# Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.2

### 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,

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free classeswhich will definitely help them in making a wise career choice in the future.So, = (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 74σ

^{2 }= (1/8) × 74= 9.2

Therefore, Mean = 9 and Variance = 9.25

### Question 2. First n natural numbers

**Solution:**

= ((n(n + 1))2)/n

= (n + 1)/2

On substituting the value of mean,

Substituting values of SummationOn extracting common values, we have,

σ

^{2}= (n^{2}– 1)/12Mean = (n + 1)/2 and Variance = (n

^{2}– 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,

So, = (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= (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 |

= 760/40

= 19

Also,

= (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 |

= 2200/22

= 100

= (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:**

Where A = 64, h = 1

So, = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, variance,

σ

^{2}= (1^{2}/100^{2}) [100(286) – 0^{2}]= (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 |

= 3210/30

= 107

= (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 |

= 1350/50

= 27

= (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 |

Where, A = 92.5, h = 5

So, = 92.5 + ((6/60) × 5)

= 92.5 + 0.5

= 92.5 + 0.5

= 93

Then, Variance,

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 | f_{i} | x_{i} | Y_{i} = (X_{i}-A)/h | Y_{i}^{2} | f_{i}y_{i} | fiyi^{2} |
---|---|---|---|---|---|---|

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 |

Where, A = 42.5, h = 4

= 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

σ

^{2}= (4^{2}/100^{2})[100(199) – 25^{2}]On solving, we get,

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553