Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.5  Set 1
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:
(i)
Value (x) 3 4 5 6 7 8 9 Frequency (f) 4 2 5 2 2 1 2 Therefore, mode = 5 because 5 occurs the maximum number of times.
(ii)
Value (x) 3 4 5 6 7 8 9 Frequency (f) 5 2 4 2 2 1 2 Therefore, mode = 3 because 3 occurs the maximum number of times.
(iii)
Value (x) 8 15 18 19 20 24 25 Frequency (f) 1 4 1 1 1 2 1 Therefore, mode = 15 because 15 occurs the maximum number of times.
Question 2. The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows:
Shirt size:  37  38  39  40  41  42  43  44 
Number of persons:  15  25  39  41  36  17  15  12 
Find the model shirt size worn by the group.
Solution:
From the data present in the table we conclude that
Model shirt size = 40
Because shirt size 40 occurred for the maximum number of times.
Question 3. Find the mode of the following distribution.
(i)
Class interval:  0 – 10  10 – 20  20 – 30  30 – 40  40 – 50  50 – 60  60 – 70  70 – 80 
Frequency:  5  8  7  12  28  20  10  10 
Solution:
From the given table we conclude that
The maximum frequency = 28
So, the model class = 40 – 50
and,
l = 40, h = 50 40 = 10, f = 28, f1 = 12, f2 = 20
Using the formula of mode
Mode =
= 40 + 160/ 24
= 40 + 6.67
= 46.67
Hence, the mode = 46.67
(ii)
Class interval  10 – 15  15 – 20  20 – 25  25 – 30  30 – 35  35 – 40 
Frequency  30  45  75  35  25  15 
Solution:
From the given table we conclude that
The maximum frequency = 75
So, the modal class = 20 – 25
And,
l = 20, h = 25 – 20 = 5, f = 75, f_{1} = 45, f_{2} = 35
Using the formula of mode
Mode =
= 20 + 150/70
= 20 + 2.14
= 22.14
Hence, the mode = 22.14
(iii)
Class interval  25 – 30  30 – 35  35 – 40  40 – 45  45 – 50  50 – 55 
Frequency  25  34  50  42  38  14 
Solution:
From the given table we conclude that
The maximum frequency = 50
So, the modal class = 35 – 40
And,
l = 35, h = 40 – 35 = 5, f = 50, f_{1} = 34, f_{2} = 42
Using the formula of mode
Mode =
= 35 + 80/24
= 35 + 3.33
= 38.33
Hence, the mode = 38.33
Question 4. Compare the modal ages of two groups of students appearing for an entrance test:
Age in years  16 – 18  18 – 20  20 – 22  22 – 24  24 – 26 
Group A  50  78  46  28  23 
Group B  54  89  40  25  17 
Solution:
For Group A:
From the given table we conclude that
The maximum frequency = 78.
So, the model class = 18 – 20
And,
l = 18, h = 20 – 18 = 2, f = 78, f_{1} = 50, f_{2} = 46
Using the formula of mode
Mode =
= 18 + 56/60
= 18 + 0.93
= 18.93 years
For Group B:
From the given table we conclude that
The maximum frequency = 89
The modal class = 18 – 20
And,
l = 18, h = 20 – 18 = 2, f = 89, f_{1} = 54, f_{2} = 40
Using the formula of mode
Mode =
= 18 + 70/84
= 18 + 0.83
= 18.83 years
After finding the mode of both A and B group we conclude that
the modal age of the Group A is greater than Group B.
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.
Marks  0 – 10  10 – 20  20 – 30  30 – 40  40 – 50  50 – 60  60 – 70  70 – 80  80 – 90  90 – 100 
Frequency  3  5  16  12  13  20  5  4  1  1 
Solution:
From the given table we conclude that
The maximum frequency = 20
The modal class = 50 – 60
And,
l = 50, h = 60 – 50 = 10, f = 20, f_{1} = 13, f_{2} = 5
Using the formula of mode
Mode =
= 50 + 70/22
= 50 + 3.18
= 53.18
Hence, the mode = 53.18
Question 6. The following is the distribution of height of students of a certain class in a city:
Height (in cm)  160 – 162  163 – 165  166 – 168  169 – 171  172 – 174 
No. of students:  15  118  142  127  18 
Find the average height of maximum number of students.
Solution:
Heights (exclusive)  160 – 162  163 – 165  166 – 168  169 – 171  172 – 174 
Heights (inclusive)  159.5 – 162.5  162.5 – 165.5  165.5 – 168.5  168.5 – 171.5  171.5 – 174.5 
No of students  15  118  142  127  18 
From the given table we conclude that
The maximum frequency = 142
The modal class = 165.5 – 168.5
And,
l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f_{1} = 118, f_{2} = 127
Using the formula of mode
Mode =
= 165.5 + 72/39
= 165.5 + 1.85
= 167.35 cm
Hence, the average height of maximum number of students = 167.35 cm
Question 7. The following table shows the ages of the patients admitted in a hospital during a year:
Ages (in years):  5 – 15  15 – 25  25 – 35  35 – 45  45 – 55  55 – 65 
No of students:  6  11  21  23  14  5 
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
For mean:
Let us considered mean (A) = 30
Age (in years) Number of patients f_{i} Class marks x_{i} d_{i} = x_{i} – 275 f_{i}d_{i} 5 – 15 6 10 20 120 15 – 25 11 20 10 110 25 – 35 21 30 0 0 35 – 45 23 40 10 230 45 – 55 14 50 20 280 55 – 65 5 60 30 150 N = 80 From the table we get
Σf_{i} = N = 80 and Σf_{i} d_{i} = 430.
Using the formula of mean
= 30 + 430/80
= 30 + 5.375
= 35.375
= 35.38
Therefore, the mean = 35.38. It represents the average age of the patients = 35.38 years.
For mode:
From the given table we conclude that
The maximum class frequency = 23
So, modal class = 35 – 45
and
l = 35, f = 23, h = 10, f_{1 }= 21, f_{2 }= 14
Using the formula of mode
Mode
= 35 + 1.81 = 36.8
Hence, the mode = 36.8. It represents the maximum number of patients admitted in hospital of age 36.8 years.
Therefore, mode is greater than mean
Question 8. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes (in hours):  0 – 20  20 – 40  40 – 60  60 – 80  80 – 100  100 – 120 
No. of components:  10  35  52  61  38  29 
Determine the modal lifetimes of the components.
Solution:
From the given table we conclude that
The maximum class frequency = 61
So, modal class = 60 – 80
and
l = 60, f = 61, h = 20, f_{1 }= 52, f_{2 }= 38
Using the formula of mode
Mode
= 60 + 5.625 = 65.625
Hence, the modal lifetime of electrical components = 65.625 hours
Question 9. The following table gives the daily income of 50 workers of a factory:
Daily income  100 – 120  120 – 140  140 – 160  160 – 180  180 – 200 
Number of workers  12  14  8  6  10 
Find the mean, mode, and median of the above data.
Solution:
Class interval  Mid value (x)  Frequency (f)  fx  Cumulative Frequency 
100 – 120  110  12  1320  12 
120 – 140  130  14  1820  26 
140 – 160  150  8  1200  34 
160 – 180  170  6  1000  40 
180 – 200  190  10  1900  50 
N = 50 
Finding Mean:
From the table we get
N = 50, fx = 7260
So using mean formula, we get
Mean = Σfx / N
= 7260/ 50
= 145.2
Hence, the mean = 145.2
Finding Median:
N/2 = 50/2 = 25
So, the cumulative frequency just greater than N/2 = 26,
The median class = 120 – 140
Such that l = 120, h = 140 – 120 = 20, f = 14, F = 12
By using the formula of median we get
Median =
= 120 + 260/14
= 120 + 18.57
= 138.57
Hence, the median = 138.57
Finding Mode:
From the table we get
The maximum frequency = 14,
So the modal class = 120 – 140
And,
l = 120, h = 140 – 120 = 20, f = 14, f_{1} = 12, f_{2} = 8
By using the formula of mode we get
Mode =
= 120 + 5
= 125
Hence, the mode = 125
Question 10. The following distribution gives the statewise teachersstudents ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures:
Number of students per teacher  Number of states/U.T

15 – 20  3 
20 – 25  8 
25 – 30  9 
30 – 35  10 
35 – 40  3 
40 – 45  0 
45 – 50  0 
50 – 55  2 
Solution:
From the given table we conclude that
The maximum class frequency = 10
So, modal class = 30 – 35
and
l = 30, h = 5, f = 10, f_{1} = 9, f_{2} = 3
By using the formula of mode we get
Mode = l + f – f_{1} 2f – f_{1} – f_{2} × hl +
= 30 + 120 – 12 × 530 +
= 30 + 120 – 12 × 530 +
= 30 + 5/8
= 30.625
Hence, the mode = 30.6 and it represents that most of states/ U.T have a teacherstudents ratio = 30.6
Now we are going to find class marks using the following formula
Class mark =
Let us considered mean(a) = 32.5, and now we are going to find d_{i}, u_{i}, and f_{i}u_{i} as following
Number of students per teacher Number of states/ U.T (f_{i}) x_{i} d_{i} = x_{i} – 32.5 U_{i} f_{i}u_{i} 15 – 20 3 17.5 15 3 9 20 – 25 8 22.5 10 2 16 25 – 30 9 27.5 5 1 9 30 – 35 10 32.5 0 0 0 35 – 40 3 37.5 5 1 3 40 – 45 0 42.5 10 2 0 45 – 50 0 47.5 10 2 0 50 – 55 2 52.5 20 4 8 Total 35 23 Using the mean formula, we get
= 32.5 – 23/7
= 32.5 – 3.28
= 29.2
Hence, the mean = 29.2 and it represents that on an average teacherstudent ratio = 29.2.
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