 # 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

### Access Other Exercise Solutions of Class 11 Maths Chapter 15 Statistics

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., xi – 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., xi – 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 = 6th = 13

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

= 7th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |xi – 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 = 5th = 46

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

= 6th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |xi – 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.

 xi 5 10 15 20 25 fi 7 4 6 3 5

Solution:-

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

 Xi fi fixi |xi – x̅| fi |xi – x̅| 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 25 350 158

The sum of calculated data, The absolute values of the deviations from the mean, i.e., |xi – x̅|, as shown in the table. 6.

 xi 10 30 50 70 90 fi 4 24 28 16 8

Solution:-

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

 Xi fi fixi |xi – x̅| fi |xi – x̅| 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 80 4000 1280  Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

 xi 5 7 9 10 12 15 fi 8 6 2 2 2 6

Solution:-

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

 Xi fi c.f. |xi – M| fi |xi – M| 5 8 8 2 16 7 6 14 0 0 9 2 16 2 4 10 2 18 3 6 12 2 20 5 10 15 6 26 8 48

Now, N = 26, which is even.

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

Then,

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

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table. 8.

 xi 15 21 27 30 35 fi 3 5 6 7 8

Solution:-

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

 Xi fi c.f. |xi – M| fi |xi – M| 15 3 3 15 45 21 5 8 9 45 27 6 14 3 18 30 7 21 0 0 35 8 29 5 40

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

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table. Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

 Income per day in ₹ 0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700 700 – 800 Number of persons 4 8 9 10 7 5 4 3

Solution:-

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

 Income per day in ₹ Number of persons fi Mid – points xi fixi |xi – x̅| fi|xi – x̅| 0 – 100 4 50 200 308 1232 100 – 200 8 150 1200 208 1664 200 – 300 9 250 2250 108 972 300 – 400 10 350 3500 8 80 400 – 500 7 450 3150 92 644 500 – 600 5 550 2750 192 960 600 – 700 4 650 2600 292 1160 700 – 800 3 750 2250 392 1176 50 17900 7896 10.

 Height in cms 95 – 105 105 – 115 115 – 125 125 – 135 135 – 145 145 – 155 Number of boys 9 13 26 30 12 10

Solution:-

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

 Height in cms Number of boys fi Mid – points xi fixi |xi – x̅| fi|xi – x̅| 95 – 105 9 100 900 25.3 227.7 105 – 115 13 110 1430 15.3 198.9 115 – 125 26 120 3120 5.3 137.8 125 – 135 30 130 3900 4.7 141 135 – 145 12 140 1680 14.7 176.4 145 – 155 10 150 1500 24.7 247 100 12530 1128.8 11. Find the mean deviation about median for the following data:

 Marks 0 -10 10 -20 20 – 30 30 – 40 40 – 50 50 – 60 Number of girls 6 8 14 16 4 2

Solution:-

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

 Marks Number of Girls fi Cumulative frequency (c.f.) Mid – points xi |xi – Med| fi|xi – Med| 0 – 10 6 6 5 22.85 137.1 10 – 20 8 14 15 12.85 102.8 20 – 30 14 28 25 2.85 39.9 30 – 40 16 44 35 7.15 114.4 40 – 50 4 48 45 17.15 68.6 50 – 60 2 50 55 27.15 54.3 50 517.1

The class interval containing Nth/2 or 25th 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:

 Age (in years) 16 – 20 21 – 25 26 – 30 31 – 35 36 – 40 41 – 45 46 – 50 51 – 55 Number 5 6 12 14 26 12 16 9

[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.

 Age Number fi Cumulative frequency (c.f.) Mid – points xi |xi – Med| fi|xi – Med| 15.5 – 20.5 5 5 18 20 100 20.5 – 25.5 6 11 23 15 90 25.5 – 30.5 12 23 28 10 120 30.5 – 35.5 14 37 33 5 70 35.5 – 40.5 26 63 38 0 0 40.5 – 45.5 12 75 43 5 60 45.5 – 50.5 16 91 48 10 160 50.5 – 55.5 9 100 53 15 135 100 735

The class interval containing Nth/2 or 50th 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 