Chapter 14 Statistics and Probability solved by expert teachers as per NCERT (CBSE) Book guidelines. All the chapter wise questions with solutions to help you to revise the complete CBSE syllabus and score more marks in Your board examinations.

NCERT Exemplar Class 9 Maths Solutions Statistics and Probability

Exercise 14.1: Multiple Choice Questions (MCQs)

Question 1:
The class mark of the class 90-120 is
(a) 90                    (b) 105                    (c) 115                        (d) 120
Solution:
(b) In a given class 90-120, upper class = 120
and lower class = 90

Question 2:
The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is
(a) 10                      (b) 15                      (c) 18                      (d) 26
Solution:
(d) In a given data, maximum value = 32 and minimum value = 6
We know, range of the data = maximum value – minimum value
= 32 – 6 = 26
Hence, the range of the given data is 26.

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
Solution:
(b) Let x and y be the upper and lower class limit in a frequency distribution.
Now, mid value of a class  (x + y )/2=10  [given]
=> x + y = 20 …(i)
Also, given that, width of class  x- y = 6 …(ii)
On adding Eqs. (i) and (ii), we get
2x =20+ 6
=> 2x =26 => x = 13
On putting x = 13 in Eq. (i), we get
13+y = 20=> y = 7
Hence, the lower limit of the class is 7.

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
Solution:
(c) Let x and y be the upper and lower class limit of frequency distribution.
Given, width of the class = 5
=> x-y= 5 …(i)
Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get
x – 10= 5 => x = 15 So, the upper class limit of the lowest class is 15.
Hence, the upper class limit of the highest class
=(Number of continuous classes x Class width + Lower class limit of the lowest class)
= 5 x 5+10 = 25+10=35
Hence,’the upper class limit of the highest class is 35.
Alternate Method
After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.
Thus, the highest class is 30-35,
Hence, the upper limit of this class is 35.

Question 5:
If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is
(a) 2 m+l                      (b) 2 m-l                    (c) m-l                           (d) m-2l
Solution:
(b) Let x and y be the lower and upper class limit of a continuous frequency distribution.
Now, mid-point of a class = (x + y)/2 = m [given]
=> x + y = 2 m =x + l = 2m
[∴ y = l = upper class limit (given)]
=> x = 2 m-l
Hence, the lower class limit of the class is 2m – l.

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
Solution:
(b) Since, the difference between mid value is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option (c) and (d) are wrong.
Since, the mid value is 20 which can get only, if we take option (b)

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
Solution:
(b) Since, the class interval 10-20 is the first interval of frequency distribution and 20-30 is the next one but the number 20 is present in both intervals. We know that, the presence of 20 in the interval 10-20 is not fully 100% while in the next interval 20-30, presence of it fully 100%.

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 interval is constructed for the following data
268, 220, 368, 258, 242, 310, 272, 342,
310, 290, 300, 320, 319, 304, 402, 318,
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
Solution:
(c) Here, we arrange the given data into groups like 210-230, 230-250 390-410.
(since, our data is from 210 to 406). The class width in this case is 20.
Now, the given data can be arrange in tabular form as follows

Hence, the frequency of the c ass 310-330 is 6.

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
Solution:
(b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is
from 14 to 112). The class width in this case is 9.
Now, the given data can be arranged in tabular form as follows

Hence, the number of classes in distribution will be 10.

Question 10:
To draw a histogram to represent the following frequency distribution.

The adjusted frequency for the class 25-45 is
(a) 6                          (b) 5                       (c) 3                             (d) 2
Thinking Process
(i) Adjust the class interval by using the formula

Solution:

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
Solution:

Question 12:
If the mean of the observations x, x + 3, x + 5, x + 7and x + 10 is 9, then mean of the last three observations is
(a) 10 (1/3)               (b) 10 (2/3)                     (c) 11 (1/3)                  (d) 11  (2/3)
Solution:

Question 13:
If x represents the mean of n observations x₁, x₂, ……,x_n, then value of

(a) 1             (b) 0                (c) 2              (d) 3
Solution:

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
Thinking Process
First, add 5 on each term of an observation, then use the formula for mean

Further, simplify it to get the required result.
Solution:

Question 15:
Let x be the mean of x₁, x₂,….,x_n and y be the mean of y₁, y₂, ……,y_n the mean of z is x₁, x₂,….,x_n , y₁, y₂, ……,y_n then z is equal to

Solution:

Question 16:

Solution:

Question 17:
If x₁, x₂, x₃,…, x_n are the means of n groups with n₁, n₂,…, n_n number of observations, respectively, then the mean x of all the groups taken together is given by

Solution:

Question 18:
The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be
(a) 50.5                       (b) 51                      (c) 51.5                        (d) 52
Solution:

Question 19:
There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. The mean of the given number is
(a) 46.5                   (b) 49.5                   (c) 53.5                     (d) 56.5
Solution:

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
Solution:
(b) Given, mean of 25 observations = 36
∴ Sum of 25 observations = 36 x 25 = 900 Now, the mean of first 13 observations = 32
∴ Sum of first 13 observations = 13 x 32 = 416 and the mean of last 13 observations = 40
∴ Sum of last 13 observations = 40 x 13 = 520
So, 13th observation = (Sum of last 13 observations + Sum of first 13 observations) – (Sum of 25 observations)
=(520 + 416)-900 = 936 – 900 = 36
Hence, the 13th observation is 36.

Question 21:
The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is
(a) 45                  (b) 49.5                      (c) 54                        (d) 56
Solution:

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 proceeding classes
Solution:
(c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

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
Thinking Process
First, arrange the given data in ascending order. Then, use the following formula for median.

Solution:

Question 24:
The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is
(a) 14                     (b) 15                       (c) 16                      (d) 17
Thinking Process
Find the maximum number of repeated observation to get the value of mode.
Solution:
(b) We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times.
Hence, the mode of the given data is 15.

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
Solution:z
(d) The total number of people in sample study, n(S) = 642.
The number of people who have high school certificate, n(E) = 514.
So, the probability that the person selected has a high school certificate
= n(E)/n(S)=514/642 = 0.8
Hence, the probability that the person has a high school certificate is 0.8.

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
Solution:
(c) Total number of survey children’s age from 19-36 months, n(S) = 364 In those of them 91 out of them liked to eat potato chips.
∴ Number of children who do not like to eat potato chips, n(E) = 364 – 91 = 273
∴ Probability that he/she does not like to eat potato chips  = n(E)/n(S) = 273/364 = 0.75
Hence, the probability that he/she does not like to eat potato chips is 0.75.

Question 27:
In a medical examination of students of a class, the following blood groups are recorded.

Solution:
(c) Total number of students in a medical examination, n(S) = 40
Number of persons who have B blood group, n(E) = 12
∴ Probability that he/she has blood group B = n(E)/n(S) = 12/40 = 3/10
Hence, the probability that he/she has blood group B is 3/10.

Question 28:
Two coins are tossed 1000 times and the outcomes are recorded as below

Solution:
(c) The total number of coins tossed, n(S) = 1000
Number of outcomes in which atmost one head, n(E) = 550 + 250 = 800
=n(E)/n(S) = 800/1000 = 4/5
Hence, the probability for atmost one head is 4/5 .

Question 29:
80 bulbs are selected at random from a lot and their life time (in hours) is recorded in the form of a frequency table given below

Solution:
(c) Total bulb in a lot, n (S) = 80
Number of bulbs whose life time is 1150, n(E) = 0
Probability that its life time is 1150 h = n(E)/n(S) = 0/80 = 0
Hence, the probability that its life time is 1150 is 0.

Question 30:
Refer to Q. 29. The probability that bulbs selected randomly from the lot has life less than 900 h, is
(a) 11/40                  (b) 5/16                 (c) 7/16                (d) 9/16
Solution:
(d) Total number of bulbs in a lot, n(S) = 80
Number of bulbs whose life time is less than 900 h, n(E) = 10 + 12 + 23 = 45
Probability that bulbs has life time less than 900 h =n(E)/n(S) = 45/80 = 9/16
Hence, the probability that bulb has life time less than 900 is 9/16.

Exercise 14.2: Very Short Answer Type Questions

Question 1:
The frequency distributionhas been represented graphically as follows.

Do you think this representation is correct? Why?
Solution:
No, here the widths of the rectangles are varying, so we need to make certain modifications in the length of the rectangles so that the areas are proportional to the frequencies. We proceed as follows

1. Select a class interval with the minimum class size, here the minimum class size is 20.
2. The length of the rectangles are then modified to be proportionate to the class size 20. Now, we get the following modified table
   
   So, the correct histogram with varying width is given below
   

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 and 44.
Which average will be a good representative of the above data and why?
Solution:
Median will be a good representative of the data, because

1. each value occurs once.
2. the data is influenced by extreme values.

Question 3:
A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median?
Solution:
The child does not understand, that data has to be arranged in ascending or descending 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 and 9. Since, the number of matches is 10 (an even number), therefore

Is it is correct answer why?
Solution:
No. It is not the correct answer, because the data have to be arranged in ascending or descending order before finding the median.
Arranging the data in ascending order 1,1,2, 3, 4, 5, 6, 7, 8, 9.
Here, number of observations is 10, which is even.

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.
Solution:
It is not correct. Because in a histogram, the area of each rectangle is proportional to the corresponding 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 and 1.64.
Is it correct to say the last interval will be 1.55-1.73? Justify your answer.
Solution:
It is not correct. Because the difference between two consecutive class marks should be equal to the class size. Here, difference between two consecutive marks is 0.1 and class size of 1.55-1.73 is 0.18, which are not equal.

Question 7:
30 children were asked about the number of hours they watched TV programmes last week. The result are recorded as under

Can we say that the number of children who watched TV for 10 or more hour in a week is 22? Justify your answer.
Solution:
No, infact the number of children who watched TV for 10 or more hour in a week is 4 + 2 i.e.,6.

Question 8:
Can the experimental probability of an event be a negative number? If not, why?
Solution:
No, since the number of trials in which the event can happen cannot be negative and the total number of trials is always positive.

Question 9:
Can the experimental probability of an event be greater than 1? Justify your answer.
Solution:
No, since the number of trials in which the event can happen cannot be greater than the total number of trials.

Question 10:
As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be ½. Is it correct? If not, write the correct one.
Solution:
No, since the number of coin increases, the ratio of the number of heads to the total number of tosses will be nearer to ½ but not exactly ½.

Exercise 14.3: Short Answer Type Questions

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.
Solution:
The number of students who have a certain type of blood group is called the frequency of . those blood groups. A frequency distribution table for the given data is given below

Question 2:
The value of n upto 35 decimal places is given below 3. 14159265358979323846264338327950288 Make a frequency distribution of the digits 0 to 9 after the decimal point.
Solution:
The number of repeated digit is called the frequency of those digits. A frequency distribution table for the given data is given below.

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.
Solution:
The number of students who have the same marks in mathematics is called the frequency of that mark.
A frequency distribution table for the given data is given below

Question 4:
Prepare a continuous grouped frequency distribution from the following data

Also, find the size of class intervals.
Thinking Process

1. Firstly, find the difference between two mid-points. Then, using the mid-point formula
   
   a is lower limit of the class and 70 is the difference between 2 the two mid-points.
2. Further, prepare a continuous grouped frequency distribution table. Here, we see that the difference between two mid-points is 15-5 i.e., 10. It means the width of the class interval is 10.

Solution:

Question 5:
Convert the given frequency distribution into a continuous grouped frequency distribution

In which intervals would 153.5 and 157.5 be included?
Solution:
It is clear that, the given table is in inclusive (discontinuous) form. So, we first convert it into exclusive form.
Now, consider the classes 150-153,154-157 Lower limit of 154-157 = 154 and upper limit of 150-153=153 Required difference = 154 -153 = 1
So, half the difference = ½ = 0.5
So, we subtract 0.5 from each lower limit and add 0.5 to each upper limit.
The table for continuous grouped frequency distribution is given below

Thus, 153.5 and 157.5 woukd use in the class intervals 153.5-157.5 and 157.5-161.5, respectively.

Question 6:
The expenditure of a family of different heads in a month is given below

Draw a bar graph to represent the data above.
Solution:
We draw a bar graph of this data in the following steps.
Step I We represent the heads (variable) on the horizontal axis choosing any scale, as the width of the bar is not important. But for clarity, we take equal widths for all bars and maintain equal gaps in between them. Let one head be represented by one unit.
Step II We represent the expenditure on the vertical axis. Since, the maximum expenditure is Rs. 4000, we can choose the scale as 1 unit = Rs. 500.
Step III To represent our first head i.e.,food, we draw a rectangular bar with width 1 unit and height 8 units.
Step IV Similarly, other heads are represented by leaving a gap of ½ unit in between two consecutive bars.
The bar graph for given data is shown below

Question 7:
Expenditure on education of a country during a five years period (2002-2006), in crore of rupees, is given below

Represent the information above by a bar graph.
Solution:
We draw bar graph of this data in the following steps
Step I We represent the education of a country (variable) on the horizontal axis choosing any scale, as the width of the bar is not important. But for clarity, we take equal widths for all bars and maintain equal gaps in between them. Let one head be represented by one unit.
Step II We represent the expenditure on the vertical axis. Since, the maximum expenditure is Rs. 240 crore, we can choose the scale as 1 unit = Rs. 25 crore.
Step III To represent our first education of a country i.e.,elementary education, we draw a rectangular bar with width 1 unit and height 9.6 units.
Step IV Similarly, other heads are represented by leaving gap of ~ unit in between two consecutive bars.
The bar graph for given data is shown below

Question 8:
The following table gives the frequencies of most commonly used letters o, e, i, o, r, t, u, from a page of a book

Represent the information above by a bar graph.
Solution:
We draw bar graph of this data in the following steps
Step I We represent the letters (variable) on the horizontal axis choosing any scale, as the width of the bar is not important. But for clarity, we take equal widths for all bars and maintain equal gaps in between them. Let one letter be represented by one unit.
Step II We represent the letters on the vertical axis. Since, the maximum frequency is 125, we can choose the scale as 1 unit = 15 frequency.
Step III To represent our first letter i.e., a, we draw a rectangular bar with width 1 unit and height 5 units.
Step IV Similarly, other heads are represented by leaving a gap of ½ unit in between two consecutive bars.
The bar graph for given data is shown below

Question 9:
If the mean of the following data is 20.2, then find the value of p.

Solution:

Question 10:
Obtain the mean of the following distribution.

Solution:

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.
Solution:
Given, the number of girls out of 50 students = 30
∴The number of boys out of 50 students = 50 – 30 = 20 Now, the marks obtained by 30 girls = 30 x 73 = 2190 and the marks obtained by 20 boys = 20 x 71 = 1420
[mean of marks scored by girls in a test is 73 and that of boys is 71]

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.
Solution:
Given, the mean of 50 observations = 80.4
∴ Sum of 50 observations = 80.4 x 50 = 4020 But later, we observe that 96 was misread as 69 at one place.

Question 13:
Ten observations 6, 14,15, 17, x + 1, 2x – 13, 30, 32, 34 and 43 are written in an ascending order. The median of the data is 24. Find the value of x.
Solution:
Given observations are 6,14,15,17, x + 1,2x -13, 30, 32, 34, 43 Here, total number of given observations, n = 10 (even)
Since, n is even, so we use the formula for median,

Question 14:
The points scored by a basket ball 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.
Solution:
First, arrange the given points scored in ascending order
2, 5, 7, 7, 8,10, 10, 10, 14, 17,18, 24, 25, 27, 28, 48 Here, number of given observations is n = 16 (even)
Since, n is even, so we use the formula for median,

We know that, mode is the observation which is repeated maximum number of times. Here, we see in the given data, 10 is repeated 3 times.
So, mode of a given data is 10.
Hence, median and mode for the given data are 12 and 10, respectively.

Question 15:
In the figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.

Solution:
It is clear from the histogram that class interval start from 150-200, 200-250, having width
The frequency distribution table is given below

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 owning 2 televisions.
(iii) of a household not having any television.
Solution:

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, then 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?
Solution:

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 2-6?
(iii) defective bulbs less than 4?
Solution:

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) atleast one defective part,
(iii) not more than 5 defective parts,
(iv) more than 13 defective parts.
Solution:
Total number of working days, n(S) = 200
(i) Number of days in which no defective part is, n(E₁) = 50
Probability that no defective part = n(E₁)/n(S) = 50/200 = ¼ = 0.25

Question 20:
A recent survey found that the ages of workers in a factory as follows.

If a person is selected at random, find the probability that the person is

(i) 40 yr or more.                                  (ii) under 40 yr.
(iii) having age from 30-39 year.      (iv) under 60 but over 39 year.
Solution:

Exercise 14.4: Long Answer Type Questions

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.
Solution:
First, arranging the given data in ascending order.
4, 5, 7, 7,13,15,16,16,17,17,19, 24, 25, 26, 27, 28, 30, 31,31,34, 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, 86, 92, 95, 97.
We arrange the given data into groups like 0-9, 10-19, 20-29, … The class width in each case is 10.
The frequency distribution of the given data is given below

Question 2:
Refer to Q.1 above. Construct a grouped frequency distribution table with width 10 of each class, in such a way that one of the class is 10-20 (20 not included).
Solution:
We arrange the given data into groups like 0-10, 10-20, 20-30 in which upper class
limit is not included in that class. The class width in each case is 10.
The frequency distribution of the given data is given below

Question 3:
Draw a histogram of the following distribution

Solution:
Clearly, the given frequency distribution is in exclusive form.
Along the horizontal axis, we represent the class intervals of heights on some suitable scale. The corresponding frequencies of number of students are represented along the vertical axis on a suitable scale.
Since, the given intervals start with 150-153. It means that there is some break (vw) is indicated near the origin to signify the graph is drawn with a scale beginning at 150.
A histogram of the given distribution is given below.

Question 4:
Draw a histogram to represent the following grouped frequency distribution

Thinking Process

1. Firstly, convert the given data into continuous form by adjusting minus 0.5 in lower limit and plus 0.5 in upper limit
2. Further, we take height in horizontal axis and frequency in vertical axis. Corresponding frequencies of the height are plotted in the form of rectangular and get the required histogram.

Solution:
The given frequency distribution is in inclusive form. So, first we convert it into exclusive form.
Now, consider the class 20-24, 25-29.
Lower limit of 25-29 is 25.
Upper limit of 20-24 is 24.
Thus, the half of the difference is = (25- 24)/2= ½  = 0.5
So, we subtract 0.5 from each lower limit and add 0.5 to each upper limit.
The table for continuous grouped frequency distribution is given below

Thus, the given data becomes in exclusive form.
Along the horizontal axis, we represent the class intervals of ages on some suitable scale. The corresponding frequencies of number of teachers are represented along the vertical axis on a suitable scale.
Since, the given intervals start with 19.5-24.5. It means that, there is some break (vw) indicated near the origin to signify the graph is drawn with a scale beginning at 19.5.
A histogram of the given distribution is given below.

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.
Thinking Process

1. Firstly, convert the given distribution into exclusive form by using adjusting factor (127-126)/2 = 0.5
2. Further, we subtract 0.5 from each lower limit and add 0.5 to each upper limit.
3. Finally, draw a histogram using the modified data.

Solution:
The given frequency distribution is in inclusive form. So, first we convert it into exclusive form.
Now, adjusting factor = (127 -126)/2 = ½  = 0.5
So, we subtract 0.5 from each lower limit and add 0.5 to each upper limit.
The table for continuous grouped frequency distribution is given below

The table for continuous grouped frequency distribution is given below
Thus, the given data becomes in exclusive form.
Along the horizontal axis, we represent the class intervals of length on some suitable scale. The corresponding frequencies of number of leaves are represented along the Y-axis on a suitable scale.
Since, the given intervals start with 117.5-126.5. It means that, there is some break (vw) indicated near the origin to signify the graph is drawn with a scale beginning at 117.5.
A histogram of the given distribution is given below

Question 6:
The marks obtained (out of 100) by a class of 80 students are given below

Solution:
In the given frequency distribution, class sizes are different. So, we calculate the adjusted frequency for each class.
Here, minimum size = 20-10 = 10 .
We use the formula,

The modified table for frequency distribution is given by

Along the horizontal axis, we represent the class intervals marks on some suitable scale. The corresponding frequencies of number of students are represented along the vertical axis on a suitable scale.
Since, the given intervals start with 10-20. It means that, there is some break (AW) indicated near the origin to signify the graph is drawn with a scale beginning at 10.
Now, we draw rectangles with class intervals as the bases and the corresponding adjusted frequencies as heights.
A histogram of the given distribution, is given below

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.
Solution:
Clearly, the given frequency distribution is in exclusive form.
Along the horizontal axis, we represent the class intervals on some suitable scale. The corresponding frequencies are represented along the vertical axis on a suitable scale.
We construct rectangles with class intervals as the bases and the respective frequencies as the heights.
Let us draw a histogram for this data and mark the mid-points of the top of the rectangles as B, C, D, E, F, G and H, respectively. Here, the first class is 30-40 and the last class is 90-100.
Also, consider the imagined classes 20-30 and 100-110 each with frequency O. The class marks of these classes are 25 and 105 at the points A and I, respectively.

Join all these points by dotted line.
Then, the curve ABCDEFGHI is the required frequency polygon.

Question 8:
Refer to Q. 7. Draw the frequency polygon representing the above data without drawing the histogram.
Solution:
We have to draw a frequency polygon without a histogram.
Firstly, we find the class marks of the classes given that is 30-40, 40-50, 50-60, 60-70 …
The class mark = (30 + 40)/2
=> 70/2 = 35
Similarly, we can determine the class marks of the other classes.
So, table for class marks is shown below

We can draw a frequency polygon by plotting the class marks along the horizontal axis and the frequency along the vertical axis. Now, plotting all the points B (35, 3), C (45, 6), D (55, 25), E (65, 65), F (75, 50), G (85, 28), H (95,14), also plot the point corresponding to the considering classes 20-30 and 100-110 each with frequency 0. Join all these point line segments.

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?
Solution:

Question 10:
The mean of the following distribution is 50.

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

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.
Solution:

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.
Solution: