NCERT Exemplar Class 9 Maths Solutions Chapter 14 Statistics and Probability

Question:1

The class mark of the class 90-120 is :

Answer: B
Classmark is the mid-value or the central value of a class
It is calculated as follows:


The class mark of class 90-120 is 105 i.e. option B

Question:2

Answer : D
The range is the difference between the largest (maximum value) and the smallest number (minimum value) of a data

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 :

Answer : B
Let x and y be the upper and lower class limit in the frequency distribution.

Here, the mid-value of the class is given as 10.


Also, given that width of the class is 6

Now, adding equation (1) and (2),


Putting x=13 in equation (1)



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:

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:

Now we have been given that this is the lowest class. So next 4 continuous classes will be:

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 :

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:

Then, mid-point


Therefore,
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 :

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





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 :

The frequency of the class 310-330 is:

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:


: No element


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 :

The number of classes in the distribution will be :

Answer: B

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 :

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


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 :

Answer : D
Let the numbers be



Let the excluded number be a
New mean =28


Putting the value of b+c+d+e in equation (i)



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

Answer:

In the given data the mean of the observations x, x + 3, x + 5, x + 7, x + 10 is given as 9





Last three observation are


So, the mean of the last three observations

Therefore option (C) is correct.

Question:13

(A) -1
(B) 0
(C) 1
(D) n-1

Answer : B
Here we are given,








Here the value of

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 be the n observations.
old mean

Now adding to each observation, the new mean becomes.



(from eq. (i))

Hence, the new mean is increased by 5.
Therefore option (D) is correct.

Question:15

Let be the mean of and the mean of . If is the mean of , then is equal to

Answer : B
is the mean of then

is the mean of then

is the mean of




Hence, the value of is
Therefore option (B) is correct.

Question:16

,

Answer : B

Multiply both sides by a


Dividing (i) by a, we get


The mean of


from (ii) and (iii) we get



Therefore option (B) is correct.

Question:17

If are the means of n groups with number of observations respectively, then the mean of all the groups taken together is given by :

Answer : C

Sum of observations = (mean) (Number of observations).
Given : are the means of n groups with number of observations respectively.



Now all the groups are taken together
Sum of all the observations:

Mean of all observations:

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.

Here number of obervations 100
Let the sum of 50 observations be x


Given that one observation 50 is replaced by 150


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



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




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 :

Answer : B

The mean of observations is .

Sum of 25 observations =900.
For first 13 observations, mean is given as 32

Sum of last 13 observations

Now, 13th observation = sum of 13 observation+sum of last 13 observation-sum of 25 observations.
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

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:

Number of terms




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:

For example
Therefore option (C) is correct.

Question:23

Median of the following numbers : 4, 4, 5, 7, 6, 7, 7, 12, 3 is

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 –

There are number of observations.
For odd number of observations, middle 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 :

Answer : D
We will use the formula of probability:-

Total possible outcomes =642
Favourable outcomes =514
Required probability

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:

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

If a child is selected at random, the probability that he/she does not like to eat potato chips is


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:

Answer:

Answer : B

Number of students with blood group B =12
Total number of students =10+13+12+5
Required probability

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:




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

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




Answer : C

Bulb’s life is given as hours.
From the table, we can see that no bulb has life as hours.
So favourable outcomes
Required probability
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 hours is :





Answer: D

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


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:

Which ‘average’ will be a good representative of the above data and why?

Answer: Median

Given data is :
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:

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,

Total number of terms = 5( odd )

(where n is the number of terms)

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


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:

Here the total number of observations, n=10 which is even.
In case of even number of terms:





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 :

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.
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,

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.




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

Hence the frequency distribution table can be represented as:

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 :


The data can be arranged in ascending order as:


Frequency distribution table:

Question:4

Prepare a continuous grouped frequency distribution from the following data:

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 :

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:

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:
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:

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:

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:
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:

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:

Answer :
The frequency distribution table can be prepared as follows-
Mean




Hence, the value of p is

Question:10

Obtain the meaning of the following distribution:

Answer :

The frequency distribution table can be prepared as follows-
Mean

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

Now, number of boys are

The mean marks of 20 boys =71
Total score of boys

Here, the total score of whole class = Total score of all girls + Total score of all boys

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 :


Mean of 50 observations was found to be 80.4 (Incorrect mean)
The incorrect sum of all the numbers = Incorrect mean Total numbers

It was discovered that 96 was misread as 69 at one place
Hence the correct sum of all the numbers

So correct mean

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 :

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)

Median =24 (given)









Hence, the correct answer is 20.

Question:14

The points scored by a basketball team in a series of matches are as follows:

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:

Number of observation = 16 (even number)
Now using the formula of the median in case number of terms is even




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 :

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:

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

(ii) A household earning Rs 25000 and more per year and owning 2 televisions.
Favourable outcomes =760

(iii) A household not having any television.
Favourable outcomes =20+10

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:

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:

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

(ii) probability of getting a sum more than 10


(iii) probability of getting a sum less than or equal to 5
Favourable events = 14 + 30 + 42 + 55

(iv) probability of getting a sum between 8 and 12
Favourable events = 53 + 46 + 28 = 127

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:

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:

Here, total events = total cartons =700
(i) no defective bulb
Favourable outcomes =400

(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

(iii) defective bulbs less than 4 = defective bulbs equal to 0 or 1 or 2 or 3
Favourable outcomes = 400 + 180 + 48 + 41 = 669

Question:19

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

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:

Here, total events = total number of working days =200
(i) no defective part
Favourable outcomes =50 days =50

(ii) Probability that at least one defective part = 1 - the probability that no defective part


(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)



(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:

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:

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)

(ii) p(person is under 40 years) = p(person having age 20 to 29 years) + p(person having age 30 to 39 years)

Hence the different age group decided the work.

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

Question:1

The following are the marks (out of 100) of 60 students in mathematics.

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.

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.

Question:2

The following are the marks (out of 100) of 60 students in mathematics.

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.





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)

Question:3

Draw a histogram of the following distribution :

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 :

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

Now, we will subtract 0.5 from all lower limits & add 0.5 to all upper limits
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 :

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

Now, we will subtract 0.5 from all lower limits & add 0.5 to all upper limits
Hence the histogram is :

Question:6

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

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 :

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 :

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.

Represent the marks of the students of both the sections on the same graph by two frequency polygons. What do you observe?

Answer:

A frequency polygon is a graph constructed by using lines to join the midpoints of each interval.
Points where frequency is zero:
Difference
First Point
Last point
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.

Find the value of a and hence the frequencies of 30 and 70.

Answer:

The frequency distribution table is as follows:
Mean

[Cross multiply]

Question:10

The mean of the following distribution is 50.

Find the value of a and hence the frequencies of 30 and 70.

Answer:

The frequency distribution table is as follows:
Mean

[Cross multiply]

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.

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


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 :



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




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:


Number of terms, n=25


13^th term is 77
Hence, median is 77