mean}$
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 :
$$\begin{gathered} (A)23 \\ (B)36 \\ (C)38 \\ (D)40 \end{gathered}$$

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
$$\begin{gathered} (A)45 \\ (B)49.5 \\ (C)54 \\ (D)56 \end{gathered}$$

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
$=10\left(\text{even}\right)$
$\text{Median}=\text{average of }{\left(\frac{n}{2}\right)}^{th}\text{and}{(\frac{n}{2}+1)}^{th}\text{term}$
${\left(\frac{n}{2}\right)}^{th}\text{term}={\left(\frac{10}{2}\right)}^{th}=5^{th}\text{term}=54$
$\text{Median}=\frac{\text{term}+\text{6th term}}{2}=\frac{108}{2}=54$
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
Therefore option (C) is correct.

Question:23

Median of the following numbers : 4, 4, 5, 7, 6, 7, 7, 12, 3 is
$$\begin{gathered} (A)4 \\ (B)5 \\ (C)6 \\ (D)7 \end{gathered}$$

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 :
$$\begin{gathered} (A)0.5 \\ (B)0.6 \\ (C)0.7 \\ (D)0.8 \end{gathered}$$

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:
$$\begin{gathered} (A)0.25 \\ (B)0.50 \\ (C)0.75 \\ (D)0.80 \end{gathered}$$

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:

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 :

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 :

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 as$1150$ 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 :

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 :

has been represented graphically as follows :

Do you think this representation is correct? Why?

Answer:

Answer : yes
Given :

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:
$\text{Median = average of }(\frac{n}{2}{)}^{th}$ $\text{ and}$ ${(\frac{n}{2}+1)}^{th}\text{ term}$
$\text{Median}=\frac{{\left(\frac{n}{2}\right)}^{th}observation+{[\frac{n}{2}+1]}^{th}observation}{2}$
$\Rightarrow \frac{{\left(\frac{10}{2}\right)}^{th}observation+{[\frac{10}{2}+1]}^{th}observation}{2}$
$\Rightarrow \frac{5^{th}\text{observation}+6^{th}\text{observation}}{2}$
$\Rightarrow \frac{4+5}{2}=\frac{9}{2}=4.5$
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.
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 :

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.
$\text{ For example, if we toss for three times and one head is obtained, so ratio}=\frac{1}{3}\ne \frac{1}{2}$
$\text{If we toss for three times and two heads are obtained, so ratio}=\frac{2}{3}\ne \frac{1}{2}$
$\text{If we toss for four times and no head is obtained, so ratio}=\frac{0}{4}\ne \frac{1}{2}$
$\text{If we toss for four times and 2 heads are obtained, so ratio }=\frac{2}{4}=\frac{1}{2}$
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:
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: