NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.

NCERT Solutions for Class 11 Maths Chapter 15 Statistics, contains solutions for all Exercise 15.1 questions. Chapter 15 Statistics of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. These NCERT Solutions help students to clear their concepts and solve difficult problems at their own pace. Students can download the NCERT Solutions of Class 11 Maths and practise offline to improve their skills.

NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

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Exercise 15.2 Solutions: 10 Questions

Exercise 15.3 Solutions: 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

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NCERT Solutions for Class 11 Maths Chapter 15

NCERT Solutions for Class 11

Access Solutions for Class 11 Maths Chapter 15 Exercise 15.1

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Solution:-

First we have to find (x̅) of the given data

So, the respective values of the deviations from mean,

i.e., x_i – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 9 = 1, 10 – 10 = 0,

10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7

6, 3, 2, 1, 0, -2, -3, -7

Now absolute values of the deviations,

6, 3, 2, 1, 0, 2, 3, 7

MD = sum of deviations/ number of observations

= 24/8

= 3

So, the mean deviation for the given data is 3.

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:-

First we have to find (x̅) of the given data

So, the respective values of the deviations from mean,

i.e., x_i – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,

50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6

-12, 20, -2, -10, -8, 5, 13, -4, 4, -6

Now absolute values of the deviations,

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

MD = sum of deviations/ number of observations

= 84/10

= 8.4

So, the mean deviation for the given data is 8.4.

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:-

First we have to arrange the given observations in ascending order,

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

The number of observations is 12

Then,

Median = ((12/2)^th observation + ((12/2)+ 1)^th observation)/2

(12/2)^th observation = 6^th = 13

(12/2)+ 1)^th observation = 6 + 1

= 7^th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

Mean Deviation,

= (1/12) × 28

= 2.33

So, the mean deviation about the median for the given data is 2.33.

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:-

First we have to arrange the given observations in ascending order,

36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

The number of observations is 10

Then,

Median = ((10/2)^th observation + ((10/2)+ 1)^th observation)/2

(10/2)^th observation = 5^th = 46

(10/2)+ 1)^th observation = 5 + 1

= 6^th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Mean Deviation,

= (1/10) × 70

= 7

So, the mean deviation about the median for the given data is 7.

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

Solution:-

Let us make the table of the given data and append other columns after calculations.

The sum of calculated data,

The absolute values of the deviations from the mean, i.e., |x_i – x̅|, as shown in the table.

6.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Now, N = 26, which is even.

Median is the mean of the 13^th and 14^th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Then,

Median = (13^th observation + 14^th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are shown in the table.

8.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Now, N = 29, which is odd.

So 29/2 = 14.5

The cumulative frequency greater than 14.5 is 21, for which the corresponding observation is 30.

Then,

Median = (15^th observation + 16^th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are shown in the table.

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

Solution:-

Let us make the table of the given data and append other columns after calculations.

10.

Solution:-

Let us make the table of the given data and append other columns after calculations.

11. Find the mean deviation about median for the following data:

Solution:-

Let us make the table of the given data and append other columns after calculations.

The class interval containing N^th/2 or 25^th item is 20-30

So, 20-30 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 20, c = 14, f = 14, h = 10 and n = 50

Median = 20 + (((25 – 14))/14) × 10

= 20 + 7.85

= 27.85

12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:

[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Solution:-

The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.

The class interval containing N^th/2 or 50^th item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38