# Long Questions

Calculate the coefficient of variation from the data given below:
 Class-interval Frequency 0-10 10-20 20-30 30-40 40-50 5 8 15 16 6
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
0-10 5 5 25
10-20 15 8 120
20-30 25 15 375
30-40 35 16 560
40-50 45 6 270
N = 50 ∑fm = 1350

Now, mean = $\frac{\sum fm}{N}$ = $\frac{1350}{50}$ = 27

Calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
5 5-27 = -22 484 5 2420
15 15-27 = -12 144 8 1152
25 25-27 = -2 4 15 60
35 35-27 = 8 64 16 1024
55 45-27 = 18 324 6 1644
N = 50 ∑fx2 = 6600
So, Standard deviation (σ) = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{6600}{50}}$ = 11.49
Finally, Coefficient of variation = $\frac{σ}{mean}$ × 100% = 42.55%
Calculate the coefficient of variation from the data given below:
 Class-interval Frequency 0-20 20-40 40-60 60-80 80-100 2 3 4 5 6
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
0-20 10 2 20
10-20 15 3 45
20-30 25 4 100
30-40 35 5 175
40-50 45 6 270
N = 20 ∑fm = 1200
Now, mean = $\frac{\sum fm}{N}$ = $\frac{1200}{20}$ = 60
Calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
10 10-60 = -50 2500 2 5000
30 30-60 = -30 900 3 2700
50 50-60 = -10 100 4 400
70 70-60 = 10 100 5 500
90 90-60 = 30 900 6 5400
N = 20 ∑fx2 = 14000
So, Standard deviation (σ) = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{14000}{20}}$ = 26.46
Finally, Coefficient of variation = $\frac{σ}{mean}$ × 100% = $\frac{26.46}{60}$ × 100% = 44.1%
Find the standard deviation of the data given below:
 Frequency Class interval 10-20 30-30 30-40 40-50 50-60 4 10 12 8 6
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
10-20 15 4 60
20-30 25 10 150
30-40 35 12 420
40-50 45 8 360
50-60 55 6 330
N = 40 ∑fm = 1420

Now, mean = $\frac{\sum fm}{N}$ = $\frac{1420}{40}$ = 35.5

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
15 15-35.5 = -20.5 420.25 4 1681
25 25-35.5 = -10.5 110.25 10 1102.5
35 35-35.5 = -0.5 0.25 12 3
45 45-35.5 = 9.5 90.25 8 722
55 55-35.5 = 19.5 380.25 6 2281.5
N = 40 ∑fx2 = 5790
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{5790}{40}}$ = 12.03

Find the standard deviation and its coefficient from the given data:
35, 30, 15, 5, 30, 10, 25

Here, Calculation of mean,
Sum of given data (∑X) = 35 + 30 + 15 + 5 + 30 + 10 + 25 = 140
Number of items (N) = 7

Now, mean = $\frac{\sum X}{N}$ = $\frac{140}{7}$ = 20

Finally, calculation of standard deviation,
x |x - Mean| = X X2
5 5-20 = -15 225
10 10-20 = -10 100
15 15-20 = -5 25
20 20-20 = 0 0
25 25-20 = 5 25
30 30-20 = 10 100
35 35-20 = 15 225
∑X2 = 700
Hence, Standard deviation (σ) = $\sqrt{\frac{\sum X^2}{N}}$ = $\sqrt{\frac{700}{7}}$ = 10
Also, Its coefficient = $\frac{σ}{\text{Mean}}$ = $\frac{10}{20}$ = 0.5
Calculate the standard deviation from the data given below:
 Class interval Frequency 0-10 10-20 20-30 30-40 40-50 2 6 5 4 3
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
0-10 5 2 10
10-20 15 6 90
20-30 25 5 125
30-40 35 4 140
40-50 45 3 135
N = 20 ∑fm = 500

Now, mean = $\frac{\sum fm}{N}$ = $\frac{500}{20}$ = 25

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
5 5-25 = -20 400 2 800
15 15-25 = -10 100 6 600
25 25-25 = 0 5 0 0
35 35-25 = 10 100 4 400
45 45-25 = 20 400 3 1200
N = 20 ∑fx2 = 3000
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{3000}{20}}$ = 12.427

Find the coefficient and standard deviation after finding mean of the following data:
30, 33, 37, 40, 44, 50

Here, Calculation of mean,
Sum of given data (∑X) = 30 + 33 + 37 + 40 + 44 + 50 = 234
Number of items (N) = 6

Now, mean = $\frac{\sum X}{N}$ = $\frac{234}{6}$ = 39

Finally, calculation of standard deviation,
x |m - Mean| = X X2
30 30-39 = -9 81
33 33-39 = -6 36
37 37-39 = -2 4
40 40-39 = 1 1
44 44-39 = 5 25
50 50-39 = 11 121
∑X2 = 267
Hence, Standard deviation (σ) = $\sqrt{\frac{\sum X^2}{N}}$ = $\sqrt{\frac{267}{7}}$ = 6.67
Also, Its coefficient = $\frac{σ}{\text{Mean}}$ = $\frac{6.67}{39}$ = 0.17
Find the standard deviation from the given data:
 Class interval No. of students 30-40 40-50 50-60 60-70 70-80 2 3 6 5 4
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
30-40 35 2 70
40-50 45 3 135
50-60 55 6 330
60-70 65 5 325
70-80 75 4 300
N = 20 ∑fm = 1160

Now, mean = $\frac{\sum fm}{N}$ = $\frac{1160}{20}$ = 58

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
35 35-58 = -23 529 2 1058
45 45-58 = -13 169 3 507
55 55-58 = -3 9 6 54
65 65-58 = 7 49 5 245
75 75-58 = 17 289 4 1156
N = 20 ∑fx2 = 3020
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{3020}{20}}$ = 12.288
Find the standard deviation of the data given below.
 Class interval Frequency 25-35 35-45 45-55 55-65 65-75 5 4 6 7 3
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
25-35 30 5 150
35-45 40 4 160
45-55 50 6 300
55-65 60 7 420
65-75 70 3 210
N = 25 ∑fm = 1240

Now, mean = $\frac{\sum fm}{N}$ = $\frac{1240}{25}$ = 49.6

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
30 30-49.6 = -19.6 384.16 5 1920.8
40 40-49.6 = -9.6 92.16 4 368.64
50 50-49.6 = 0.4 0.16 6 0.96
60 60-49.6 = 10.4 108.16 7 757.12
70 70-49.6 = 20.4 416.16 3 1248.48
N = 25 ∑fx2 = 4296
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{4296}{25}}$ = 13.1087
Find the standard deviation of the data given below.
 Class interval Frequency 5-15 15-25 25-35 35-45 45-55 7 3 6 4 5
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
5-15 10 7 70
15-25 20 3 60
25-35 30 6 180
35-45 40 4 160
45-55 50 5 250
N = 25 ∑fm = 720

Now, mean = $\frac{\sum fm}{N}$ = $\frac{720}{25}$ = 28.8

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
10 10-28.8 = -18.8 353.44 7 2474.08
20 20-28.8 = -8.8 77.44 3 232.32
30 30-28.8 = 1.2 125.44 4 501.76
40 40-28.8 = 11.2 125.44 4 501.76
50 50-28.8 = 21.2 449.44 5 2247.2
N = 25 ∑fx2 = 5464
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{5464}{25}}$ = 14.78
Find the standard deviation and its coefficient from the data given below.
 Class interval Frequency 0-10 10-20 20-30 30-40 40-50 50-60 8 12 20 40 12 8
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
0-10 5 8 40
10-20 15 12 180
20-30 25 20 500
30-40 35 40 1400
40-50 45 12 540
50-60 55 8 440
N = 100 ∑fm = 3100

Now, mean = $\frac{\sum fm}{N}$ = $\frac{3100}{100}$ = 31

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
5 5-31 = -26 676 8 5408
15 15-31 = -16 256 12 3072
25 25-31 = -6 36 20 720
35 35-31 = 4 16 40 640
45 45-31 = 14 196 12 2352
55 24 576 8 4608
N = 100 ∑fx2 = 16800
Hence, Standard deviation (σ) = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{16800}{100}}$ = 12.96
And, its coefficient = $\frac{σ}{\text{Mean}}$ = $\frac{12.96}{31}$ = 0.42
Find the standard deviation of the data given below.
 Class interval Frequency 0-4 4-8 8-12 12-16 16-20 20-24 7 7 10 15 7 6
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
0-4 2 7 14
4-8 6 7 42
8-12 10 10 100
12-16 14 15 210
16-20 18 7 126
20-24 22 6 132
N = 52 ∑fm = 624

Now, mean = $\frac{\sum fm}{N}$ = $\frac{624}{52}$ = 12

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
2 2-12 = -10 100 7 700
6 6-12 = -6 36 7 252
10 10-12 = -2 4 10 40
14 14-12 = 2 4 15 60
18 18-12 = 6 36 7 252
22 22-12 = 10 100 6 600
N = 52 ∑fx2 = 1904
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{1904}{52}}$ = 6.05
Find the standard deviation of the data given below.
 Class interval Frequency 0-4 4-8 8-12 12-16 16-20 15 12 10 8 5
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
0-4 2 15 30
4-8 6 12 72
8-12 10 10 100
12-16 14 8 112
16-20 18 5 90
N = 50 ∑fm = 404

Now, mean = $\frac{\sum fm}{N}$ = $\frac{404}{50}$ = 8.08

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
2 2-8.08 = -6.08 36.966 15 554.496
6 6-8.08 = -2.08 4.326 12 51.917
10 10-8.08 = 1.92 3.686 10 36.864
14 14-8.08 = 5.92 35.046 8 280.371
18 18-8.08 = 9.92 98.406 5 492.032
N = 50 ∑fx2 = 1415.68
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{1415.68}{50}}$ = 5.32
Find the standard deviation and its coefficient of the data given below.
5, 10, 20, 15, 25
Here, Calculation of mean,
Sum of given data (∑X) = 5 + 10 + 20 + 15 + 25 = 75
Number of items (N) = 5

Now, mean = $\frac{\sum X}{N}$ = $\frac{75}{5}$ = 15

Finally, calculation of standard deviation,
x |m - Mean| = X X2
5 5-15 = -10 100
10 10-15 = -5 25
15 15-15 = 0 0
20 20-15 = 5 25
25 25-15 = 10 100
∑X2 = 250
Hence, Standard deviation (σ) = $\sqrt{\frac{\sum X^2}{N}}$ = $\sqrt{\frac{250}{7}}$ = 7.07
Also, Its coefficient = $\frac{σ}{\text{Mean}}$ = $\frac{7.07}{15}$ = 0.47
Find the standard deviation of the data given below.
 Class interval (C.I.) Frequency (f) 10-20 20-30 30-40 40-50 50-60 4 10 12 8 6
Here, Calculation of mean,
Class interval (C.I.) Mid value (m) Frequency (f) f × m
10-20 15 4 60
20-30 25 10 250
30-40 35 12 420
40-50 45 8 360
50-60 55 6 330
N = 40 ∑fm = 1420

Now, mean = $\frac{\sum fm}{N}$ = $\frac{1420}{40}$ = 35.5

Finally, calculation of standard deviation,
Mid value (m) |m - Mean| = X X2 Frequency (f) fX2
15 15-35.5 = -20.5 420.25 4 1681
25 25-35.5 = -10.5 110.25 10 1102.5
35 35-35.5 = 0.5 0.25 12 3
45 45-35.5 = 9.5 90.25 8 722
55 55-35.5 = 19.5 380.25 6 2281.25
N = 40 ∑fx2 = 5790
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{5790}{40}}$ = 12.03
Find the standard deviation of the given data.
 x Frequency (f) 6 8 10 12 14 16 18 20 1 14 25 27 18 9 4 2
Here, Calculation of mean,
x Frequency (f) f × x
6 1 6
8 14 112
10 25 250
12 27 324
14 18 252
16 9 144
18 4 72
20 2 40
N = 100 ∑fx = 1200

Now, mean = $\frac{\sum fm}{N}$ = $\frac{1200}{100}$ = 12

Finally, calculation of standard deviation,
x |x - Mean| = X X2 Frequency (f) fX2
6 6-12 = -6 36 1 36
8 8-12 = -4 16 14 224
10 10-12 = -2 4 25 100
12 12-12 = 0 0 27 0
14 14-12 = 2 4 18 72
16 16-12 = 4 16 9 144
18 18-12 = 6 36 4 144
20 20-12 = 8 64 2 128
N = 100 ∑fx2 = 848
Hence, Standard deviation = $\sqrt{\frac{\sum fx^2}{N}}$ = $\sqrt{\frac{848}{100}}$ = 2.912