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# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.5 | Set 1

• Last Updated : 05 Mar, 2021

### (iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

Solution:

(i)

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Therefore, mode = 5 because 5 occurs the maximum number of times.

(ii)

Therefore, mode = 3 because 3 occurs the maximum number of times.

(iii)

Therefore, mode = 15 because 15 occurs the maximum number of times.

### 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.

### (i)

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)

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, f1 = 45, f2 = 35

Using the formula of mode

Mode = = 20 + 150/70

= 20 + 2.14

= 22.14

Hence, the mode = 22.14

### (iii)

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, f1 = 34, f2 = 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:

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, f1 = 50, f2 = 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, f1 = 54, f2 = 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.

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, f1 = 13, f2 = 5

Using the formula of mode

Mode = = 50 + 70/22

= 50 + 3.18

= 53.18

Hence, the mode = 53.18

### Find the average height of maximum number of students.

Solution:

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, f1 = 118, f2 = 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

### 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

From the table we get

Σfi = N = 80 and Σfi di = 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, f1 = 21, f2 = 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

### Determine the modal lifetimes of the components.

Solution:

From the given table we conclude that

The maximum class frequency = 61

So, modal calss = 60 – 80

and

l = 60, f = 61, h = 20, f1 = 52, f2 = 38

Using the formula of mode

Mode = 60 + 5.625 = 65.625

Hence, the modal lifetime of electrical components = 65.625 hours

### Find the mean, mode, and median of the above data.

Solution:

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, f1 = 12, f2 = 8

By using the formula of mode we get

Mode = = 120 + 5

= 125

Hence, the mode = 125

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

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, f1 = 9, f2 = 3

By using the formula of mode we get

Mode = l + f – f1 2f – f1 – f2 × 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 teacher-students 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 di, ui, and fiui as following

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 teacher-student ratio = 29.2.

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