Class 11 NCERT Solutions Chapter 15 Statistics Exercise 15.2 (original) (raw)

Last Updated : 11 Sep, 2024

Chapter 15 of Class 11 NCERT focuses on Statistics a crucial branch of mathematics that deals with the data collection, analysis, and interpretation. Exercise 15.2 dives deeper into the various statistical methods including the computation of measures of the dispersion such as range, variance, and standard deviation for both the grouped and ungrouped data. These concepts form the backbone of the data analysis and are essential for students to understand real-world data behavior.

Statistics

Statistics is a mathematical science that involves methods to collect, classify, analyze, and interpret data. It helps in making decisions based on the data, predicting future trends and solving complex problems in various fields like economics, biology, and engineering. In this chapter, students learn different tools for summarizing and describing data efficiently. The measures of central tendency and dispersion are explored giving a comprehensive view of the data distribution.

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

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

**Solution:

We know,

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

So, \bar{x} = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

**x i **Deviations from mean ****(x** i **- x') ****(xi - x')** 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

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

σ2 = (1/8) × 74

= 9.2

Therefore, Mean = 9 and Variance = 9.25

Question 2. First n natural numbers

**Solution:

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = ((n(n + 1))2)/n

= (n + 1)/2

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

On substituting the value of mean,

= \frac{1}{n}\sum_{i=1}^{n}(x_i - \frac{n+1}{2})^2 \\ = \frac{1}{n}\sum_{i=1}^{n}(x_i)^2 - \frac{1}{n}\sum_{i=1}^{n}2x_i(\frac{n+1}{2})+\frac{1}{n}\sum_{i=1}^{n}(\frac{n+1}{2})^2 Substituting values of Summation \\ = \frac{1}{n}\frac{n(n+1)(2n+1)}{6}-\frac{n+1}{n}[\frac{n(n+1)}{2}]+\frac{(n+1)^2}{4n}×n

On extracting common values, we have,

= (n+1)[\frac{4n+2-3n-3}{12}] \\ = \frac{(n+1)(n-1)}{12}

σ2 = (n2 – 1)/12

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

Question 3. First 10 multiples of 3

**Solution:

The required multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

We know,

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

So, \bar{x} = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

**x i **Deviations from mean (x i **- x') ****(x** i - x') 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

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/10) × 742.5

= 74.25

Therefore, Mean = 16.5 and Variance = 74.25

Question 4.

**x i **6 **10 **14 **18 **24 **28 **30
**f i **2 **4 **7 **12 **8 **4 **3

**Solution:

**x i **f i **f i x i **x i - x' ****(x** i - x') 2 **f i (x i **- x') 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
**1736

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = 760/40

= 19

Also,

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/40) × 1736

= 43.4

Question 5.

**x i **92 **93 **97 **98 **102 **104 **109
**f i **3 **2 **3 **2 **6 **3 **3

**Solution:

**x i **f i **f i x i **x i **- x' ****(x** i - x') 2 **f i (x i - x') 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

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = 2200/22

= 100

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/22) × 640

= 29.09

Therefore, Mean = 100 and Variance = 29.09

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

**x i **60 **61 **62 **63 **64 **65 **66 **67 **68
**f i **2 **1 **12 **29 **25 **12 **10 **4 **5

**Solution:

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h

Where A = 64, h = 1

So, \bar{x} = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, variance,

\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]

σ2 = (12/1002) [100(286) – 02]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

Therefore,

Mean = 64 and Standard Deviation = 1.691

Question 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:

**Classes **f i **x i **f i x i ****(x** i **- x') ****(xi - x')** 2 **f i (xi - x') 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

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = 3210/30

= 107

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/30) × 68280

= 2276

Therefore, Mean = 107 and Variance = 2276

Question 8.

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

**Solution:

**Classes **f i **x i **f i x i ****(x** i -x') ****(x** i -x') 2 **f i (x i -x') 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

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = 1350/50

= 27

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/50) × 6600

= 132

Therefore, Mean = 27 and Variance = 132

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

**Heights 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:

**Height **f i **X i **Y i = ****(X** i -A)/h **Y i 2 **f i y i **f i y i 2
70-75 2 72.5 -4 19 -12 48
75-80 1 77.5 -3 9 -12 36
80-85 12 82.5 -2 4 -14 28
85-90 29 87.5 -1 1 -7 7
90-95 25 92.5 0 0 0 0
95-100 12 97.5 1 1 9 9
100-105 10 102.5 2 4 12 24
105-110 4 107.5 3 9 18 54
110-115 5 112.5 4 16 12 48
**115-120 **N = 60 **6 **254

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h

Where, A = 92.5, h = 5

So, \bar{x}= 92.5 + ((6/60) × 5)

= 92.5 + 0.5

= 92.5 + 0.5

= 93

Then, Variance,

\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]

Standard deviation = σ = √105.583

= 10.275

Question 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:

Height fi 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

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h

Where, A = 42.5, h = 4

\bar{x} = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]

σ2 = (42/1002)[100(199) – 252]

On solving, we get,

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553

Conclusion

Chapter 15 helps students understand how to calculate and interpret the measures of the dispersion in statistics. Exercise 15.2 focuses on practical problems related to the variance and standard deviation reinforcing the importance of data spread and variability in the real-life scenarios. Mastering these concepts is essential for analyzing data effectively which is useful in the various academic and professional disciplines.