Measure of Dispersion Exercise: 14.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

Class 11 Basic Mathematics Solution Exercise 14.1

Exercise 14.1

1. a) Find the standard deviation from the following set of observation

20, 25, 30, 36, 32, 43

Solution:

Sum of observation ($\sum x $) = 20+25+30+36+32+43 =186

And, Mean$(\bar x){\rm{ = }}\frac{{\sum x }}{N} = \frac{{186}}{6} = 31$

x	d=x-$\bar x$	d2
20	-11	121
25	-6	36
30	-1	1
36	5	25
32	1	1
43	12	144
$\sum x = 186$		$\sum {{d^2}} = 328$

Thus, Standard Deviation =$\sqrt {\frac{{\sum {{d^2}} }}{N}}$ = $\sqrt {\frac{{328}}{6}} $=3.01

 

b) Daily Expenditure of 6 families are given below:

Rs.240, Rs.180, Rs. 320, Rs.160, Rs.260, Rs.400

Find Standard Deviation.

Solution:

Sum of expenditure ($\sum x $) = 240+180+320+160+260+400 = 1560

 Mean$(\bar x){\rm{ = }}\frac{{\sum x }}{N} = \frac{{1560}}{6} = 260$

x	d=x-$\bar x$	d2
240	-20	400
180	-80	6400
320	60	3600
160	-100	10000
260	0	0
400	140	19600
$\sum x = 1560$		$\sum {{d^2}} = 40000$

Thus, Standard Deviation =$\sqrt {\frac{{\sum {{d^2}} }}{N}}$ = $\sqrt {\frac{{40000}}{6}} $=81.65

2. a) The following table presents the distribution of the bonus for 50 workers.

Bonus	5	10	15	20	25
Frequency	12	20	8	6	4

Find the mean and Standard Deviation.

Solution:

Bonus(x)	Frequency(f)	fx	x - ${\rm{\bar x}}$	(x - ${\rm{\bar x}}$)2	f(x - ${\rm{\bar x}}$)2
5 10 15 20 25	12 20 8 6 4	60 200 120 120 100	-7 -2 3 8 13	49 4 9 64 169	588 80 72 384 676
	N = 50	$\mathop \sum \nolimits^ {\rm{fx}}$ = 600			$\mathop \sum \nolimits^ {\rm{f}}{\left( {{\rm{x}} - {\rm{\bar x}}} \right)^2}$=1,800

 

We have, Mean(${\rm{\bar x}}$)= $\frac{{\mathop \sum \nolimits^ {\rm{fx}}}}{{\rm{N}}}$ = $\frac{{600}}{{50}}$ = 12.

Again, S.D.(σ) = $\sqrt {\frac{{\mathop \sum \nolimits^ {\rm{f}}{{\left( {{\rm{x}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}} $ = $\sqrt {\frac{{1800}}{{50}}} $ = $\sqrt {36} $ = 6.

b) Find the standard deviation from the following data:

X	10	20	30	40	50
f	8	12	15	9	6

Solution: 

Bonus(x)	Frequency(f)	fx	x - ${\rm{\bar x}}$	(x - ${\rm{\bar x}}$)2	f(x - ${\rm{\bar x}}$)2
10 20 30 40 50	8 12 15 9 6	80 240 450 360 300	- 18.6 -8.6 1.4 11.4 21.4	345.96 73.96 1.96 129.96 457.96	2767.68 887.52 29.4 1169.64 2747.76
	N = $\mathop \sum \nolimits^ {\rm{f}}$ = 50	$\mathop \sum \nolimits^ {\rm{fx}}$ = 1430			$\mathop \sum \nolimits^ {\rm{f}}{\left( {{\rm{x}} - {\rm{\bar x}}} \right)^2}$=7602.

We have, Mean(${\rm{\bar x}}$)= $\frac{{\mathop \sum \nolimits^ {\rm{fx}}}}{{\rm{N}}}$ = $\frac{{1430}}{{50}}$ = 28.6

Again, S.D.(σ) = $\sqrt {\frac{{\mathop \sum \nolimits^ {\rm{f}}{{\left( {{\rm{x}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}} $ = $\sqrt {\frac{{7602}}{{50}}} $ = $\sqrt {152.04} $ = 12.23.

C) Find the variance of the following data:

Variable	10	12	15	18	20
Frequency	2	7	10	8	3

Solution:

Let Assume Mean be A=15.

X	f	d=x-A	fd	fd2
10	2	-5	-10	50
12	7	-3	-21	61
15	10	0	0	0
18	8	3	24	72
20	3	5	15	75
	N=30		$\sum {fd}  = 8$	$\sum {f{d^2}}  = 258$

Standard Deviation = $\sigma  = {\rm{ }}\sqrt {\frac{{\sum {f{d^2}} }}{N} - {{\left( {\frac{{\sum {fd} }}{N}} \right)}^2}} $

=$\sigma  {\rm{ }}\sqrt {\frac{{258}}{{30}} - {{\left( {\frac{8}{{30}}} \right)}^2}} $

=$ \sqrt {8.52} $

=2.92

And Variance = ${\sigma ^2} = {2.92^2} = 8.52$

3. a) Find the standard deviation from the following frequency distribution table:

Age	2-4	4-6	6-8	8-10
Frequency	6	6	7	2

Solution:

Age	F	Mid-value(x)	fx	x - ${\rm{\bar x}}$	(x - ${\rm{\bar x}}$)2	f(x - ${\rm{\bar x}}$)2
2 – 4 4 – 6 6 – 8 8 – 10	6 5 7 2	3 5 7 9	18 25 49 18	-2.5 -0.5 1.5 3.5	6.25 0.25 2.25 12.25	37.5 1.25 15.75 24.5
	N = $\mathop \sum \nolimits^ {\rm{f}}$ = 20		$\mathop \sum \nolimits^ {\rm{fx}}$ = 110			$\mathop \sum \nolimits^ {\rm{f}}{\left( {{\rm{x}} - {\rm{\bar x}}} \right)^2}$= 79.

 

So, Mean $({\rm{\bar x}}$) = $\frac{{\mathop \sum \nolimits^ {\rm{fx}}}}{{\rm{N}}}$ = $\frac{{110}}{{20}}$ = 5.5.

Again, S.D. (σ) = $\sqrt {\frac{{\mathop \sum \nolimits^ {\rm{f}}{{\left( {{\rm{x}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{N}}}} $ = $\sqrt {\frac{{79}}{{20}}} $ = $\sqrt {3.95} $ = 1.99

 

b) Calculate the mean, standard deviation and the coefficient of variation from the following frequency distribution table:

Income	300-400	400-500	500-600	600-700	700-800
No. Of Person	8	12	20	6	4

Solution:

Income	F	x	fx	x - ${\rm{\bar x}}$	(x - ${\rm{\bar x}}$)2	f(x - ${\rm{\bar x}}$)2
300 – 400 400 – 500 500 – 600 600 – 700 700 – 800	8 12 20 6 4	350 450 550 650 750	2800 5400 11000 3900 3000	-172 -72 28 128 228	29584 5184 784 16384 51984	236672 62208 15680 98304 207936
	N = $\mathop \sum \nolimits^ {\rm{f}}$ = 50		$\mathop \sum \nolimits^ {\rm{fx}}$ = 26100			$\mathop \sum \nolimits^ {\rm{f}}{\left( {{\rm{x}} - {\rm{\bar x}}} \right)^2}$= 620800.

 

So, Mean $({\rm{\bar x}}$) = $\frac{{\mathop \sum \nolimits^ {\rm{fx}}}}{{\rm{N}}}$ = $\frac{{26100}}{{50}}$ = Rs.522.

Again, S.D. (σ) = $\sqrt {\frac{{\mathop \sum \nolimits^ {\rm{f}}{{\left( {{\rm{x}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{N}}}} $ = $\sqrt {\frac{{620800}}{{50}}} $ = $\sqrt {12416} $ = Rs. 111.43

Coefficient of variation (C.V.) = $\frac{\sigma }{{\rm{x}}}$ * 100% = $\frac{{111.43}}{{522}}$ * 100% = 21.35%.

c) Determine the mean, standard deviation and coefficient of variation from the following frequency data:

Profits	0-10	10-20	20-30	30-40	40-50
No. of Shops	8	13	16	8	5

Solution:

Income	f	x	fx	x - ${\rm{\bar x}}$	(x - ${\rm{\bar x}}$)2	f(x - ${\rm{\bar x}}$)2
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50	8 13 16 8 5	5 15 25 35 45	40 195 400 280 225	-17.8 -7.8 2.2 12.2 22.5	316.84 60.84 4.84 148.84 492.84	2534.72 790.92 77.44 1190.72 2464.2
	N = $\mathop \sum \nolimits^ {\rm{f}}$ = 50		$\mathop \sum \nolimits^ {\rm{fx}}$ = 1140			$\mathop \sum \nolimits^ {\rm{f}}{\left( {{\rm{x}} - {\rm{\bar x}}} \right)^2}$= 7058.

 

So, Mean $({\rm{\bar x}}$) = $\frac{{\mathop \sum \nolimits^ {\rm{fx}}}}{{\rm{N}}}$ = $\frac{{1140}}{{50}}$ = Rs.22.8

Again, S.D. (σ) = $\sqrt {\frac{{\mathop \sum \nolimits^ {\rm{f}}{{\left( {{\rm{x}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{N}}}} $ = $\sqrt {\frac{{7083}}{{50}}} $ = $\sqrt {141.16} $ = Rs. 11.88

Coefficient of variation (C.V.) = $\frac{\sigma }{{\rm{x}}}$ * 100% = $\frac{{11.88}}{{22.8}}$ * 100% = 52.1%.

4. a) The coefficient of variation of two series are 24% and 40% and their respective standard deviaton are 6 and 8, find the respective means.

Solution:

For I^st Series:

C.V =24%

S.D. (σ) = 6

We Know:

$\begin{array}{l}CV = \frac{\sigma }{{\bar x}} \times 100\% \\or,\bar x = \frac{6}{{24}} \times 100\%  = 25\end{array}$

For 2^nd Series:

C.V =40%

S.D. (σ) = 8

We Know:

$\begin{array}{l}CV = \frac{\sigma }{{\bar x}} \times 100\% \\or,\bar x = \frac{8}{{40}} \times 100\%  = 20\end{array}$

b) The coefficient of variation of two series is 45% and 30% and their respective means are 16 and 40, find their respective standard deviations.

Solution:

For I^st Series:

C.V =45%

Mean ${(\bar x)}$ =16

We Know:

$\begin{array}{l}CV = \frac{\sigma }{{\bar x}} \times 100\% \\i.e.\;{\rm{45\%   = }}\frac{\sigma }{{16}} \times 100\% \\or,\,\sigma  = \frac{{16 \times 45\% }}{{100}} = 7.2\end{array}$

For 2^nd Series:

C.V =30%

Mean ${(\bar x)}$ =40

We Know:

$[\begin{array}{l}CV = \frac{\sigma }{{\bar x}} \times 100\% \\i.e.\;{\rm{30\%   = }}\frac{\sigma }{{40}} \times 100\% \\or,\,\sigma  = \frac{{30 \times 40\% }}{{100}} = 12\end{array}$

5. a) Following are the information about the marks of two students A and B

	A	B
Average Marks	84	92
Variance of Marks	16	25

Examine who has got the uniform mark.

Solution:

For student A,

Or, ${\rm{\bar x}}$ = 84, σ² = 16.

So, σ = 4.

So, C.V. = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100% = $\frac{4}{{84}}$ * 100% = 4.76%.

For student B,

Or, ${\rm{\bar x}}$ = 92, σ² = 25.

So, σ = 5.

So, C.V. = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100% = $\frac{5}{{92}}$ * 100% = 5.43%.

Here, C.V. of the student (A) < C.V. of student (B).

So, A has got the uniform mark.

b) From the following information, examine which of the firm A or B has greater variability of distribution of wage.

	Firm A	Firm B
No. of Workers:	25	15
Average monthly wage:	Rs. 6400	Rs. 7500
Standard Deviation wage:	4.5	5.4

Also find the combined mean.

Solution:

For firm A,

n = 25, ${\rm{\bar x}}$ = 6400, σ = 4.5

C.V. = = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100% = $\frac{{4.5}}{{6400}}$ * 100% = 0.070%.

For firm B,

n = 15, ${\rm{\bar x}}$ = 7500, σ = 54

C.V. = = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100% = $\frac{{5.4}}{{7500}}$ * 100% = 0.072%.

Here, C.v. of firm B Is grater than C.V. of form A.

So, the distribution of wages in firm B has greater variability than firm A.

Now, combined mean $\left( {{{{\rm{\bar x}}}_{12}}} \right)$ = $\frac{{{{\rm{n}}_1}.{{{\rm{\bar x}}}_1} + {{\rm{n}}_2}{{{\rm{\bar x}}}_2}}}{{{{\rm{n}}_1} + {{\rm{n}}_2}}}$ = $\frac{{25{\rm{*}}6400 + 15{\rm{*}}7500}}{{25 + 15}}$ = $\frac{{272500}}{{40}}$ = 6812.5

c) The average weekly wage in a factory has increased from Rs.4200  to Rs.4800  and the standard deviation has increased from Rs.5  to Rs.8 . Can you conclude that the wage has increased uniformly?

Solution:

Weekly wage amount = 4800-4200 = Rs. 600

Standard Deviation = 8-5= Rs. 3

Increased Uniformly by $\frac{{600}}{3} = 200$

d)  Examine which variable X, the length (in cm) or Y, the weight (in kg) show the greater variability from the following data:

$\sum x  = 280;\;{\sum x ^2} = 835;\;\sum Y  = 565;\;{\sum Y ^2} = 3640;\,{\rm{n = 10}}$

Solution:

 

6. a) Following are the marks obtained by the students X and Y in 6 tests of 100 marks each:

Test	1	2	3	4	5	6
X	56	72	48	69	64	81
Y	63	74	45	57	82	63

If the consistency of the performance is the criteria for awarding a price who should get the price?

Solution:

Arranging the marks of X and Y in ascending order.

x(X)	d = x – 64	d2	x(Y)	d = y – 63	y + d2
48 56 64 69 72 81	-16 -8 0 5 8 17	256 64 0 25 64 289	45 57 63 63 74 82	-18 -6 0 0 11 19	324 36 0 0 121 361
	$\mathop \sum \nolimits^ {\rm{d}}$ = 6	$\mathop \sum \nolimits^ {{\rm{d}}^2}$ = 689		$\mathop \sum \nolimits^ {\rm{d}}$ = 6	$\mathop \sum \nolimits^ {{\rm{d}}^2}$ = 842.

 

For student X,

Or, ${\rm{\bar x}}$ = a + $\frac{{\mathop \sum \nolimits^ {\rm{d}}}}{{\rm{n}}}$ = 64 + $\frac{6}{6}$ = 65.

S.D.(σ) = $\sqrt {\frac{{\mathop \sum \nolimits^ {{\rm{d}}^2}}}{{\rm{n}}} - {{\left( {\frac{{\mathop \sum \nolimits^ {\rm{d}}}}{{\rm{n}}}} \right)}^2}} $ = $\sqrt {\frac{{689}}{6} - {{\left( {\frac{6}{6}} \right)}^2}} $

= $\sqrt {116.33 - 1} $ = $\sqrt {115.33} $ = 10.74

For student Y,

Or, ${\rm{\bar x}}$ = a + $\frac{{\mathop \sum \nolimits^ {\rm{d}}}}{{\rm{n}}}$ = 63 + $\frac{6}{6}$ = 64.

S.D.(σ) = $\sqrt {\frac{{\mathop \sum \nolimits^ {{\rm{d}}^2}}}{{\rm{n}}} - {{\left( {\frac{{\mathop \sum \nolimits^ {\rm{d}}}}{{\rm{n}}}} \right)}^2}} $ = $\sqrt {\frac{{842}}{6} - {{\left( {\frac{6}{6}} \right)}^2}} $

= $\sqrt {139.33} $ = 11.8.

So, C.V. (X) = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100 % = $\frac{{10.74}}{{65}}$ * 100% = 16.52%.

So, C.V. (Y) = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100 % = $\frac{{11.80}}{{64}}$ * 100 % = 18.42%.

Here, C.V.(X) < C.V.(Y).

So, X should get the prize.

b) The goals scored by two teams X and Y in the football season were as follows:

No of Goals Scored in Match	No. Of Matches
	X	Y
0	27	17
1	9	9
2	8	6
3	5	5
4	4	3

Find out which team is constituent.

Solution:

 

c) Ananda obtained samples of CFL bulbs from two suppliers. He got the samples tested in his laboratory for the lengths of the life. The results of the test are given below which supplier's bulb shows greater variability in the length of the life?

Length of Life	No. of Bulbs
	Supplier A	Supplier B
400-500	8	6
500-600	20	24
600-700	16	12
700-800	6	8

Which supplier’s bulb shows greater variability in the length of the life.

Solution:

Length of life	x	d = x – a	d2	For A(f)	fd	fd2	For B(f)	fd	fd2
400 – 500 500 – 600 600 – 700 700 – 800	450   550   650   750	-100   0   100   200	10000   0   10000   40000	8   20   16   6	-800   0   1600   1200	80000   0   160000   240000	6   24   12   8	-600   0   1200   1600	60000   0   120000 320000
				N = $\mathop \sum \nolimits^ {\rm{f}}$ = 50	$\mathop \sum \nolimits^ {\rm{fd}}$ =20000	$\mathop \sum \nolimits^ {\rm{f}}{{\rm{d}}^2}$ =400000	N = $\mathop \sum \nolimits^ {\rm{f}}$ = 50	$\mathop \sum \nolimits^ {\rm{fd}}$ =2200	$\mathop \sum \nolimits^ {\rm{f}}{{\rm{d}}^2}$ =500000

Now,

For supplier A,

Or, ${\rm{\bar x}}$ = a + $\frac{{\mathop \sum \nolimits^ {\rm{fd}}}}{{\rm{N}}}$ = 550 + $\frac{{2000}}{{50}}$ = 590.

σ(A) = $\sqrt {\frac{{\mathop \sum \nolimits^ {\rm{f}}{{\rm{d}}^2}}}{{\rm{N}}} - {{\left( {\frac{{\mathop \sum \nolimits^ {\rm{fd}}}}{{\rm{N}}}} \right)}^2}} $ = $\sqrt {\frac{{400000}}{{50}} - {{\left( {\frac{{2000}}{{50}}} \right)}^2}} $.

= $\sqrt {8000 - 1600} $ = 80.

So, C.V. (A) = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100% = $\frac{{80}}{{590}}$ * 100% = 15.56%.

For supplier B,

Or, ${\rm{\bar y}}$ = a + $\frac{{\mathop \sum \nolimits^ {\rm{fd}}}}{{\rm{N}}}$ = 550 + $\frac{{2200}}{{50}}$ = 594.

σ(B) = $\sqrt {\frac{{\mathop \sum \nolimits^ {\rm{f}}{{\rm{d}}^2}}}{{\rm{N}}} - {{\left( {\frac{{\mathop \sum \nolimits^ {\rm{fd}}}}{{\rm{N}}}} \right)}^2}} $ = $\sqrt {\frac{{500000}}{{50}} - {{\left( {\frac{{2200}}{{50}}} \right)}^2}} $.

= $\sqrt {10000 - 1936} $ = $\sqrt {8064} $ = 89.8.

So, C.V. (B) = $\frac{\sigma }{{{\rm{\bar x}}}}$ * 100% = $\frac{{89.8}}{{594}}$ * 100% = 15.11%.

Here, C.V. (A) < C.V. (B)

So, supplier B shows greater variability in the lengths of life.

7. a) The average score of 200 students in a class A is 70 with standard deviation 12 and the average score of 300 students of collage B is observed to be 60 with standard deviation 15. What are the combined standard deviation of scores of two colleges combined together?

Solution:

For College A n1=200 $\begin{array}{l}{{\bar x}_1} = 70\\{\sigma _1} = 12\end{array}$

For College B n2=300 $\begin{array}{l}{{\bar x}_2} = 6\\{\sigma _2} = 15\end{array}$

$Combined{\rm{  Mean (}}{{\bar x}_{12}}) = \frac{{{n_1}{{\bar x}_1} + {n_2}{{\bar x}_2}}}{{{n_1} + {n_2}}} = \frac{{200 \times 70 + 300 \times 60}}{{200 + 300}} = 64$

$\begin{array}{l}{\sigma _{12}} = \sqrt {\frac{{{n_1}({\sigma _1} + {d_1}^2) + {n_2}({\sigma _2} + {d_2}^2)}}{{{n_1} + {n_2}}}} \\where,\;{d_1} = {{\bar x}_1} - {{\bar x}_{12}} = 70 - 64 = 6\\{d_2} = {{\bar x}_2} - {{\bar x}_{12}} = 60 - 64 =  - 4\\Thus,\\{\sigma _{12}} = \sqrt {\frac{{200({{12}^2} + {6^2}) + 300({{15}^2} + 16)}}{{200 + 300}}}  = 14.72\end{array}$

b) The monthly wage paid to the workers of the firm A and B belonging to same industry have presented below.

Firm A	Firm B
No.of workers	50	40
Average monthly wages	Rs 63	Rs. 54
Variance of wages	81	36

Determine the combined mean and combined standard deviation of the combined group of 90 workers.

Solution:

Firm A	Firm B
$\begin{array}{l}{n_2} = 50\\{{\bar x}_2} = 63\\{\sigma _2} = 9\end{array}$	$\begin{array}{l}{n_2} = 40\\{{\bar x}_2} = 54\\{\sigma _2} = 6\end{array}$

$Combined{\rm{ Mean (}}{{\bar x}_{12}}) = \frac{{{n_1}{{\bar x}_1} + {n_2}{{\bar x}_2}}}{{{n_1} + {n_2}}} = \frac{{50 \times 63 + 40 \times 54}}{{50 + 40}} = 59$

$\begin{array}{l}{\sigma _{12}} = \sqrt {\frac{{{n_1}({\sigma _1} + {d_1}^2) + {n_2}({\sigma _2} + {d_2}^2)}}{{{n_1} + {n_2}}}} \\where,\;{d_1} = {{\bar x}_1} - {{\bar x}_{12}} = 63 - 59 = 4\\{d_2} = {{\bar x}_2} - {{\bar x}_{12}} = 54 - 59 =  - 5\\Thus,\\{\sigma _{12}} = \sqrt {\frac{{50({9^2} + {4^2}) + 40({6^2} + {5^2})}}{{50 + 40}}}  = 9\end{array}$

8.a) The arithmetic mean and the standard deviation of a series of 20 items as calculated, by a student were 20 cm and 5 cm respectively. But while calculating an item 13 were misread as 30. Find the correct mean and standard deviation.

Solution:

 Given:

n = 20, ${\bar x}$ = 20, σ_x = 5

Correct value = 13, wrong value = 30

$\begin{array}{l}\bar x = \frac{{\sum {{x_i}} }}{n}\\i.e.20 = \frac{{\sum {{x_i}} }}{{20}}\\or,\sum {{x_i}}  = 400\end{array}$

Corrected ∑x_i = 400 – (wrong value) + (correct value)

= 400 – 30 + 13

= 383

Corrected Mean = $\frac{{Corrected{\rm{ }}\sum {{x_i}} }}{n} = \frac{{383}}{{20}} = 19.15$

$\begin{array}{l}{\sigma _x}^2 = \frac{{\sum {{x_i}^2} }}{n} - {{\bar x}^2}\\\therefore 25 = \frac{{\sum {{x_i}^2} }}{{20}} - {20^2}\\\therefore 425 = \frac{{\sum {{x_i}^2} }}{{20}}\\\therefore \sum {{x_i}^2}  = 8500\end{array}$

Corrected ∑x_i² = 8500 – (wrong value)² + (correct value)²

= 8500 – 900 + 169

= 7769

Corrected variance = $\begin{array}{l} = \frac{{corrected\sum {{x_i}^2} }}{n} - {(Corrected{\rm{ }}Mean)^2}\\ = \frac{{7769}}{{20}} - {(19.5)^2}\\ = 21.73\end{array}$

∴ Corrected S.D. = $\sqrt {21.73} $ = 4.66

Hence, the corrected mean and S.D. are 19.15 cm and 4.66 cm respectively.

b) The arithmetic mean and standard deviation of 21 observations are 40 and 8 respectively. At the time of calculation, if one item 35 is wrongly recorded. Find the new standard deviation of the items when the wrongly recorded item is omitted.

Solution:

Number of observations (n)=21
Incorrect mean =40
Incorrect standard deviation =8

\[\bar x = \frac{1}{n}\sum\limits_{i = 1}^{21} {{x_i}} \]

\[or,\;{\rm{40  = }}\frac{1}{{21}}\sum\limits_{i = 1}^{21} {{x_i}} \]

\[or,\sum\limits_{i = 1}^{21} {{x_i}}  = 840\]

So, the incorrect sum of observations =840
Correct sum of observation =840−35=805

Correct Mean = $\frac{{Correct{\rm{ Sum }}}}{{21 - 1}} = \frac{{805}}{{20}} = 40.25$ [since one is omitted.]