 # 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

### 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.

 Xi Deviations from mean (xi – x̅) (xi – x̅)2 6 6 – 9 = -3 9 7 7 – 9 = -2 4 10 10 – 9 = 1 1 12 12 – 9 = 3 9 13 13 – 9 = 4 16 4 4 – 9 = – 5 25 8 8 – 9 = – 1 1 12 12 – 9 = 3 9 74

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.

 Xi Deviations from mean (xi – x̅) (xi – x̅)2 3 3 – 16.5 = -13.5 182.25 6 6 – 16.5 = -10.5 110.25 9 9 – 16.5 = -7.5 56.25 12 12 – 16.5 = -4.5 20.25 15 15 – 16.5 = -1.5 2.25 18 18 – 16.5 = 1.5 2.25 21 21 – 16.5 = – 4.5 20.25 24 24 – 16.5 = 7.5 56.25 27 27 – 16.5 = 10.5 110.25 30 30 – 16.5 = 13.5 182.25 742.5

Then, Variance = (1/10) × 742.5

= 74.25

∴Mean = 16.5 and Variance = 74.25

4.

 xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3

Solution:-

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

 Xi fi fixi Deviations from mean (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 6 2 12 6 – 19 = 13 169 338 10 4 40 10 – 19 = -9 81 324 14 7 98 14 – 19 = -5 25 175 18 12 216 18 – 19 = -1 1 12 24 8 192 24 – 19 = 5 25 200 28 4 112 28 – 19 = 9 81 324 30 3 90 30 – 19 = 11 121 363 N = 40 760 1736 5.

 xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3

Solution:-

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

 Xi fi fixi Deviations from mean (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 92 3 276 92 – 100 = -8 64 192 93 2 186 93 – 100 = -7 49 98 97 3 291 97 – 100 = -3 9 27 98 2 196 98 – 100 = -2 4 8 102 6 612 102 – 100 = 2 4 24 104 3 312 104 – 100 = 4 16 48 109 3 327 109 – 100 = 9 81 243 N = 22 2200 640 6. Find the mean and standard deviation using short-cut method.

 xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5

Solution:-

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

We obtain the following table from the given data.

 Xi Frequency fi Yi = (xi – A)/h Yi2 fiyi fiyi2 60 2 -4 16 -8 32 61 1 -3 9 -3 9 62 12 -2 4 -24 48 63 29 -1 1 -29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 0 286

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.

 Classes 0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180 180 – 210 Frequencies 2 3 5 10 3 5 2

Solution:-

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

 Classes Frequency fi Mid – points xi fixi (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 0 – 30 2 15 30 -92 8464 16928 30 – 60 3 45 135 -62 3844 11532 60 – 90 5 75 375 -32 1024 5120 90 – 120 10 105 1050 -2 4 40 120 – 150 3 135 405 28 784 2352 150 – 180 5 165 825 58 3364 16820 180 – 210 2 195 390 88 7744 15488 N = 30 3210 68280 8.

 Classes 0 – 10 10 – 20 20 – 30 30 – 40 40 –50 Frequencies 5 8 15 16 6

Solution:-

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

 Classes Frequency fi Mid – points xi fixi (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 0 – 10 5 5 25 -22 484 2420 10 – 20 8 15 120 -12 144 1152 20 – 30 15 25 375 -2 4 60 30 – 40 16 35 560 8 64 1024 40 –50 6 45 270 18 324 1944 N = 50 1350 6600  9. Find the mean, variance and standard deviation using short-cut method

 Height in cms 70 – 75 75 – 80 80 – 85 85 – 90 90 – 95 95 – 100 100 – 105 105 – 110 110 – 115 Frequencies 3 4 7 7 15 9 6 6 3

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.

 Height (class) Number of children Frequency fi Midpoint Xi Yi = (xi – A)/h Yi2 fiyi fiyi2 70 – 75 3 72.5 -4 16 -12 48 75 – 80 4 77.5 -3 9 -12 36 80 – 85 7 82.5 -2 4 -14 28 85 – 90 7 87.5 -1 1 -7 7 90 – 95 15 92.5 0 0 0 0 95 – 100 9 97.5 1 1 9 9 100 – 105 6 102.5 2 4 12 24 105 – 110 6 107.5 3 9 18 54 110 – 115 3 112.5 4 16 12 48 N = 60 6 254

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:

 Diameters 33 – 36 37 – 40 41 – 44 45 – 48 49 – 52 No. of circles 15 17 21 22 25

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.

 Height (class) Number of children (Frequency fi) Midpoint Xi Yi = (xi – A)/h Yi2 fiyi fiyi2 32.5 – 36.5 15 34.5 -2 4 -30 60 36.5 – 40.5 17 38.5 -1 1 -17 17 40.5 – 44.5 21 42.5 0 0 0 0 44.5 – 48.5 22 46.5 1 1 22 22 48.5 – 52.5 25 50.5 2 4 50 100 N = 100 25 199

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.