# Chapter – 15: Statistics

# 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., x_{i} – 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., x_{i} – 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 into 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 = 6^{th} = 13

(12/2)+ 1)^{th} observation = 6 + 1

= 7^{th} = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – 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

= (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 into 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 = 5^{th} = 46

(10/2)+ 1)^{th} observation = 5 + 1

= 6^{th} = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – 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.**

x_{i} | 5 | 10 | 15 | 20 | 25 |

f_{i} | 7 | 4 | 6 | 3 | 5 |

**Solution: –**

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

X_{i} | f_{i} | f_{i}x_{i} | |x_{i} – x̅| | f_{i} |x_{i} – 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., |x_{i} – x̅|, as shown in the table.

**6.**

x_{i} | 10 | 30 | 50 | 70 | 90 |

f_{i} | 4 | 24 | 28 | 16 | 8 |

**Solution: –**

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

X_{i} | f_{i} | f_{i}x_{i} | |x_{i} – x̅| | f_{i} |x_{i} – 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.**

x_{i} | 5 | 7 | 9 | 10 | 12 | 15 |

f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |

**Solution: –**

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

X_{i} | f_{i} | c.f. | |x_{i} – M| | f_{i} |x_{i} – 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 13^{th} and 14^{th} observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Then,

Median = (13^{th} observation + 14^{th }observation)/2

= (7 + 7)/2

= 14/2

= 7

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

**8.**

x_{i} | 15 | 21 | 27 | 30 | 35 |

f_{i} | 3 | 5 | 6 | 7 | 8 |

**Solution: –**

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

X_{i} | f_{i} | c.f. | |x_{i} – M| | f_{i} |x_{i} – 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 = (15^{th} observation + 16^{th }observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – 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 f_{i} | Mid – points x_{i} | f_{i}x_{i} | |x_{i}– x̅| | f_{i}|x_{i}– 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 f_{i} | Mid – points x_{i} | f_{i}x_{i} | |x_{i} – x̅| | f_{i}|x_{i} – 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 f_{i} | Cumulative frequency (c.f.) | Mid – points x_{i} | |x_{i} – Med| | f_{i}|x_{i} – 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 N^{th}/2 or 25^{th} 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 f_{i} | Cumulative frequency (c.f.) | Mid – points x_{i} | |x_{i} – Med| | f_{i}|x_{i}– 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 N^{th}/2 or 50^{th} 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

# 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.

X_{i} | Deviations from mean (x_{i} – x̅) | (x_{i} – 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,

σ^{2} = (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) = a^{2} – b^{2}

σ^{2} = (n^{2} – 1)/12

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

X_{i} | Deviations from mean (x_{i} – x̅) | (x_{i} – 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.**

x_{i} | 6 | 10 | 14 | 18 | 24 | 28 | 30 |

f_{i} | 2 | 4 | 7 | 12 | 8 | 4 | 3 |

**Solution: –**

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

X_{i} | f_{i} | f_{i}x_{i} | Deviations from mean (x_{i} – x̅) | (x_{i} – x̅)^{2} | f_{i}(x_{i} – 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.**

x_{i} | 92 | 93 | 97 | 98 | 102 | 104 | 109 |

f_{i} | 3 | 2 | 3 | 2 | 6 | 3 | 3 |

**Solution: –**

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

X_{i} | f_{i} | f_{i}x_{i} | Deviations from mean (x_{i} – x̅) | (x_{i}– x̅)^{2} | f_{i}(x_{i}– 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.**

x_{i} | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |

f_{i} | 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.

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

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,

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

∴ 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 f_{i} | Mid – points x_{i} | f_{i}x_{i} | (x_{i}– x̅) | (x_{i} – x̅)^{2} | f_{i}(x_{i} – 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 f_{i} | Mid – points x_{i} | f_{i}x_{i} | (x_{i}– x̅) | (x_{i}– x̅)^{2} | f_{i}(x_{i}– 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 f_{i} | Midpoint X_{i} | Y_{i} = (x_{i} – A)/h | Y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} |

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,

σ^{2} = (5^{2}/60^{2}) [60(254) – 6^{2}]

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

Mean,

Where, A = 42.5, h = 4

So, x̅ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

σ^{2} = (4^{2}/100^{2})[100(199) – 25^{2}]

= (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.

# Exercise 15.3

**1. From the data given below state which group is more variable, A or B?**

Marks | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |

Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |

Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |

**Solution: –**

For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

Co-efficient of variation (C.V.) = (σ/ x̅) × 100

Where, σ = standard deviation, x̅ = mean

For Group A

Marks | Group A f_{i} | Mid-point X_{i} | Y_{i} = (x_{i}– A)/h | (Y_{i})^{2} | f_{i}y_{i} | f_{i}(y_{i})^{2} |

10 – 20 | 9 | 15 | ((15 – 45)/10) = -3 | (-3)^{2} = 9 | – 27 | 81 |

20 – 30 | 17 | 25 | ((25 – 45)/10) = -2 | (-2)^{2} = 4 | – 34 | 68 |

30 – 40 | 32 | 35 | ((35 – 45)/10) = – 1 | (-1)^{2} = 1 | – 32 | 32 |

40 – 50 | 33 | 45 | ((45 – 45)/10) = 0 | 0^{2} | 0 | 0 |

50 – 60 | 40 | 55 | ((55 – 45)/10) = 1 | 1^{2} = 1 | 40 | 40 |

60 – 70 | 10 | 65 | ((65 – 45)/10) = 2 | 2^{2} = 4 | 20 | 40 |

70 – 80 | 9 | 75 | ((75 – 45)/10) = 3 | 3^{2} = 9 | 27 | 81 |

Total | 150 | -6 | 342 |

Where A = 45,

and y_{i} = (x_{i} – A)/h

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ^{2} = (10^{2}/150^{2}) [150(342) – (-6)^{2}]

= (100/22500) [51,300 – 36]

= (100/22500) × 51264

= 227.84

Hence, standard deviation = σ = √227.84

= 15.09

∴ C.V for group A = (σ/ x̅) × 100

= (15.09/44.6) × 100

= 33.83

Now, for group B.

Marks | Group B f_{i} | Mid-point X_{i} | Y_{i} = (x_{i}– A)/h | (Y_{i})^{2} | f_{i}y_{i} | f_{i}(y_{i})^{2} |

10 – 20 | 10 | 15 | ((15 – 45)/10) = -3 | (-3)^{2} = 9 | – 30 | 90 |

20 – 30 | 20 | 25 | ((25 – 45)/10) = -2 | (-2)^{2} = 4 | – 40 | 80 |

30 – 40 | 30 | 35 | ((35 – 45)/10) = – 1 | (-1)^{2} = 1 | – 30 | 30 |

40 – 50 | 25 | 45 | ((45 – 45)/10) = 0 | 0^{2} | 0 | 0 |

50 – 60 | 43 | 55 | ((55 – 45)/10) = 1 | 1^{2} = 1 | 43 | 43 |

60 – 70 | 15 | 65 | ((65 – 45)/10) = 2 | 2^{2} = 4 | 30 | 60 |

70 – 80 | 7 | 75 | ((75 – 45)/10) = 3 | 3^{2} = 9 | 21 | 63 |

Total | 150 | -6 | 366 |

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ^{2} = (10^{2}/150^{2}) [150(366) – (-6)^{2}]

= (100/22500) [54,900 – 36]

= (100/22500) × 54,864

= 243.84

Hence, standard deviation = σ = √243.84

= 15.61

∴ C.V for group B = (σ/ x̅) × 100

= (15.61/44.6) × 100

= 35

By comparing C.V. of group A and group B.

C.V of Group B > C.V. of Group A

So, Group B is more variable.

**2. From the prices of shares X and Y below, find out which is more stable in value:**

X | 35 | 54 | 52 | 53 | 56 | 58 | 52 | 50 | 51 | 49 |

Y | 108 | 107 | 105 | 105 | 106 | 107 | 104 | 103 | 104 | 101 |

**Solution: –**

From the given data,

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

X (x_{i}) | Y (y_{i}) | X_{i}^{2} | Y_{i}^{2} |

35 | 108 | 1225 | 11664 |

54 | 107 | 2916 | 11449 |

52 | 105 | 2704 | 11025 |

53 | 105 | 2809 | 11025 |

56 | 106 | 8136 | 11236 |

58 | 107 | 3364 | 11449 |

52 | 104 | 2704 | 10816 |

50 | 103 | 2500 | 10609 |

51 | 104 | 2601 | 10816 |

49 | 101 | 2401 | 10201 |

Total = 510 | 1050 | 26360 | 110290 |

We have to calculate Mean for x,

Mean x̅ = ∑x_{i}/n

Where, n = number of terms

= 510/10

= 51

Then, Variance for x =

= (1/10^{2})[(10 × 26360) – 510^{2}]

= (1/100) (263600 – 260100)

= 3500/100

= 35

WKT Standard deviation = √variance

= √35

= 5.91

So, co-efficient of variation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have to calculate Mean for y,

Mean ȳ = ∑y_{i}/n

Where, n = number of terms

= 1050/10

= 105

Then, Variance for y =

= (1/10^{2})[(10 × 110290) – 1050^{2}]

= (1/100) (1102900 – 1102500)

= 400/100

= 4

WKT Standard deviation = √variance

= √4

= 2

So, co-efficient of variation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of X and Y.

C.V of X > C.V. of Y

So, Y is more stable than X.

**3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:**

Firm A | Firm B | |

No. of wages earners | 586 | 648 |

Mean of monthly wages | Rs 5253 | Rs 5253 |

Variance of the distribution of wages | 100 | 121 |

**(i) Which firm A or B pays larger amount as monthly wages?**

**(ii) Which firm, A or B, shows greater variability in individual wages?**

**Solution: –**

(i) From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 × 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 × 5253

= Rs 34,03,944

So, firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that, standard deviation (σ)= √100

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=√(121 )

=11

Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.

**4. The following is the record of goals scored by team A in a football session:**

No. of goals scored | 0 | 1 | 2 | 3 | 4 |

No. of matches | 1 | 9 | 7 | 5 | 3 |

**For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?**

**Solution: –**

From the given data,

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

Number of goals scored x_{i} | Number of matches f_{i} | f_{i}x_{i} | X_{i}^{2} | f_{i}x_{i}^{2} |

0 | 1 | 0 | 0 | 0 |

1 | 9 | 9 | 1 | 9 |

2 | 7 | 14 | 4 | 28 |

3 | 5 | 15 | 9 | 45 |

4 | 3 | 12 | 16 | 48 |

Total | 25 | 50 | 130 |

Since C.V. of firm B is greater

∴ Team A is more consistent.

**5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:**

**Which is more varying, the length or weight?**

**Solution: –**

First we have to calculate Mean for Length x,

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