CLASS 11 MATH CHAPTER 15 NCERT SOLUTION
CH15 – STATISTICS
Exercise 15.1 Page: 360
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., xi – 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., xi – 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 = 6th = 13
(12/2)+ 1)th observation = 6 + 1
= 7th = 14
Median = (13 + 14)/2
= 27/2
= 13.5
So, the absolute values of the respective deviations from the median, i.e., |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
Mean Deviation
= (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 = 5th = 46
(10/2)+ 1)th observation = 5 + 1
= 6th = 49
Median = (46 + 49)/2
= 95
= 47.5
So, the absolute values of the respective deviations from the median, i.e., |xi – 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.
Solution:-
Let us make the table of the given data and append other columns after calculations.
The sum of calculated data,
Solution:-
Let us make the table of the given data and append other columns after calculations.
Find the mean deviation about the median for the data in Exercises 7 and 8.
7.
Solution:-
Let us make the table of the given data and append other columns after calculations.
Xi fi c.f. |xi – M| fi |xi – 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 13th and 14th observations. Both of these observations lie in the cumulative frequency of 14, for which the corresponding observation is 7.
Then,
Median = (13th observation + 14th observation)/2
= (7 + 7)/2
= 14/2
= 7
So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.
8.
xi 15 21 27 30 35
fi 3 5 6 7 8
Solution:-
Let us make the table of the given data and append other columns after calculations.
Xi fi c.f. |xi – M| fi |xi – 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 = (15th observation + 16th observation)/2
= (30 + 30)/2
= 60/2
= 30
So, the absolute values of the respective deviations from the median, i.e., |xi – 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 fi Midpoints
xi fixi |xi – x̅| fi|xi – 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 fi Midpoints
xi fixi |xi – x̅| fi|xi – 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 fi Cumulative frequency (c.f.) Mid -points
xi |xi – Med| fi|xi – 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 Nth/2 or 25th 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 fi Cumulative frequency (c.f.) Midpoints
xi |xi – Med| fi|xi – 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 Nth/2 or 50th 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
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Exercise 15.2 Page: 371
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.
Xi Deviations from mean
(xi – x̅) (xi – 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 the Variance,
σ2 = (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 substituting the value of x̅, we get
WKT, (a + b)(a – b) = a2 – b2
σ2 = (n2 – 1)/12
∴Mean = (n + 1)/2 and Variance = (n2 – 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.
Xi Deviations from mean
(xi – x̅) (xi – 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, the Variance
= (1/10) × 742.5
= 74.25
∴Mean = 16.5 and Variance = 74.25
4.
xi 6 10 14 18 24 28 30
fi 2 4 7 12 8 4 3
Solution:-
Let us make the table of the given data and append other columns after calculations.
Xi fi fixi Deviations from mean
(xi – x̅) (xi – x̅)2 fi(xi – 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.
xi 92 93 97 98 102 104 109
fi 3 2 3 2 6 3 3
Solution:-
Let us make the table of the given data and append other columns after calculations.
Xi fi fixi Deviations from mean
(xi – x̅) (xi – x̅)2 fi(xi – 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.
xi 60 61 62 63 64 65 66 67 68
fi 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.
Xi Frequency fi Yi = (xi – A)/h Yi2 fiyi fiyi2
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, the variance,
σ2 = (12/1002) [100(286) – 02]
= (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 fi Midpoints
xi fixi (xi – x̅) (xi – x̅)2 fi(xi – 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 fi Midpoints
xi fixi (xi – x̅) (xi – x̅)2 fi(xi – 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 the 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 fi Midpoint
Xi Yi = (xi – A)/h Yi2 fiyi fiyi2
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, the Variance,
σ2 = (52/602) [60(254) – 62]
= (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 fi) Midpoint
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
Mean,
Where, A = 42.5, h = 4
So, x̅ = 42.5 + (25/100) × 4
= 42.5 + 1
= 43.5
Then, the Variance,
σ2 = (42/1002)[100(199) – 252]
= (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.
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Exercise 15.3 Page: 375
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
fi Midpoint
Xi Yi = (xi – A)/h (Yi)2 fiyi fi(yi)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 02 0 0
50 – 60 40 55 ((55 – 45)/10) = 1 12
= 1 40 40
60 – 70 10 65 ((65 – 45)/10) = 2 22
= 4 20 40
70 – 80 9 75 ((75 – 45)/10) = 3 32
= 9 27 81
Total 150 -6 342
Where A = 45,
and yi = (xi – A)/h
Here h = class size = 20 – 10
h = 10
So, x̅ = 45 + ((-6/150) × 10)
= 45 – 0.4
= 44.6
σ2 = (102/1502) [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
fi Midpoint
Xi Yi = (xi – A)/h (Yi)2 fiyi fi(yi)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 02 0 0
50 – 60 43 55 ((55 – 45)/10) = 1 12
= 1 43 43
60 – 70 15 65 ((65 – 45)/10) = 2 22
= 4 30 60
70 – 80 7 75 ((75 – 45)/10) = 3 32
= 9 21 63
Total 150 -6 366
Where A = 45,
h = 10
So, x̅ = 45 + ((-6/150) × 10)
= 45 – 0.4
= 44.6
σ2 = (102/1502) [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 the 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 (xi) Y (yi) Xi2 Yi2
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 = ∑xi/n
Where, n = number of terms
= 510/10
= 51
Then, Variance for x =
= (1/102)[(10 × 26360) – 5102]
= (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 = ∑yi/n
Where, n = number of terms
= 1050/10
= 105
Then, Variance for y =
= (1/102)[(10 × 110290) – 10502]
= (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 a 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 team B, the 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 xi Number of matches fi fixi Xi2 fixi2
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
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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Miscellaneous Exercise Page: 380
1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Solution:-
2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.
Solution:-
3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Solution:-
We know that,
4. Given that x̅ is the mean and σ2 is the variance of n observations x1, x2, …,xn . Prove that the mean and variance of the observations ax1, ax2, ax3, …., axn are ax̅ and a2σ2, respectively, (a ≠ 0).
Solution:-
From the question, it is given that, n observations are x1, x2,…..xn
Mean of the n observation = x̅
Variance of the n observation = σ2
As we know,
5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12
Solution:-
(i) If the wrong item is omitted,
From the question, it is given that
The number of observations, i.e., n = 20
The incorrect mean = 20
The incorrect standard deviation = 2
(ii) If it is replaced by 12,
From the question, it is given that
The number of incorrect sum observations, i.e., n = 200
The correct sum of observations n = 200 – 8 + 12
n = 204
Then, correct mean = correct sum/20
= 204/20
= 10.2
6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry, are given below.
Subject Mathematics Physics Chemistry
Mean 42 32 40.9
Standard deviation 12 15 20
Which of the three subjects shows the highest variability in marks, and which shows the lowest?
Solution:-
From the question, it is given that
Mean of Mathematics = 42
Standard deviation of Mathematics = 12
Mean of Physics = 32
Standard deviation of physics = 15
Mean of Chemistry = 40.9
Standard deviation of chemistry = 20
As we know,
7. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on, it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Solution:-
From the question, it is given that
The total number of observations (n) = 100
Incorrect mean, (x̅) = 20
And, Incorrect standard deviation (σ) = 3
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