Exercise 15.1
Q1. Calculate the mean deviation about the mean for the given data
5, 8, 9, 10, 11, 13, 14, 18
Sol:
The given data is,
5, 8, 9, 10, 11, 13, 14, 18
Mean,
x¯=5+8+9+10+11+13+14+188 x¯=888 x¯=11
The deviations of the respective observations from the mean x¯,
xi–x¯ are -6, -3, -2, -1, 0, 2, 3, 7
The absolute values of the deviations,
|xi–x¯| are 6, 3, 2, 1, 0 , 2, 3, 7
The required mean deviation about the mean is,
M.D.(x¯)=∑8i=1∣∣xi–x¯∣∣8 M.D.(x¯)=6+3+2+1+0+2+3+78 M.D.(x¯)=248 = 3
Q2. Calculate the mean deviation about the mean for the given data
39, 71, 49, 41, 43, 56, 64, 47, 55, 45
Sol:
The given data is,
39, 71, 49, 41, 43, 56, 64, 47, 55, 45
Mean,
x¯=39+71+49+41+43+56+64+47+55+4510 x¯=51010 x¯=51
The deviations of the respective observations from the mean x¯,
xi–x¯ are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6.
The absolute values of the deviations,
|xi–x¯| are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.
The required mean deviation about the mean is,
M.D.(x¯)=∑10i=1∣∣xi–x¯∣∣10 M.D.(x¯)=12+20+2+10+8+5+13+4+4+610 M.D.(x¯)=8410 = 8.4
Q3. Calculate the mean deviation about the median for the given data:
13, 16, 11, 10, 11, 12, 17, 14, 13, 16, 18, 17
Sol:
The given data is,
13, 16, 11, 10, 11, 12, 17, 14, 13, 16, 18, 17
There are 12 observations which is even number.
Arrange the data in ascending order:
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Median,
M=(122)thobservation+(122+1)thobservation2
= 6thobservation+7thobservation2
= 13+142
= 272
= 13.5
The deviations of the respective observations from the median x¯,
xi–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
The absolute values of the deviations,
|xi–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
The required mean deviation about the median is,
M.D.(M)=∑12i=1|xi–M|12
= 3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.512
= 2812
= 2.33
Q4. Calculate the mean deviation about the median for the given data:
36, 46, 60, 53, 51, 72, 42, 45, 46, 49
Sol:
The given data is,
36, 46, 60, 53, 51, 72, 42, 45, 46, 49
There are 10 observations which is even number.
Arrange the data in ascending order:
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Median,
M=(102)thobservation+(102+1)thobservation2
= 5thobservation+6thobservation2
= 46+492
= 952
= 47.5
The deviations of the respective observations from the median x¯,
xi–M are –11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5
The absolute values of the deviations,
|xi–M| are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
The required mean deviation about the median is,
M.D.(M)=∑10i=1|xi–M|10
= 11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.510
= 7010
= 7
Q5. Calculate the mean deviation about the mean for the given data
xi 5 10 15 20 25
fi 7 4 6 3 5
Sol:
xi xi xi |xi–x¯| fi|xi–x¯|
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
N=∑5i=1fi=25
∑5i=1fixi=350
Therefore,
x¯=1N∑5i=1fixi
= 125×350
= 14
MD(x¯)=1N∑5i=1fi|xi–x¯|
= 125×158
= 6.32
Q6. Calculate the mean deviation about the mean for the given data
xi 10 10 15 20 25
fi 7 4 6 3 5
Sol:
xi xi xi |xi–x¯| fi|xi–x¯|
10 4 35 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
N=∑5i=1fi=80
∑5i=1fixi=4000
Therefore,
x¯=1N∑5i=1fixi
= 180×4000
= 50
MD(x¯)=1N∑5i=1fi|xi–x¯|
= 180×1280
= 16
Q7. Calculate the mean deviation about the median for the given data
xi 10 10 15 20 25
fi 7 4 6 3 5
xi 5 7 9 10 12 15
fi 8 6 2 2 2 6
Sol:
The observations given are in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:
xi fi c.f.
5 8 8
7 6 14
9 2 16
10 2 18
12 2 20
15 6 26
Here, N = 26, which is even.
Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
Median=13thobservation+14thobservation2
= 7+72
= 7
The absolute values of the deviations from median,
|xi–M| 2 0 2 3 5 8
fi 8 6 2 2 2 6
fi|xi–M| 16 0 4 6 10 48
∑6i=1fi=26
∑6i=1fi|xi–M|=84
Therefore,
MD(M)=1N∑6i=1fi|xi–M|
= 126×84
= 3.23
Q8. Calculate the mean deviation about the median for the given data
xi 15 21 27 30 35
fi 3 5 6 7 8
Sol:
The observations given are in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:
xi fi c.f.
15 3 3
21 5 8
27 6 14
30 7 21
35 8 29
Here, N = 29, which is odd.
Median=(29+12)thobservation
= 15th observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
Therefore,
Median = 30
The absolute values of the deviations from median,
|xi–M| 15 9 3 0 5
fi 3 5 6 7 8
fi|xi–M| 45 45 18 0 40
∑5i=1fi=29
∑5i=1fi|xi–M|=148
Therefore,
MD(M)=1N∑5i=1fi|xi–M|
= 129×148
= 5.1
Q9. Calculate the mean deviation about the mean for the given data:
Income per day Number of persons
0 – 100 4
100 – 200 8
200 – 300 9
300 – 400 10
400 – 500 7
500 – 600 5
600 – 700 4
700 – 800 3
Sol:
The given table is formed.
Income per day Number of persons fi Mid – point xi fi xi |xi–x¯| fi|xi–x¯|
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 1168
700 – 800 3 750 2250 392 1176
50 17900 7896
N=∑8i=1fi=50
∑8i=1fixi=17900
Therefore,
x¯=1N∑8i=1fixi
= 150×17900
= 358
MD(x¯)=1N∑8i=1fi|xi–x¯|
= 150×7896
= 175.92
Q10. Calculate the mean deviation about the mean for the given data:
Height in cm Number of boys
95 – 105 9
105 – 115 13
115 – 125 26
125 – 135 30
135 – 145 12
145 – 155 10
Sol:
The given table is obtained:
Height in cm Number of boys fi Mid – point xi fi xi |xi–x¯| fi|xi–x¯|
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
N=∑6i=1fi=100
∑6i=1fixi=12530
Therefore,
x¯=1N∑6i=1fixi
= 1100×12530
= 125.3
MD(x¯)=1N∑6i=1fi|xi–x¯|
= 1100×1128.8
= 11.28
Q11. Calculate the mean deviation about the median for the given data:
Marks Number of girls
0 – 10 6
10 – 20 8
20 – 30 14
30 – 40 16
40 – 50 4
50 – 60 2
Sol:
The given table is obtained:
Marks Number of girls fi Cumulative frequency (c.f.) Mid – point xi |xi–M| fi|xi–x¯|
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 (N2)th or 25th item is 20 – 30.
Therefore, 20 -30 is the median class.
Now,
Median=l+N2–Cf×h
Where, l = 20
C = 14
f = 14
h = 10
N = 50
Median=20+25–1414×10
= 20+11014
= 20 + 7.85
= 27.85
Mean deviation about the median is,
MD(M)=1N∑6i=1fi|xi–M|
= 150×517.1
= 10.34
Q12. Calculate the mean deviation about the median for the given data:
Age Number
16 – 20 5
21 – 25 6
26 – 30 12
31 – 35 14
36 – 40 26
41 – 45 12
46 – 50 16
51 – 55 9
Sol:
The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
The table is as given below:
Age Number fi Cumulative frequency (c.f.) Mid – point xi |xi–M| fi|xi–x¯|
16 – 20 5 5 18 20 100
21 – 25 6 11 23 15 90
26 – 30 12 23 28 10 120
31 – 35 14 37 33 5 70
36 – 40 26 63 38 0 0
41 – 45 12 75 43 5 60
46 – 50 16 91 48 10 160
51 – 55 9 100 53 15 135
100
The class interval containing (N2)th or 50th item is 35.5 – 40.5. Therefore, 35.5 – 40.5 is the median class.
Now,
Median=l+N2–Cf×h
Where, l = 35.5
C = 37
f = 26
h = 5
N = 100
Median=35.5+50–3726×5
= 35.5+13×526
= 35.5 + 2.5
= 38
Mean deviation about the median is,
MD(M)=1N∑8i=1fi|xi–M|
= 1100×735
= 7.35
Exercise 15.2
Q1. Calculate mean and variance for the given data 6, 7, 10, 12, 13, 4, 8, 12
Sol:
6, 7, 10, 12, 13, 4, 8, 12
Mean,
x¯=∑8i=1xin
= 6+7+10+12+13+4+8+128
= 728
= 9
The given table is obtained,
xi (xi–x¯) (xi–x¯)2
6 -3 9
7 -2 4
10 -1 1
12 3 9
13 4 16
4 -5 25
8 -1 1
12 3 9
74
Variance,
(σ2)=1n∑8i=1(xi–x¯)2
= 18×74
= 9.25
Q2. Calculate the mean and variance for the first n natural numbers.
Sol:
The mean of first n natural number is,
Mean=SumofallobservationsNumberofobservations Mean=n(n+1)2n
= n+12
Variance,
(σ2)
= 1n∑ni=1(xi–x¯)2
= 1n∑ni=1[xi–(n+12)]2
= 1n∑ni=1x2i–1n∑ni=12(n+12)xi+1n∑ni=1(n+12)2
= 1nn(n+1)(2n+1)6–(n+1n)[n(n+1)2]+(n+1)24n×n
= (n+1)(2n+1)6–(n+1)22+(n+1)24
= (n+1)(2n+1)6–(n+1)24
= (n+1)[4n+2–3n–312]
= (n+1)(n–1)12
= n2–112
Q3. Calculate the mean and variance for the first 10 multiples of 3.
Sol:
The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Number of observations,
n = 10
Mean,
x¯=∑10i=110
= 16510
= 16.5
The given table is obtained,
xi (xi–x¯) (xi–x¯)2
3 -13.5 182.25
6 -10.5 110.25
9 -7.5 56.25
12 -4.5 20.25
15 -1.5 2.25
18 1.5 2.25
21 4.5 20.25
24 7.5 56.25
27 10.5 110.25
30 13.5 182.25
742.5
Variance,
(σ2)
= 1n∑10i=1(xi–x¯)2
= 110×742.5
= 74.25
Q4. Calculate the mean and variance for the data
xi 6 10 14 18 24 28 30
fi 2 4 7 12 8 4 3
Sol:
The data obtained is given in the tabular form:
xi xi fixi xi–x¯ (xi−x¯)2 fi(xi−x¯)2
6 2 12 -13 169 338
10 4 40 -9 81 324
14 7 98 -5 25 175
18 12 216 -1 1 12
24 8 192 5 25 200
28 4 112 9 81 324
30 3 90 11 121 363
40 760 1736
Here,
N = 40
∑7i=1fixi = 760
x¯=∑7i=1fixiN
= 76040
= 19
Variance,
(σ2)
= 1n∑7i=1(xi–x¯)2
= 140×1736
= 43.4
Q5. Calculate the mean and variance for the data
xi 92 93 97 98 102 104 109
fi 3 2 3 2 6 3 3
Sol:
The data obtained is given in the tabular form:
xi fi fixi xi–x¯ (xi−x¯)2 fi(xi−x¯)2
92 3 276 -8 64 192
93 2 186 -7 49 98
97 3 291 -3 9 27
98 2 196 -2 4 8
102 6 612 2 4 24
104 3 312 4 16 48
109 3 327 9 81 243
22 2200 640
Here,
N = 22
∑7i=1fixi = 2200
x¯=∑7i=1fixiN
= 220022
= 100
Variance,
(σ2)
= 1n∑7i=1(xi–x¯)2
= 122×640
= 29.09
Q6. Calculate 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
Sol:
The data is obtained in tabular form as follows.
xi fi fi=xi−641 y2i fiyi fiy2i
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
100 220 0 286
Mean,
x¯=A∑9i=1fiyiN×h
= 64+0100×1
= 64 + 0
= 64
Variance,
(σ2)
= h2N2[N∑9i=1fiy2i–(∑9i=1fiyi)2]
= 11002[100×286–0]
= 2.86
Standard deviation,
σ=2.86−−−−√
= 1.69
Q7. Calculate the mean and variance for the given frequency distribution.
Classes 0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180 180 – 210
Frequencies 2 3 5 10 3 5 2
Sol:
Class fi xi yi=xi−10530 y2i fiyi fiy2i
0 – 30 2 15 -3 9 -6 18
30 – 60 3 45 -2 4 -6 12
60 – 90 5 75 -1 1 -5 5
90 – 120 10 105 0 0 0 0
120 – 150 3 135 1 1 3 3
150 – 180 5 165 2 4 10 20
180 – 210 2 195 3 9 6 18
30 2 76
Mean,
x¯
= A+∑7i=1fiyiN×h
= 105+230×30
= 105 + 2
= 107
Variance,
(σ2)
= h2N2[N∑7i=1fiy2i–(∑7i=1fiyi)2]
= 302302[30×76–(2)2]
= 2280 – 4
= 2276
Q8. Calculate the mean and variance for the following frequency distribution.
Classes 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Frequencies 5 8 15 16 6
Sol:
Class Frequency fi Mid – point xi yi=xi−2510 y2i fiyi fiy2i
0 – 10 5 5 -2 4 -10 20
10 – 20 8 15 -1 1 -8 8
20 – 30 15 25 0 0 0 0
30 – 40 16 35 1 1 16 16
40 – 50 6 45 2 4 12 24
50 10 68
Mean,
x¯
= A+∑5i=1fiyiN×h
= 25+1050×10
= 25 + 2
= 27
Variance,
(σ2)
= h2N2[N∑5i=1fiy2i–(∑5i=1fiyi)2]
= 102502[50×68–(10)2]
= 125[3400–100]
= 330025
= 132
Q9. Calculate the mean, variance and standard deviation using short – cut method:
Height in cm Number of children
70 – 75 3
75 – 80 4
80 – 85 7
85 – 90 7
90 – 95 15
95 – 100 9
100 – 105 6
105 – 110 6
110 – 115 3
Sol:
Class fi xi yi=xi−92.55 y2i fiyi fiy2i
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
60 6 254
Mean,
x¯
= A+∑9i=1fiyiN×h
= 92.5+660×5
= 92.5 + 0.5
= 93
Variance,
(σ2)
= h2N2[N∑9i=1fiy2i–(∑9i=1fiyi)2]
= 52602[60×254–(6)2]
= 253600[155204]
= 105.58
Standard deviation,
(σ)
= 105.58−−−−−√
= 10.27
Q10. The diameters of circles (in mm) drawn in a design are given below:
Diameters Number of Children
33 – 36 15
37 – 40 17
41 – 44 21
45 – 48 22
49 – 52 25
Sol:
Class fi xi yi=xi−42.54 y2i fiyi fiy2i
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
100 25 199
N = 100
h = 4
Let us assume that mean A be 42.5
Mean,
x¯
= A+∑5i=1fiyiN×h
= 42.5+25100×4
= 43.5
Variance,
(σ2)
= h2N2[N∑5i=1fiy2i–(∑5i=1fiyi)2]
= 421002[100×199–(25)2]
= 1610000[19900–625]
= 1610000×19275
= 30.84
Standard deviation,
(σ)
= 30.84−−−−√
= 5.55
Exercise 15.3
Q1. 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
Sol:
The standard deviation of group A is calculated as given in the table below.
Marks fi xi yi=xi−4510 y2i fiyi fiy2i
10 – 20 9 15 -3 9 -27 81
20 – 30 17 25 -2 4 -34 68
30 – 40 32 35 -1 1 -32 32
40 – 50 33 45 0 0 0 0
50 – 60 40 55 1 1 40 40
60 – 70 10 65 2 4 20 40
70 – 80 9 75 3 9 27 81
150 -6 342
N = 150
h = 10
A = 45
Mean,
x¯
= A+∑7i=1xiN×h
= 45+(−6)150×10
= 45 – 0.4
= 44.6
Variance,
(σ21)
= h2N2[N∑7i=1fiy2i–(∑7i=1fiyi)2]
= 1021102[150×342–(−6)2]
= 10022500[51264]
= 1225×51264
= 227.84
Standard deviation,
(σ1)
= 227.84−−−−−√
= 15.09
The standard deviation of group A is calculated as given in the table below.
Marks fi xi yi=xi−4510 y2i fiyi fiy2i
10 – 20 10 15 -3 9 -30 90
20 – 30 20 25 -2 4 -40 80
30 – 40 30 35 -1 1 -30 30
40 – 50 25 45 0 0 0 0
50 – 60 43 55 1 1 43 43
60 – 70 15 65 2 4 30 60
70 – 80 7 75 3 9 21 63
150 -6 366
N = 150
h = 10
A = 45
Mean,
x¯
= A+∑7i=1xiN×h
= 45+(−6)150×10
= 45 – 0.4
= 44.6
Variance,
(σ22)
= h2N2[N∑7i=1fiy2i–(∑7i=1fiyi)2]
= 1021102[150×366–(−6)2]
= 10022500[54864]
= 1225×54864
= 243.84
Standard deviation,
(σ1)
= 243.84−−−−−√
= 15.61
As the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.
Q2. 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
Sol:
The prices of the shares X are 35, 54, 52, 53, 56, 58, 52, 50, 51, and 49.
Here, the number of observations, N = 10
Mean,
x¯
= 1N∑10i=1xi
= 110×510
= 51
The following table is obtained corresponding to shares X.
xi (xi−x¯) (xi−x¯)2
35 -16 256
54 3 9
52 1 1
53 2 4
56 5 25
58 7 49
52 1 1
50 -1 1
51 0 0
49 -2 4
350
Variance,
(σ21)
= 1N∑10i=1(xi–x¯)2
= 110×350
= 35
Standard deviation,
(σ1)
= 35−−√
= 5.91
C.V.
σ1x×100
= 5.9151×100
= 11.58
The prices of share Y are 108, 107, 105, 105, 106, 107, 104, 103, 104, and 101.
Mean,
y¯
= 1N∑10i=1yi
= 110×1050
= 105
The following table is obtained corresponding to shares Y.
yi (yi−y¯) (yi−y¯)2
108 3 9
107 2 4
105 0 0
105 0 0
106 1 1
107 2 4
104 -1 1
103 -2 4
104 -1 1
101 -4 16
40
Variance,
(σ22)
= 1N∑10i=1(yi–y¯)2
= 110×40
= 4
Standard deviation,
(σ2)
= 4–√
= 2
C.V.
σ2y×100
= 2105×100
= 1.9
C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Therefore, the prices of shares Y are more stable than the prices of shares X.
Q3. 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 wage 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?
Sol:
(i)
Monthly wages of firm A = Rs. 5253
Number of wage earners in firm A = 586
Therefore,
Total amount paid = Rs. 5253 × 586
Monthly wages of firm B = Rs. 5253
Number of wage earners in firm B = 648
Therefore,
Total amount paid = Rs. 5253 × 648
Hence, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.
(ii)
Variance of the distribution of wages in firm A,
A(σ21) = 100
Therefore,
Standard deviation of the distribution of wages in firm A,
A(σ1)=100−−−√ = 10
Variance of the distribution of wages in firm B,
B(σ22) = 121
Therefore,
Standard deviation of the distribution of wages in firm B,
B(σ2)=121−−−√ = 11
The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.
Therefore, firm B has greater variability in the individual wages.
Q4. The following is the record of goals scored by team A in a football session:
Number of goals scored 0 1 2 3 4
Number of matches 1 9 7 5 3
For the team B, mean number of goals scored per match was 2 with standard deviation 1.25 goals. Find which team may be considered more consistent?
Sol:
The mean and the standard deviation of goals scored by team A are calculated as follows.
No. of goals scored No. of matches fixi x2i fix2i
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
23 50 130
Mean,
∑5i=1fixi∑5i=1fi
= 5025
= 2
Therefore, the mean of both the teams is same.
Standard deviation,
σ
= 1NN∑fix2i–(∑fixi)2−−−−−−−−−−−−−−−−√
= 12525×130–(50)2−−−−−−−−−−−−√
= 125750−−−√
= 125×27.38
= 1.09
The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same that is 2. Therefore, the team with lower standard deviation will be more consistent.
Therefore, team A is more consistent than team B.
Q5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are as given below:
∑50i=1xi=212
∑50i=1x2i=902.8
∑50i=1yi=261
∑50i=1y2i=1457.6
Which is more varying, the length or weight?
Sol:
Length
∑50i=1xi=212 ∑50i=1x2i=902.8
Here, N = 50
Mean,
x¯
= ∑50i=1yiN
= 21250
= 4.24
Variance,
σ21
= 1N∑50i=1(xi–x¯)2
= 150∑50i=1(xi–4.24)2
= 150∑50i=1[x2i–8.48xi+17.97]
= 150[∑50i=1x2i–8.48∑50i=1xi+17.97×50]
= 150[902.8–8.48×(212)+898.5]
= 150[1801.3–1797.76]
= 150×3.54
= 0.07
Standard deviation,
σi(Length)
= 0.07−−−−√
= 0.26
C.V. (Length)
= StandarddeviationMean×100
= 0.264.24×100
= 6.13
Weight
∑50i=1yi=261 ∑50i=1y2i=1457.6
Mean,
y¯
= ∑50i=1yiN
= 26150
= 5.22
Variance,
σ22
= 1N∑50i=1(yi–y¯)2
= 150∑50i=1(yi–5.22)2
= 150∑50i=1[y2i–10.44yi+27.24]
= 150[∑50i=1y2i–10.44∑50i=1yi+27.24×50]
= 150[1457.6–10.44×(261)+1362]
= 150[2819.6–2724.84]
= 150×94.76
= 1.89
Standard deviation,
σ2(Weight)
= 1.89−−−−√
= 1.37
C.V. (Weight)
= StandarddeviationMean×100
= 1.375.22×100
= 26.24
Therefore, C.V. of weights is greater than the C.V. of lengths. Hence, weights vary more than the lengths.
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