NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

*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 15.2 questions. Chapter 15, Statistics of Class 11 Maths, is categorised under the CBSE Syllabus for the sessions 2023-24. By practising these questions, students can build their confidence level. NCERT Solutions also help the students to excel in their skills in statistics. Download these NCERT Solutions of Class 11 Maths now and start practising for the board exam.

NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

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Access Other Exercise Solutions of Class 11 Maths Chapter 15 Statistics

Exercise 15.1 Solutions : 12 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 NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2

Find the mean and variance for each of the data in Exercise 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12

Solution:-

So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

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

We know that Variance,

σ² = (1/8) × 74

= 9.2

∴Mean = 9 and Variance = 9.25

2. First n natural numbers

Solution:-

We know that Mean = Sum of all observations/Number of observations

∴Mean, x̅ = ((n(n + 1))2)/n

= (n + 1)/2

and also WKT Variance,

By substitute that value of x̅ we get,

WKT, (a + b)(a – b) = a² – b²

σ² = (n² – 1)/12

∴Mean = (n + 1)/2 and Variance = (n² – 1)/12

3. First 10 multiples of 3

Solution:-

First we have to write the first 10 multiples of 3,

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

So, x̅ = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

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

Then, Variance

= (1/10) × 742.5

= 74.25

∴Mean = 16.5 and Variance = 74.25

4.

Solution:-

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

5.

Solution:-

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

6. Find the mean and standard deviation using short-cut method.

Solution:-

Let the assumed mean A = 64. Here h = 1

We obtain the following table from the given data.

Mean,

Where A = 64, h = 1

So, x̅ = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, variance,

σ² = (1²/100²) [100(286) – 0²]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

∴ Mean = 64 and Standard Deviation = 1.691

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

7.

Solution:-

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

8.

Solution:-

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

9. Find the mean, variance and standard deviation using short-cut method

Solution:-

Let the assumed mean, A = 92.5 and h = 5

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

Mean,

Where, A = 92.5, h = 5

So, x̅ = 92.5 + ((6/60) × 5)

= 92.5 + ½

= 92.5 + 0.5

= 93

Then, Variance,

σ² = (5²/60²) [60(254) – 6²]

= (1/144) [15240 – 36]

= 15204/144

= 1267/12

= 105.583

Hence, standard deviation = σ = √105.583

= 10.275

∴ Mean = 93, variance = 105.583 and Standard Deviation = 10.275

10. The diameters of circles (in mm) drawn in a design are given below:

Calculate the standard deviation and mean diameter of the circles.

[Hint first make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

Solution:-

Let the assumed mean, A = 42.5 and h = 4

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

Mean,

Where, A = 42.5, h = 4

So, x̅ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

σ² = (4²/100²)[100(199) – 25²]

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553

∴ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.