# RD Sharma Class 10 Solutions Chapter 7 Statistics

## RD Sharma Class 10 Solutions Chapter 7 Statistics

### RD Sharma Class 10 Solutions Statistics Exercise 7.1

Question 1.
Calculate the mean for the following distribution :

 X 5 6 7 8 9 f 4 8 14 11 3

Solution:

Question 2.
Find the mean of the following data:

 X 19 21 23 25 27 29 31 f 13 15 16 18 16 15 13

Solution:

Question 3.
If the mean of the following data is 20.6. Find the value of p. (C.B.S.E. 1997)

 X 10 15 p 25 35 y 3 10 25 7 5

Solution:

Question 4.
If the mean of the following data is 15, find p. (C.B.S.E. 1992C)

 X 5 10 15 20 25 f 6 P 6 10 5

Solution:

Question 5.
Find the value of p for the following distribution whose mean is 16.6.

 X 8 12 15 P 20 25 30 f 12 16 20 24 16 8 4

Solution:
Mean = 16.6

Question 6.
Find the missing value of p for the following distribution whose mean is 12.58. (C.B.S.E. 1992C)

 X 5 8 10 12 P 20 25 f 2 5 8 22 7 4 2

Solution:

Question 7.
Find the missing frequency (p) for the following distribution whose mean is 7.68.

 X 3 5 7 9 11 13 f 6 8 15 P 8 4

Solution:

Question 8.
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

 Age (in years) 15 16 17 18 19 20 No. of students 3 8 10 10 5 4

Solution:

Question 9.
Candidates of four schools appear in a mathematics test. The data were as follows :

 Schools No. of Candidates Average Score I 60 75 II 48 80 III Not available 55 IV 40 50

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Solution:

Question 10.
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

 No. of heads per toss No. of tosses 0 38 1 144 2 342 3 287 4 164 5 25 Total 1000

Solution:

Question 11.
The arithmetic mean of the following data is 14, find the value of k. (C.B.S.E. 2002C)

 X 5 10 15 20 25 f 7 k 8 4 5

Solution:
Mean=14

⇒ 14 (24 + k) = 360 + 10k
⇒ 336 + 14k = 360 + 10k
⇒ 14k- 10k- 360 -336 24
⇒ 4k = 24
⇒ k= $\frac { 24 }{ 4 }$= 6 4
Hence k = 6

Question 12.
The arithmetic mean of the following data is 25, find the value of k. (C.B.S.E. 2001)

 X 5 15 25 35 45 f 3 k 3 6 2

Solution:
Mean =25

⇒ 25 (14 + k) = 390 + 15k
⇒ 350 + 25k= 390 + 15k
⇒ 25k- 15k = 390 -350
⇒ 10k = 40 ⇒ k = $\frac { 40 }{ 10 }$= 4
Hence k = 4

Question 13.
If the mean of the following data is 18.75. Find the value of p.

 X 10 15 P 25 30 f 5 10 7 8 2

Solution:

⇒ 460 + 7p = 32 (18.75)
⇒ 460 + 7p = 600
⇒ 7p = 600 – 460 = 140
⇒ p =  $\frac { 140 }{ 7 }$= 20
∴ p = 20

Question 14.
Find the value of p, if the mean of the following distribution is 20.

 X 15 17 19 20 + p 23 f 2 3 4 5p 6

Solution:

⇒ 5p2 + 100p + 295 = 20 (15 + 5p)
⇒ 5p2 + 100p + 295 = 300 + 100p
⇒ 5p2 + 100p – 100p = 300 – 295
⇒  5p2 = 5 ⇒  p2  =  $\frac { 5 }{ 5 }$  = 1
⇒ P= ±1
P = -1 i s not possible
∴ p= 1

Question 15.
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

Solution:

## RD Sharma Class 10 Solutions Chapter 7 Statistics Ex 7.2

### RD Sharma Class 10 Solutions Statistics Exercise 7.2

Question 1.
The. number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:

Compute the mean number of calls per interval.
Solution:
Let assumed mean (A) = 4

Hence mean number of calls per interval = 3.54

Question 2.
Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss

Solution:
Let assumed means (A) = 3

Hence mean number of tosses per head = 2.47

Question 3.
The following table gives the number of branches and number of plants in the garden of a school.

Calculate the average number of branches per plant.
Solution:
Let assumed mean (A) = 4

∴Mean number of branches per plant = 3.62

Question 4.
The following table gives the number of children of 150 families in a village

Find the average number of children per family.
Solution:
Let assumed mean (A) = 3

= 3 – 0.65 = 2.35
Hence mean number of children per family = 2.35

Question 5.
The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:

Find the average number of marks.
Solution:

Question 6.
The number of students absent in a class were recorded every day for 120 days and the

Solution:
Let assumed mean = 3

= 3 + 0.525 = 3.525 = 3.53 (approx)

Question 7.
In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained:

Find the average number of misprints per page.
Solution:
Let assumed mean (A) = 2

= 2 – 127 = 0.73
∴ Average of number of misprints per page = 0.73

Question 8.
The following distribution gives the number of accidents met by 160 workers in a factory during a month.

Find the average number of accidents per worker.
Solution:
Let assumed mean = 2

= 2 – 1.168 = 2 – 1.17 = 0.83 (approx)
∴ Average number of accidents per worker = 0.83

Question 9.
Find the mean from the following frequency distribution of marks at a test in statistics

Solution:
Let assumed mean = 25

∴Average of marked obtained per student = 22.075

## RD Sharma Class 10 Solutions Chapter 7 Statistics Ex 7.3

### RD Sharma Class 10 Solutions Statistics Exercise 7.3

Question 1.
The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Find the average expenditure ( in rupees ) per household.
Solution:

Question 2.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why ?
Solution:
Let assumed mean (A) = 7

Question 3.
Consider the following distribution of daily wages of 50 workers of a facotry.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

= 150 – 4.80 = 145.20
Mean daily wages per worker = Rs. 145.20

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Find the mean of each of the following frequency distribution (5 – 14)
Solution:

Hence heart beats per minute = 75.9

Question 5.
Find the mean of each the following frequency distributions : (5 – 14)

Solution:
Let Assumed mean (A) =15

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:
Let assumed mean = 20

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.
For the following distribution, calculate mean using all suitable methods.

Solution:
Let assumed mean (A) = 12.5

Question 16.
The weekly observation on cost of living index in a certain city for the year 2004 – 2005 arc given below. Compute the weekly cost of living index.

Solution:
Let assumed mean (A)= 1650

= 1650 + 13.46 = 1663.46

Question 17.
The following table shows the marks scored by 140 students in an examination of a certain paper:

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Solution:
(i) Direct Method :

(ii) Shortcut Method:

Question 18.
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency / and (C.B.S.E. 2004)

Solution:

Question 19.
The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs. 18.00. Find out the missing frequency.

Solution:

⇒ 752 + 20p = 792 + 18p
⇒ 20p- 18p = 792 – 752
⇒2p = 40
⇒p = $\frac { 40 }{ 2 }$
Hence missing frequency = 20

Question 20.
If the mean of the following distribution is 27, find the value of p.

Solution:
Mean = 27

⇒ 27 (43 +p) = 1245 + 15p
⇒ 1161 + 21p = 1245 + 15p
⇒ 27p -15p= 1245 – 1161
⇒ 12p = 84
⇒ p = $\frac { 84 }{ 12 }$
Hence p = 1

Question 21.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?
Solution:
We shall apply the assumed mean deviation method
Let assumed mean (A) = 57
We shall choose the method of assumed mean deviation :

= 57 + 3 x $\frac { 25 }{ 100 }$
= 57 + $\frac { 3 }{ 16 }$
= 57 + 0.1875 = 57.1875 = 57.19

Question 22.
The table below shows the daily expenditure on food of 25 households in a locality

Solution:
Let assumed mean (A) = 225

∴ Mean expenditure on food = Rs. 211

Question 23.
To find out the concentration of S02 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below :

Find the mean concentration of S02 in the air.
Solution:
Let assumed mean (A) = 0.10

= 0.10 – 0.00133 = 0.09867 = 0.099 (approx)

Question 24.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days student was absent.

Solution:

∴ Mean number of days a students was absent = 12.475

Question 25.
The following table gives the literacy rate (in percentage) of 3§ cities. Find the mean

Solution:

Question 26.
The following is the cummulative frequency distribution (of less than type) of 1000 persons each of age 20 years and above. Determine the mean age.[NCERT Exemplar]

Solution:
First, we make the frequency distribution of the given data and then proceed to calculate mean by computing class marks (xi), ui’s and fiui‘s as follows:

= 45 + 6.3 = 51.3
Thus, the mean age is 51.3 years.

Question 27.
If the mean of the following frequency distribution is 18, find the missing frequency.

Solution:

Question 28.
Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.

Solution:

⇒ 4+ f2 -f1 = 0
⇒ -f2+ f1 = 4 ……..(ii)
On adding Eqs. (i) and (ii), we get
⇒ 2f1 = 56
⇒ f1= 28
Put the value of f1 in Eq. (i), we get
f2 = 52-28
⇒ f2 = 24
Hence, f1 = 28 and f2 = 24

Question 29.
The daily income of a sample of 50 employees are tabulated as follows:

Find the mean daily income of employees. [NCERT Exemplar]
Solution:
Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Now we first, find the class mark xt of each class and then proceed as follows:

∴ Assumed mean, a = 300.5
Class width, h = 200
and total observation, N = 50
By step deviation method,

= 300.5 + 200 x $\frac { 1 }{ 50 }$x 14
= 300.5 + 56 = 356.5

## RD Sharma Class 10 Solutions Chapter 7 Statistics Ex 7.4

### RD Sharma Class 10 Solutions Statistics Exercise 7.4

Question 1.
Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median :
715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719
Solution:
Arranging in ascending order, we get
694, 696, 699, 705, 710, 712, 715, 716, 719, 724, 725, 728, 729, 734, 745 Here N = 15 which is odd

Question 2.
The following is the distribution of height of students of a certain class in a certain city :

Find the median height.
Solution:
Arranging the classes in exclusive form and then forming its cumulative frequency table as given

Question 3.
Following is the distribution of I.Q. of 100 students. Find the median I.Q.

Solution:
Arranging the classes in exclusive form and then forming it in cumulative frequency table as shown below :

Here N = 100
$\frac { N }{ 2 }$= $\frac { 100 }{ 2 }$= 50
∴ Corresponding median class is = 94.5-104.5
l = 94.5, f= 33, F = 34 and h = 10

Question 4.
Calculate the median from the following data :

Solution:
Writing the given data in cumulative frequency table as shown

Question 5.
Calculate the median from the following data :

Solution:

Question 6.
Calculate the missing frequency from the following distribution, it being given that the median of the distribution

Solution:
Median = 24, let p be the missing frequency.

Question 7.
The following table gives the frequency distribution of married women by age at marriage:

Calculate the median and interpret the results.
Solution:
Writing classes in exclusive form,

Question 8.
The following table gives the distribution of the life time of 400 neon lamps :

Find the median life.
Solution:

= 3000 + 406.98 = 3406.98

Question 9.
The distribution below gives the weight of 30 students in a class. Find the median weight of students :

Solution:
Here  N =30 ,$\frac { N }{ 2 }$$\frac { 30 }{ 2 }$which  lies in the class 55 – 60 ( ∵ 13 < 15 < 19)

Question 10.
Find the missing frequencies and the median for the following distribution if the mean is 1.46.

Solution:
Mean = 1.46, N= 200
Let p1 and p2 be the missing frequencies

⇒ p1+ 2p2= 292 – 140 = 152
114-p2 + 2p2= 152
⇒ p2 = 152- 114 = 38
∴ p1= 114-p2= 114-38 = 76
Hence missing frequencies are 76 and 38
Median = Here $\frac { N }{ 2 }$= $\frac { 200 }{ 2 }$= 100
∴ cf of 2nd class is 46 + 76 = 122
∴ Median = 1

Question 11.
An incomplete distribution is given below :

You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
Solution:
Let p1, and p2 be the missing frequencies
Median = 46 and N = 230

∴ 150 + p1 + p2 = 230

⇒ p1+p2 = 230 – 150 = 80
∴ p2 = 80-p1 ….(i)
∵ Median = 46 which lies in the class 40-50
∴ I = 40, f= 65, F = 42 +p1, h = 10

⇒ 39 = 73 – p1
⇒ p1 = 73 -39 = 34
∴ p2 – 80 – p1 = 80 – 34 = 46
Hence missing frequencies are 34, and 46
Mean Let assumed mean (A) = 45

= 45 + 0.8695
= 45 + 0.87
= 45.87

Question 12.
If the median of the following frequency distribution is 28.5 find the missing frequencies:

Solution:
Mean = 28.5 , N = 60

17 = 25 -f1
⇒ f1= 25 -17 = 8
and f2 = 15-f1 = 15-8 = 7
Hence missing frequencies are 8 and 7

Question 13.
The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:

Solution:
Median = 525, N = 100

⇒ 525 – 500 = (14 -f1) x 5
⇒ 25 = 70- 5f1
⇒ 5f1 = 70 – 25 = 45
⇒ f1 = $\frac { 45 }{ 5 }$= 9
and f2 = 24 – f1 = 24 – 9 = 15
Hence f1 = 9, f2 = 15

Question 14.
If the median of the following data is 32.5, find the missing frequencies.

Solution:
Mean = 32.5 and N= 40

⇒ 2.5 x 12 = 60 – 10f1
⇒ 30 = 60 – 10f1
⇒ 10f1 = 60-30 = 30
⇒ f1 = $\frac { 30 }{ 10 }$=3
∴ f2 = 9 – f1 = 9-3 = 6
Hence f1 = 3, f2= 6

Question 15.
Compute the median for each of the following data:

Solution:
(i) Less than

Here N= 100=
$\frac { N }{ 2 }$= $\frac { 100 }{ 2}$= 50 which lies in the class 70-90 (∵ 50 < 65 and > 43)
∴ l = 70, F =43 , f = 22 ,h = 20

(ii) Greater than

N = 150, $\frac { N }{ 2 }$= $\frac { 150 }{ 2 }$= 75 which lies in the class 110-120 (∵ 75 > 105 and 75 > 60)
∴ l = 110, F = 60 , f=45, h= 10

Question 16.
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and
the following data was obtained.

Find the median height.
Solution:

Here $\frac { N }{ 2 }$= $\frac { 51 }{ 2 }$= 25.5 or 26 which lies in the class 145-150
l= 145, F= 11, f= 18, h= 5

= 145 + 4.03 = 149.03

Question 17.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.

Solution:

Here N = 100, $\frac { N }{ 2 }$= $\frac { 100 }{ 2 }$= 50 which lies in the class 35-40 ( ∵ 45 > 50> 78)
l = 35, F = 45, f= 33, h = 5

= 35 + 0.76 = 35.76

Question 18.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the mean length of leaf.
Solution:

N = 40, $\frac { N }{ 2 }$= $\frac { 40 }{ 2 }$= 20 which lies in the class 144.5-153.5 as 17 < 20 < 29
∴ l= 144.5, F= 17, f= 12, h = 9

= 144.5 + 2.25 = 146.75

Question 19.
An incomplete distribution is given as follows :

You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.
Solution:
Median = 25 and ∑f= N = 170
Let p1 and p2 be two missing frequencies

∴ 110 + p1 +p2 = 170
⇒ p1+ p2 = 170 – 110 = 60
Here N = 170, $\frac { N }{ 2 }$= $\frac { 170 }{ 2 }$= 85
∴ Median = 35 which lies in the class 30-40
Here l = 30, f= 40, F = 30 + p1 and h = 10

20 = 55 – p1
⇒ p1 = 55 – 20 = 35
But p1+ p2 = 60
∴ p2 = 60 -p1 = 60 – 35 = 25
Hence missing frequencies are 35 and 25

Question 20.
The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

Solution:

It is given that n = 20
So, 10 + x + y – 20, i.e., x+y= 10 …(i)
It is also given that median = 14.4
Which lies in the class interval 12-18
So, l = 12, f= 5, cf = 4 + x, h = 6
Using the formula,

Question 21.
The median of the following data is SO. Find the values of p and q, if the sum of all the frequencies is 90.

Solution:

Given, N = 90
$\frac { N }{ 2 }$= $\frac { 90 }{ 2 }$= 45
Which lies in the interval 50-60
Lower limit, l = 50, f= 20, cf= 40 + p, h = 10

∴ P = 5
Also, 78 +p + q = 90
⇒ 78 + 5 + q = 90
⇒ q = 90-83
∴ q = 7

## RD Sharma class 10 Solutions Chapter 7 Statistics Ex 7.5

### RD Sharma Class 10 Solutions Statistics Exercise 7.5

Question 1.
Find the mode of the following data :
(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
Solution:

We see that 5 occurs in maximum times which is 5
∴ Mode = 5

we see that 3 occurs in maximum times i.e. 5
∴ Mode = 3

Here we see that 15 occurs in maximum times i.e. 54
∴ Mode = 15

Question 2.
The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows :

Find the model shirt size worn by the group.
Solution:

We see that frequency of 40 is maximum which is 41
∴Mode = 40

Question 3.
Find the mode of the following distribution.

Solution:

Question 4.
Compare the modal ages of two groups of students appearing for an entrance test :

Solution:
(i) For group A

We see that class 18-20 has the maximum frequency
∴ It is a modal class

(ii)  For group B

We see that class 18-20 has the maximum frequency
∴ It is a modal class

Question 5.
The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science.

Solution:

We see that class 50-60 has the maximum frequency 20
∴ It is a modal class

Question 6.
The following is the distribution of height of students of a certain class in a certain city :

Find the average height of maximum number of students.
Solution:
Writing the classes in exclusive form,

Here model class is 165.5 – 168.5
and l = 165.5, h = 3, f= 142, f1= 118, f2= 127

Question 7.
The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:

(i) We see that class 35-45 has the maximum frequency 23
∴ It is a modal class

= 35.375 = 35.37 years
We see that mean is less than its mode

Question 8.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.
Solution:

We see that class 60-80 has the maximum frequency 61
∴ It is the modal class
Here l = 60, f= 6, f1 = 52, f2 = 38,h = 20
∴ Modal of life time (in hrs.)

Question 9.
The following table gives the daily income of 50 workers of a factory :

Find the mean, mode and median of the above data. (C.B.S.E. 2009)
Solution:

Here i = 20 and AM = 150

(ii) Max frequency  = 14
∴ Model class = 120 – 140

Question 10.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:

Solution:
Let assumed mean (A) = 32.5

(i)  We see that the class 30-35 has the maximum frequency
∴ It is the modal class
Here l = 30,f = 10, f1 = 9, f2 = 3, h = 5

Question 11.
Find the mean, median and mode of the following data:

Solution:
Let assumed mean (A) =175

(i) Here N = 25, $\frac { 5 }{ 3 }$= $\frac { 25 }{ 2 }$= 12.5 or 13 which lies in the class 150-200
l= 150, F= 10, f= 6, h = 50

Question 12.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Solution:

We see that class 40-50 has the maximum frequency
∴ It is a modal class

Question 13.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them:

Solution:
Let assumed mean (A) = 135

Here N = 68 , $\frac { N }{ 2 }$= $\frac { 68 }{ 2 }$= 34

Question 14.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, And the modal size of the surnames.
Solution:
Let Assumed mean (A) = 8.5

(i) Here N = 100, $\frac { N }{ 2 }$= $\frac { 100 }{ 2 }$= 50 which lies in the class 7-10
Here l = 7, F = 36, f= 40, h = 3

Question 15.
Find the mean, median and mode of the following data (C.B.S.E. 2008)

Solution:
Let assumed mean A = 70

(i) Here N = 50, then $\frac { N }{ 2 }$= $\frac { 50 }{ 2 }$= 25 which lies in the class 60-80
∴ l= 60, F= 24, f = 12 , h = 20

Question 16.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. .Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

(i) We see that the c ass 1500-2000 has maximum frequency 40
∴ It is a modal class
Here l = 1500, f = 40 , f1 = 24 , f2 = 33 , h =500

Question 17.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Solution:

We see that class 4000-5000 has the maximum frequency 18
∴It is a modal class

Question 18.
The frequency distribution table of agriculture holdings in a village is given below:

Find the modal agriculture holdings of the village.
Solution:
Here the maximum class frequency is 80,
and the class corresponding to this frequency is 5-7.
So, the modal class is 5-7.
l (lower limit of modal class) = 5
f1 (frequency of the modal class) = 80
f0 (frequency of the class preceding the modal class) = 45
f2(frequency of the class succeeding the modal class) = 55
h (class size) = 2

Hence, the modal agricultural holdings of the village is 6.2 hectares.

Question 19.
The monthly income of 100 families are given as below:

Solution:
In a given data, the highest frequency is 41, which lies in the interval 10000-15000.
Here, l = 10000,f1 = 41, f0 = 26,f2 = 16 and h = 5000

## RD Sharma class 10 Solutions Chapter 7 Statistics Ex 7.6

### RD Sharma Class 10 Solutions Statistics Exercise 7.6

Question 1.
Draw an ogive by less than method for the following data :

Solution:

Take numbers of rooms along the x-axis and c.f along the y-axis
Plot the points (1, 4), (2, 13), (3, 35), (4, 63), (5, 87), (6, 99), (7, 107), (8, 113), (9, 118) and (10, 120) on the graph and join them and with free hand to get an ogive as shown. This is the less than ogive.

Question 2.
The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

Prepare a cumulative frequency table by less than method and draw on ogive.
Solution:

Take marks along ,t-axis and no. of students (c.f) along 3 -axis. Now plot the points (640, 16), (680, 61), (720, 217), (760, 501), (800, 673), (840, 732) and (880, 750) on the graph and join them with free hand. This is the less than ogive.

Question 3.
Draw an ogive to represent the following frequency distribution:

Solution:
Representing the classes in exclusive form :

Represent class intervals along x-axis and c.f. along y-axis. Now plot the points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23) and (24.5, 26) on the graph and join then in free hand to get an ogive as shown.

Question 4.
The monthly profits (in Rs.) of 100 shops are distributed as follows:

Draw the frequency polygon for it.
Solution:

Represent profits per shop along x-axis and no. of shop (c.f.) along y-axis.
Plot the points (50, 12), (100, 30), (150, 57), (200, 77), (250, 94) and (300, 100) on the graph and join them with ruler. This is the cumulative polygon as shown.

Question 5.
The following distribution gives the daily income of 50 workers of a factory:

Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive.
Solution:

Now plot the points (120, 12), (140, 26), (160, 34), (180, 40) and (200,50) on the graph and join them with free hand to get an ogive which is less than.

Question 6.
The following table gives production yield per hectare of wheat of 100 farms of a village:

Draw ‘less than’ ogive and ‘more than’ ogive.
Solution:
(i) Less than

Now plot the points (55, 2), (60, 10), (65, 22), (70, 46), (75, 84) and (80, 100) on the graph and join them in free hand to get a less than ogive.
(ii) More than

Now plot the points (50, 100), (55, 84), (60, 46), (65, 22), (70, 10), (75, 2) and (80, 0) on the graph and join than in free hand to get a more than ogive as shown below :

Question 7.
During the medical check-up of 35 students of a class, their weights were recorded as follows :

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula. (C.B.S.E. 2009)
Solution:

Plot the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35) on the graph and join them in free hand to get an ogive as shown.
Here N = 35 which is odd
$\frac { N }{ 2 }$= $\frac { 25 }{ 2 }$= 17.5
From 17.5 on y-axis draw a line parallel to x-axis meeting the curve at P. From P, draw PM ⊥ x-axis
∴ Median which is 46.5 (approx)
Now N = 17.5 lies in the class 46 – 48 (as 14 < 17.5 < 28)
∴ 46-48 is the median class
Here l= 46, h = 2,f= 14, F= 14

Question 8.
The annual rainfall record of a city for 66 days is given in the following tab

Calculate the median rainfall using ogives of more than type and less than type. [NCERT Exemplar]
Solution:
We observe that, the annual rainfall record of a city less than 0 is 0. Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0-10. So, the total annual rainfall record of a city for less than 10 cm is 0 + 22 = 22 days. Continuing in this manner, we will get remaining less than 20, 30, 40, 50 and 60.
Also, we observe that annual rainfall record of a city for 66 days is more than or equal to 0 cm. Since, 22 days lies in the interval 0-10. So, annual rainfall record for 66 – 22 = 44 days is more than or equal to 10 cm. Continuing in this manner we will get remaining more than or equal to 20, 30 , 40, 50, and 60.
Now, we construct a table for less than and more than type.

To draw less than type ogive we plot the points (0,0), (10,22), (20,32), (30, 40), (40, 55), (50, 60), (60, 66) on the paper and join them by free hand.
To draw the more than type ogive we plot the points (0, 66), (10, 44), (20, 34), (30, 26), (40, 11), (50, 6) and (60, 0) on the graph paper and join them by free hand.

∵ Total number of days (n) = 66
Now, $\frac { n }{ 2 }$= 33
Firstly, we plot a line parallel to X-axis at intersection point of both ogives, which further intersect at (0, 33) on Y- axis. Now, we draw a line perpendicular to X-axis at intersection point of both ogives, which further intersect at (21.25, 0) on X-axis. Which is the required median using ogives.
Hence, median rainfall = 21.25 cm.

Question 9.
The following table gives the height of trees:

Draw ‘less than’ ogive and ‘more than’ ogive.
Solution:
(i) First we prepare less than frequency table as given below:

Now we plot the points (7, 26), (14, 57), (21, 92), (28, 134), (35, 216), (42, 287), (49, 341), (56, 360) on the graph and join then in a frequency curve which is ‘less than ogive’
(ii) More than ogive:
First we prepare ‘more than’ frequency table as shown given below:

Now we plot the points (0, 360), (7, 341), (14, 287), (21, 216), (28, 134), (35, 92), (42, 57), (49, 26), (56, 0) on the graph and join them in free hand curve to get more than ogive.

Question 10.
The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:

Draw both ogives for the above data and hence obtain the median.
Solution:

Now plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) on the graph and join them to get a more than curve.
Less than curve:

Now plot the points (10, 3), (15, 7), (20, 10), (25, 14), (30, 16), (35, 28) and (40, 30) on the graph and join them to get a less them ogive. The two curved intersect at P. From P, draw PM 1 x-axis, M is the median which is 22.5
∴ Median = Rs. 22.5 lakh

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