Class 11 - Maths - Statistics

Exercise 15.1

Question 1:

Find the mean deviation about the mean for the data

4, 7, 8, 9, 10, 12, 13, 17

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data, X = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)/10 = 80/10 = 8

The deviations of the respective observations from the mean X i.e. xi – X are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e. |xi – X|, are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

MD(X) = (Σi=18|xi – X|)/8 = (6 + 3 + 2 + 1 + 0 + 2 + 3 + 7)/8 = 24/8 = 3

Question 2:

Find the mean deviation about the mean for the data

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the data, X = (38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44)/10 = 500/10 = 50

The deviations of the respective observations from the mean X i.e. xi – X are

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. |xi – X|, are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

MD(X) = (Σi=110|xi – X|)/10 = (12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6)/10 = 84/10 = 8.4

Question 3:

Find the mean deviation about the median for the data.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Median, M = [(12/2)th observation + (12/2 + 1)th observation]/2

= [6th observation + 7th observation]/2

= (13 + 14)/2

= 27/2 = 13.5

The deviations of the respective observations 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

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

MD(X) = (Σi=112|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 + 4.5)/12

= 28/12

= 2.33

Question 4:

Find the mean deviation about the median for the data.

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the numbers of observations are 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Median, M = [(10/2)th observation + (10/2 + 1)th observation]/2

= [5th observation + 6th observation]/2

= (46 + 49)/2

= 95/2 = 47.5

The deviations of the respective observations 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

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

MD(X) = (Σi=110|xi – M|)/10

= (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5)/10 = 70/10 = 7

Question 5:

Find the mean deviation about the mean for the data.

 xi 5 10 15 20 25 fi 7 4 6 3 5

N = Σi=15 fi = 25

Σi=15 fi xi = 350

Now, Mean x = Σi=15 fi xi/ Σi=15 fi

= 350/25

= 14

So, MD(x) = (1/N) * Σi=15 fi|xi – x|= 158/25 = 6.32

Question 6:

Find the mean deviation about the mean for the data.

 xi 10 30 50 70 90 fi 4 24 28 16 8

N = Σi=15 fi = 80

Σi=15 fi xi = 4000

Now, Mean x = Σi=15 fi xi/ Σi=15 fi

= 4000/80

= 50

So, MD(x) = (1/N) * Σi=15 fi |xi – x|= 1280/80 = 16

Question 7:

Find the mean deviation about the median for the data.

 xi 5 7 9 10 12 15 yi 8 6 2 2 2 6

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we get the

following table:

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.

So, Median = (13th observations + 14th observations)/2

= (7 + 7)/2 = 14/2 = 7

The absolute values of the deviations from median, i.e. |xi - M|are

Σi=16 fi = 26

Σi=16 fi |xi – M|= 84

So, MD(X) = (1/N) * Σi=16 fi |xi – M|= 84/26 = 3.23

Question 8:

Find the mean deviation about the median for the data.

 xi 15 21 27 30 35 yi 3 5 6 7 8

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we get the

following table:

Here, N = 29, which is odd.

Now, Median = (29 + 1)/2 th observations = 15th observations

This observation lies in the cumulative frequency 21,

for which the corresponding observation is 30.

So, Median = 30

The absolute values of the deviations from median, i.e. |xi - M|are

Σi=15 fi = 29

Σi=15 fi |xi – M|= 148

So, MD(X) = (1/N) * Σi=15 fi |xi – M|= 148/29 = 5.1

Question 9:

Find the mean deviation about the mean for the 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

The following table is formed:

Here, N = Σi=18 fi = 50

Σi=18 fi xi = 17900

Now, x = (1/N) * Σi=18 fi xi = 17900/50 = 358

So, MD(x) = (1/N) * Σi=18 fi |xi – x|= 7896/50 = 157.92

Question 10:

Find the mean deviation about the mean for the data:

 Heights in cms Number of boys 95-105 9 105-115 13 115-125 26 125-135 30 135-145 12 145-155 10

The following table is formed:

Here, N = Σi=16 fi = 100

Σi=16 fi xi = 12530

Now, x = (1/N) * Σi=16 fi xi = 12530/100 = 125.3

So, MD(x) = (1/N) * Σi=16 fi |xi – x|= 1128.8/100 = 11.28

Question 11:

Find the mean deviation about median for the following data:

 Marks Number of girls 0-10 6 10-20 8 20-30 14 30-40 16 40-50 4 50-60 2

The following table is formed:

The class interval containing the (N/2)th or 25th item is 20 – 30.

Therefore, 20 – 30 is the median class.

We know that

Median M = l + {(N/2 - C)/f} * h

Here, l = 20, C = 14, f = 14, h = 10, and N = 50

Median, M = 20 + {(25 - 14)/14} * 10

=> M = 20 + 110/14

=> M = 20 + 7.85

=> M = 27.85

Thus, mean deviation about the median is given by,

MD(X) = (1/N) * Σi=16 fi |xi – M|= 517.1/50 = 10.34

Question 12:

Calculate the mean deviation about median age for the age distribution of 100 persons given below:

 Age 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

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 formed as follows:

The class interval containing the (N/2)th or 50th item is 35.5 – 40.5

Therefore, 35.5 – 40.5 is the median class.

We know that

Median M = l + {(N/2 - C)/f} * h

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

Median, M = 35.5 + {(50 - 37)/26} * 5

=> M = 35.5 + (13 * 5)/26

=> M = 35.5 + 2.5

=> M = 38

Thus, mean deviation about the median is given by,

MD(X) = (1/N) * Σi=16 fi |xi – M|= 735/100 = 7.35

Exercise 15.2

Question 1:

Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12

Given data is: 6, 7, 10, 12, 13, 4, 8, 12

Mean x = Σi=18 xi = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

The following table is obtained:

Now, varience σ2 = (1/N) * Σi=18 (xi – x)2 = 74/8 = 9.25

Question 2:

Find the mean and variance for the first n natural numbers.

The mean of first n natural numbers is calculated as follows:

Mean = Sum of all observations/Number of observations

=> Mean, X = {n(n + 1)/2}/n = (n + 1)/2

Now, varience σ2 = (1/n) * Σi=1n (xi – X)2

= (1/n) * Σi=1n xi2 – (1/n) * Σi=1n 2 * {(n + 1)/2}xi + (1/n) * Σi=1n {(n + 1)/2}2

= (1/n){n(n + 1)(2n + 1)/6} – {(n + 1)/n}[n(n + 1)/2] + {(n + 1)2/4n} * n

= (n + 1)(2n + 1)/6 – (n + 1)2/2 + (n + 1)2/4

= (n + 1)(2n + 1)/6 – (n + 1)2/4

= (n + 1)[(4n + 2 – 3n - 3)/12]

= {(n + 1)(n - 1)}/12

= (n2 - 1)/12

Question 3:

Find the mean and variance for the first 10 multiples of 3

The first 10 multiples of 3 are

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Here, number of observations, n = 10

Mean x = Σi=110 xi = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

The following table is obtained:

Now, varience σ2 = (1/N) * Σi=110 (xi – x)2 = 742.5/10 = 74.25

Question 4:

Find the mean and variance for the data

 xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3

The data is obtained in tabular form as follows:

Here, N = 40, Σi=17 fi xi = 760

Mean x = (1/N) * Σi=17 fi xi = 760/40 = 19

Now, varience σ2 = (1/N) * Σi=17 (xi – x)2 = 1736/19 = 43.4

Question 5:

Find the mean and variance for the data

 xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3

The data is obtained in tabular form as follows:

Here, N = 22, Σi=17 fi xi = 2200

Mean x = (1/N) * Σi=17 fi xi = 2200/22 = 100

Now, varience σ2 = (1/N) * Σi=17 (xi – x)2 = 640/22 = 29.09

Question 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

The data is obtained in tabular form as follows:

Mean x = A + [Σi=19 fi yi /N] * h = 64 + 0/100 * 1 = 64 + 0 = 64

Now, varience σ2 = (h2/N2) * [N * Σi=19 fi yi2 – (Σi=19 fi yi)2]

= (1/1002)[100 * 286 - 0]

= 286/100

= 2.86

So, standard deviation (σ) = √(2.86) = 1.69

Question 7:

Find the mean and variance for the following 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

The data is obtained in tabular form as follows:

Mean x = A + [Σi=17 fi yi /N] * h = 105 + (2/30) * 30

= 105 + 2 = 107

Now, varience σ2 = (h2/N2) * [N * Σi=17 fi yi2 – (Σi=17 fi yi)2]

= (302/302)[30 * 76 - 22]

= 2284 - 4

= 2276

Question 8:

Find 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

The data is obtained in tabular form as follows:

Mean x = A + [Σi=15 fi yi /N] * h = 25 + (10/50) * 10

= 25 + 2 = 27

Now, varience σ2 = (h2/N2) * [N * Σi=15 fi yi2 – (Σi=15 fi yi)2]

= (102/502)[50 * 68 - 102]

= (3400 - 100)/25

= 3300/25

= 132

Question 9:

Find the mean, varience 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 No. of children 3 4 7 7 15 9 6 6 3

The data is obtained in tabular form as follows:

Mean x = A + [Σi=19 fi yi /N] * h = 92.5 + (6/60) * 5

= 92.5 + 0.5

= 93

Now, varience σ2 = (h2/N2) * [N * Σi=19 fi yi2 – (Σi=19 fi yi)2]

= (52/602)[60 * 254 - 62]

= (25/3600) * 15204

= 105.58

So, standard deviation (σ) = √(105.58) = 10.27

Question 10:

The diameters of circles (in mm) drawn in a design are given below:

 Diameter 33-36 37-40 41-44 45-48 49-52 No. of children 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.]

Here, N = 100, h = 4

The data is obtained in tabular form as follows:

Let the assumed mean, A, be 42.5

Mean x = A + [Σi=19 fi yi /N] * h = 42.5 + (25/100) * 4

= 42.5 + 1 = 43.5

Now, varience σ2 = (h2/N2) * [N * Σi=19 fi yi2 – (Σi=19 fi yi)2]

= (16/10000)[100 * 199 - 252]

= (16/10000)[19900 - 625]

= (16/10000) * 19275

= 30.84

So, standard deviation (σ) = √(30.84) = 5.55

Exercise 15.3

Question 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

Firstly, the standard deviation of group A is calculated as follows:

Here, h = 10, N = 150, A = 45

Mean x = A + [Σi=17 xi /N] * h

= 45 + (-6/150) * 10

= 45 - 0.4

= 44.6

Now, varience σ12 = (h2/N2) * [N * Σi=17 fi yi2 – (Σi=17 fi yi)2]

= (100/22500)[150 * 342 – (-6)2]

= (1/225) * 51264

= 227.84

So, standard deviation (σ) = √(227.84) = 15.09

The standard deviation of group B is calculated as follows:

Mean x = A + [Σi=17 xi /N] * h

= 45 + (-6/150) * 10

= 45 - 0.4

= 44.6

Now, varience σ22 = (h2/N2) * [N * Σi=17 fi yi2 – (Σi=17 fi yi)2]

= (100/22500)[150 * 366 – (-6)2]

= (1/225) * 54864

= 243.84

So, standard deviation (σ) = √(243.84) = 15.61

Since 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.

Question 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

The prices of the shares X are:

35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Here, the number of observations, N = 10

Mean x = Σi=110 xi /N

= 510/10

= 51

The following table is obtained corresponding to shares X.

Now, varience σ12 = (1/N) * Σi=110 (xi – x)2 = 350/10 = 35

So, Standard deviation (σ1) = √35 = 5.91

CV (Shares X) = (σ1/x) * 100 = (5.91/51) * 100 = 11.58

The prices of share Y are:

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

Mean y = Σi=110 yi /N

= 1050/10

= 105

The following table is obtained corresponding to shares Y.

Now, varience σ22 = (1/N) * Σi=110 (yi – y)2 = 40/10 = 4

So, Standard deviation (σ2) = √4 = 2

CV (Shares X) = (σ2/y) * 100 = (2/105) * 100 = 1.9 < 11.58

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Thus, the prices of shares Y are more stable than the prices of shares X.

Question 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 earner 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?

(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

So, total amount paid = Rs 5253 * 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

So, total amount paid = Rs 5253 * 648

Thus, 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 (σ12) = 100

Standard deviation of the distribution of wages in firm

A (σ1) = √100 = 10

Variance of the distribution of wages in firm B (σ22) = 121

Standard deviation of the distribution of wages in firm

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.

Thus, firm B has greater variability in the individual wages.

Question 4:

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

 No. of goal 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?

The mean and the standard deviation of goals scored by team A are calculated as follows:

Mean = Σi=15 fi xi / Σi=15 fi

= 50/2

= 2

Thus, the mean of both the teams is same.

Now, Standard deviation of team A is

σ1 = (1/N)√[N Σfi xi2 – (Σ fi xi)2]

= (1/25)√[25 * 130 – (50)2]

= √750/25

= 27.38/25

= 1.09

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same i.e., 2.

Therefore, the team with lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.

Question 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:

Σi=150 xi = 212, Σi=150 xi2 = 902.8, Σi=150 yi = 261, Σi=150 yi2 = 1457.6

Which is more varying, the length or weight?

Given, Σi=150 xi = 212, Σi=150 xi2 = 902.8

Here, N = 50

So, Mean, x = Σi=150 xi/N = 212/50 = 4.24

Variance σ12 = (1/N)* Σi=150 (xi – x)2

= (1/50) * Σi=150 (xi – 4.24)2

= (1/50) * Σi=150 [xi 2 – 8.48 xi + 17.97]

= (1/50) * [Σi=150 xi 2 – 8.48 Σi=150 xi + 17.97 * 50]

= (1/50) * [902.8 – 8.48 * 212 + 898.5]

= (1/50) * [1801.3 – 1797.76]

= 3.54/50

= 0.07

So, standard deviation (σ1) = √0.07 = 0.26

CV (Length) = (standard deviation/Mean) * 100

= (0.26/4.24) * 100

= 6.13

Again, given Σi=150 yi = 261, Σi=150 yi2 = 1457.6

Here, N = 50

So, Mean, y = Σi=150 yi/N = 261/50 = 5.22

Variance σ22 = (1/N)* Σi=150 (yi – y)2

= (1/50) * Σi=150 (yi – 5.22)2

= (1/50) * Σi=150 [yi 2 – 10.44 yi + 27.24]

= (1/50) * [Σi=150 yi 2 – 10.44 Σi=150 yi + 27.24 * 50]

= (1/50) * [1457.6 – 10.44 * 261 + 1362]

= (1/50) * [2819.6 – 2724.84]

= 94.76/50

= 1.89

So, standard deviation (σ2) = √1.89 = 1.37

CV (Length) = (standard deviation/Mean) * 100

= (1.37/5.22) * 100

= 26.24

Thus, C.V. of weights is greater than the C.V. of lengths.

Therefore, weights vary more than the lengths.

Miscellaneous Exercise of Chapter 12

Question 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.

Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

Mean, x = (6 + 7 + 10 + 12 + 12 + 13 + x + y)/8 = 9

=> 60 + x + y = 9 * 8

=> 60 + x + y = 72

=> x + y = 72 – 69

=> x + y = 12    ………………1

Variance = (1/n)* Σi=18 (xi – x)2 = 9.25

=> 9.25 = (1/8)[(-3)2 + (-2)2 + 12 + 32 + 32 + 42 + x2 + y2 – 2 * 9 *(x + y) + 2 * 92]

=> 9.25 = (1/8)[9 + 4 + 1 + 9 + 9 + 16 + x2 + y2 – 18 * 12 + 162]

=> 9.25 = (1/8)[48 + x2 + y2 – 216 + 162]

=> 9.25 = (1/8)[x2 + y2 – 6]

=> x2 + y2 – 6 = 74

=> x2 + y2 = 80     ………………..2

From equation 1, we get

x2 + y2 + 2xy = 144     ...........3

From equation 2 and 3, we get

80 + 2xy = 144

=> 2xy = 64    ......................4

Subtracting equation 4 from 2, we get

x 2 + y2 – 2xy = 80 – 64 = 16

=> (x – y)2 = 16

=> x – y = ± 4   ...............5

Therefore, from equation 1 and 5, we get

x = 8 and y = 4, when x – y = 4

x = 4 and y = 8, when x – y = –4

Thus, the remaining observations are 4 and 8.

Question 2:

The mean and variance of 7 observations are 8 and 16, respectively.

If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

Mean, x = (2 + 4 + 10 + 12 + 14 + x + y)/7 = 8

=> 42 + x + y = 8 * 7

=> 42 + x + y = 56

=> x + y = 56 – 42

=> x + y = 14    ………………1

Variance = (1/n)* Σi=17 (xi – x)2 = 16

=> 16 = (1/7)[(-6)2 + (-4)2 + 22 + 42 + 62 + x2 + y2 – 2 * 8 *(x + y) + 2 * 82]

=> 16 = (1/7)[36 + 16 + 4 + 16 + 36 + x2 + y2 – 16 * 14 + 128]

=> 16 = (1/7)[108 + x2 + y2 – 224 + 128]

=> 16 = (1/7)[x2 + y2 + 12]

=> x2 + y2 + 12 = 112

=> x2 + y2 = 100     ………………..2

From equation 1, we get

x2 + y2 + 2xy = 196     ...........3

From equation 2 and 3, we get

100 + 2xy = 196

=> 2xy = 96    ......................4

Subtracting equation 4 from 2, we get

x 2 + y2 – 2xy = 100 – 96

=> (x – y)2 = 4

=> x – y = ± 2   ...............5

Therefore, from equation 1 and 5, we get

x = 8 and y = 6, when x – y = 2

x = 6 and y = 8, when x – y = –2

Thus, the remaining observations are 6 and 8.

Question 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.

Let the observations be x1, x2, x3, x4, x5, and x6.

It is given that mean is 8 and standard deviation is 4.

Mean, x = (x1 + x2 + x3 + x4 + x5 + x6)/6 = 8    ………………1

If each observation is multiplied by 3 and the resulting observations are yi, then

yi = 3xi

i.e. xi = yi/3 for i = 1 to 6

So, New Mean, y = (y1 + y2 + y3 + y4 + y5 + y6)/6 = 8

= 3(x1 + x2 + x3 + x4 + x5 + x6)/6

= 3 * 8

= 24

Standard deviation σ = √[(1/n) * Σi=16 (xi – x)2]

=> 42 = Σi=16 (xi – x)2/6

=> Σi=16 (xi – x)2 = 96   ………….2

From equation 1 and 2, it can be observed that,

y = 3x

=> x = y/3

Substituting the values of xi and x in equation 2, we obtain

=> Σi=16 (yi /3  – y/3)2 = 96

=> Σi=16 (yi  – y)2 = 864

Therefore, variance of new observations = 864/6 = 144

Hence, the standard deviation of new observations is √144 = 12

Question 4:

Given that x 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 ax and a2σ2 respectively (a ≠ 0).

The given n observations are x1, x2 ... xn.

Mean = x

Variance = σ2

So, σ2 = (1/n) * Σi=1n (xi – x)2  …………1

If each observation is multiplied by a and the new observations are yi, then

yi = axi

=> xi = yi/a

So, y = (1/n) * Σi=1n xi = (1/n) * Σi=1n axi = (a/n) * Σi=1n xi  = ax

Therefore, mean of the observations, ax1, ax2 ... axn, is ax.

Substituting the values of xi and x in equation 1, we obtain

σ2 = (1/n) * Σi=1n (yi /a – y/a)2

=> a2 σ2 = (1/n) * Σi=1n (yi – y)2

Thus, the variance of the observations ax1, ax2 ... axn, is a2 σ2

Question 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.

(i) Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

Mean, x = (1/n) * Σi=120 xi

=> 10 = (1/20) * Σi=120 xi

=> Σi=120 xi = 200

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

Correct mean = correct sum/19 = 192/19 = 10.1

Standard deviation σ = √[(1/n) * Σi=1n xi2 – (Σi=1n xi)2 /n2]

=> 2 = √[(1/n) * incorrect Σi=1n xi2 – (x)2]

=> 2 = √[(1/20) * incorrect Σi=1n xi2 – (10)2]

=> 4 = (1/20) * incorrect Σi=1n xi2 – 100

=> incorrect Σi=1n xi2 = 2080

So, correct Σi=1n xi2 = incorrect Σi=1n xi2 – 82

= 2080 – 64

= 2016

So, Correct Standard deviation = √[correct Σ xi2/n – (correct mean)2]

= √[2016/19 – (10.1)2]

= √[106.1 – 102.01]

= √4.09

= 2.02

(ii) When 8 is replaced by 12,

Incorrect sum of observations = 200

So, Correct sum of observations = 200 – 8 + 12 = 204

Correct mean = correct sum/20 = 204/20 = 10.2

Standard deviation σ = √[(1/n) * Σi=1n xi2 – (Σi=1n xi)2 /n2]

=> 2 = √[(1/n) * incorrect Σi=1n xi2 – (x)2]

=> 2 = √[(1/20) * incorrect Σi=1n xi2 – (10)2]

=> 4 = (1/20) * incorrect Σi=1n xi2 – 100

=> (1/20) * incorrect Σi=1n xi2 = 104

=> incorrect Σi=1n xi2 = 2080

So, correct Σi=1n xi2 = incorrect Σi=1n xi2 – 82 + 122

= 2080 – 64 + 144

= 2160

So, Correct Standard deviation = √[correct Σ xi2/n – (correct mean)2]

= √[2160/20 – (10.2)2]

= √[108 – 104.04]

= √3.96

= 1.98

Question 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?

Standard deviation of Mathematics = 12

Standard deviation of Physics = 15

Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) = (Standard Deviation/Mean) * 100

CV (in Mathematics) = (12/42) * 100 = 28.57

CV (in Physics) = (15/32) * 100 = 46.87

CV (in Chemistry) = (20/40.9) * 100 = 48.89

The subject with greater C.V. is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is

in Mathematics.

Question 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.

Number of observations (n) = 100

Incorrect mean (x) = 20

Incorrect standard deviation (σ) = 3

=> 20 = (1/20) * Σi=1100 xi

=> Σi=1100 xi = 20 * 100 = 2000

So, incorrect sum of observations = 2000

Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

Correct mean = correct sum/(100 - 3) = 1940/97 = 20

Standard deviation σ = √[(1/n) * Σi=1n xi2 – (Σi=1n xi)2 /n2]

=> 3 = √[(1/n) * incorrect Σ xi2 – (x)2]

=> 3 = √[(1/100) * incorrect Σ xi2 – (20)2]

=> 9 = (1/100) * incorrect Σ xi2 – 400

=> incorrect Σ xi2 = 100(9 + 400)

=> incorrect Σ xi2 = 40900

So, correct Σi=1n xi2 = incorrect Σi=1n xi2 – 212 - 212 - 182

= 40900 – 441 – 441 - 324

= 39694

So, Correct Standard deviation = √[correct Σ xi2/n – (correct mean)2]

= √[39694/97 – (20)2]

= √[409.216 – 400]

= √9.216

= 3.036