5 Marks Questions

Question 15 Marks

The mean and standard deviation of a group of $100$ observation 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.

Answer

Here $n=100, \bar{x}=20$ and $\sigma=3$
$\therefore \bar{x}=\frac{1}{n} \Sigma x_i \Rightarrow \Sigma x_i=n \times \bar{x}=100 \times 20=2000$
$\therefore \text { Incorrect } \Sigma x_{i}=2000$
Now $\frac{1}{n} \Sigma x_i^2-(\bar{x})=\sigma^2$
$\Rightarrow \frac{1}{100} \Sigma x_i^2-(20)^2=9 \Rightarrow \Sigma x_i^2=40900$
When wrong items 21,21 and 18 are omitted from the data, we have 97 observations.
$\text { Correct } \Sigma x_i=\text { Incorrect } \Sigma x_i-21-21-18$
$=2000-21-21-18=1940$
$\therefore \text { Correct mean }=\frac{1940}{97}=20$
Also correct $\Sigma x_i^2=\operatorname{lncorrect} \Sigma x_i^2-(21)^2-(21)^2-(18)^2$
$=40900-441-441-324=39694$
$\therefore$ Correct variance $=\frac{1}{97}$ (correct $\left.\Sigma x_i^2\right)-(\text { correct mean })^2$
$=\frac{1}{97} \times 39694-(20)^2$
$=409.22-400=9.22$
Correct S.D. $=\sqrt{9.22}=3.036$

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Question 25 Marks

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 these three subjects shows the highest variability in marks and which shows the lowest?

Answer

Given, n = 50
For Mathematics
$\overline x$ = 42 and $\sigma$ = 12
Coefficient of variation (CV)= $\frac{\sigma}{\overline{x}} \times$ 100 = $\frac{12}{42}$ $\times$ 100
= $\frac{2}{7}$ $\times$ 100 = $\frac{200}{7}$ = 28.57 ...(i)
For Physics
$\overline x$ = 32 and $\sigma$ = 15
Coefficient of variation (CV) = $\frac{\sigma}{\overline{x}} \times$ 100 $=\frac{15}{32}$ $\times$ 100
= $\frac{1500}{32}$ = 46.87 ....(ii)
For Chemistry
$\overline x$ = 40.9 and $\sigma$ = 20
Coefficient of variation (CV) = $\frac{\sigma}{\overline{x}} \times$ 100 $=\frac{20}{40.9}$ $\times$ 100
= $\frac{2000}{40.9}$ = 48.89 ....(iii)
From equations (i), (ii) and (iii),
$\Rightarrow$ CV of Chemistry > CV of Physics > CV of Mathematics
$\therefore$ Chemistry shows the highest variability while Mathematics shows the least variability.

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Question 35 Marks

Given that $\bar{x}$ is the mean and $\sigma^2$ is the variance of n observations $x_1, x_2, \ldots x_n$ Prove that the mean and variance of the observation $ax _1, ax _2, \ldots . ax _{ n }$ are $a \bar{x}$ and $a ^2 \sigma^2$ respectively $(a \neq 0)$

Answer

Here $\bar x = \frac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n} = \frac{{\Sigma x}}{n}$
Also $\frac{{x_1^2 + x_2^2 + x_3^2 + ... + x_n^2}}{n} = \frac{{\Sigma {x^2}}}{n}$
New mean $= \frac{{a{x_1} + a{x_2} + a{x_3} + ... + a{x_n}}}{n} = \frac{{a\left( {{x_1} + {x_2} + {x_3} + ... + {x_n}} \right)}}{n} = a\bar x$
Also ${\sigma ^2} = \frac{{n(x_1^2 + x_2^2 + x_3^2 + ... + x_n^2) - {{({x_1} + {x_2} + {x_3} + ...+{x_n})}^2}}}{{{n^2}}}$
New variance $ = \frac{{n({a^2}x_1^2 + {a^2}x_2^2 + {a^2}x_3^2 + ... + {a^2}x_n^2) - {{(a{x_1} + a{x_2} + a{x_3} + ... + a{x_n})}^2}}}{{{n^2}}}$
$ = {a^2}\left[ {\frac{{n(x_1^2 + x_2^2 + x_3^2 + ... + x_n^2) - {{({x_1} + {x_2} + {x_3} + ... + {x_n})}^2}}}{{{n^2}}}} \right]$
$= {a^2}{\sigma ^2}$

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Question 45 Marks

The mean and standard deviation of six observation are $8$ and $4$ respectively. If each observation is multiplied by $3$, find the new mean and new standard deviation of the resulting observations.

Answer

Let $x_{1,} x_{2,} x_{3,} x_{4,} x_{5,} x_6$ be six observations, then
$\frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8$
$\Rightarrow x_1+x_2+x_3+x_4+x_5+x_6=48$
Now if each observation is multiplied by 3 then
New mean $= \frac{{3{x_1} + 3{x_2} + 3{x_3} + 3{x_4} + 3{x_5} + 3{x_6}}}{6}$
$ = \frac{{3\left( {{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}} \right)}}{6} = \frac{1}{2} \times 48$ = 24
Also $\frac{1}{6}\left( {x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2} \right) - {(8)^2} = 16$
$ \Rightarrow \;x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 = 480$
If each observation multiplied by 3 then
New variance $ = \frac{1}{6}\left( {9x_1^2 + 9x_2^2 + 9x_3^2 + 9x_4^2 + 9x_5^2 + 9x_6^2} \right) - {(24)^2}$
$= \frac{9}{6} \times 480 - 576 = 720 - 576 = 144$
$\therefore$ New S.D. $= \sqrt {144}$ = 12

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Question 55 Marks

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.

Answer

Let two remaining observations be x and y. Then
$\frac{{2 + 4 + 10 + 12 + 14 + x + y}}{7} = 8$
$\therefore 42+x+y=56$
$\Rightarrow x+y=14$
Also $\frac{1}{7}\left(2^2+4^2+10^2+12^2+14^2+x^2+y^2\right)-(8)^2=16$
$\Rightarrow \frac{1}{7}\left(4+16+100+144+196+x^2+y^2\right)-64=16$
$\Rightarrow 460+x^2+y^2=560$
$\Rightarrow x^2+y^2=100 \ldots \text { (ii) }$
$\operatorname{Now}(x+y)^2+(x-y)^2=2\left(x^2+y^2\right)$
$\Rightarrow(14)^2+(x-y)^2=2 \times 100$
$\Rightarrow(x-y)^2=200-196$
$\Rightarrow(x-y)^2=4$
$\Rightarrow x-y= \pm 2$
When $x-y=2$
Solving $x+y=14$ and $x-y=2$ we get $x=8$ and $y=6$
When $x-y=-2$
Solving $x+y=14$ and $x-y=-2$ we get $x=6$ and $y=8$

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Question 65 Marks

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.

Answer

Let two remaining observations be $x$ and $y$. Then
$\frac{6+7+10+12+12+13+x+y}{8}=9$
$\therefore 60+x+y=72$
$\Rightarrow x+y=12 \ldots \text { (i) }$
Also $\frac{1}{8}\left(6^2+7^2+10^2+12^2+12^2+13^2+x^2+y^2\right)-(9)^2=9.25$
$\Rightarrow \frac{1}{8}\left(36+49+100+144+144+169+x^2+y^2\right)-81=9.25$
$\Rightarrow 642+x^2+y^2=722$
$\Rightarrow x^2+y^2=80 \ldots \text { (ii) }$
Now $(x+y)^2+(x-y)^2=2\left(x^2+y^2\right)$
$\Rightarrow(12)^2+(x-y)^2=2 \times 80$
$\Rightarrow(x-y)^2=160-144$
$\Rightarrow(x-y)^2=16$
$\Rightarrow x-y= \pm 4$
When $x-y=4$
Solving $x+y=12$ and $x-y=4$ we get $x=8$ and $y=4$
When $x-y=-4$
Solving $x+y=12$ and $x-y=-4$ we get $x=4$ and $y=8$

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Question 75 Marks

Find the mean, variance and standard deviation using short cut method.

Height in cm	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

Answer

Height in cms.	Mid values $x_i$	$f_i$	$u = \frac{{x - 92.5}}{5}$	$fu$	$fu^2$
70-75	72.5	3	- 4	- 12	48
75-80	77.5	4	- 3	- 12	36
80-85	82.5	7	- 2	- 14	28
85-90	87.5	7	- 1	- 7	7
90-95	92.5	15	0	0	0
95-100	97.5	9	1	9	9
100-105	102.5	6	2	12	24
105-110	107.5	6	3	18	54
110-115	112.5	3	4	12	48
		60		6	254

Mean $(\bar x) = A + \frac{{\Sigma fu}}{N} \times h = 92.5 + \frac{6}{{60}} \times 5$ = 92.5 + 0.5 = 93
Variance $({\sigma ^2}) = \frac{{{h^2}}}{{{N^2}}}[N\Sigma f{u^2} - {(\Sigma fu)^2}]$
$= \frac{{{{(5)}^2}}}{{{{(60)}^2}}}[60 \times 254 - {(6)^2}]$
$ = \frac{{25}}{{3600}}[15240 - 36] = \frac{{25}}{{3600}} \times 15204 = 105.58$
Standard deviation $(\sigma ) = \sqrt {105.58} = 10.27$

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Question 85 Marks

Find the mean and variance for the following frequency distribution

Classes	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequencies	$5$	$8$	$15$	$16$	$6$

Answer

Classes	Mid values $x_i$	$f_i$	$u = \frac{{x - 25}}{{10}}$	$fu$	$fu^2$
$0-10$	$5$	$5$	$- 2$	$- 10$	$20$
$10-20$	$15$	$8$	$- 1$	$- 8$	$8$
$20-30$	$25$	$15$	$0$	$0$	$0$
$30-40$	$35$	$16$	$1$	$16$	$16$
$40-50$	$45$	$6$	$2$	$12$	$24$
		$50$		$10$	$68$

Mean $(\bar x) = A + \frac{{\Sigma fu}}{N} \times h$
$ = 25 + \frac{{10}}{{50}} \times 10$ = 25 + 2 = 27
Variance $({\sigma ^2}) = \frac{{{h^2}}}{{{N^2}}}[N\Sigma f{u^2} - {(\Sigma fu)^2}]$
$ = \frac{{{{(10)}^2}}}{{{{(50)}^2}}}[50 \times 68 - {(10)^2}]$
$ = \frac{{100}}{{2500}}[3400 - 100]$
$ = \frac{1}{{25}} \times 3300 = 132$

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Question 95 Marks

Find the mean and variance for the following frequency distribution

Class	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequencies	2	3	5	10	3	5	2

Answer

Classes	Mid values $x_i$	$f_i$	$u = \frac{{x - 105}}{{30}}$	$fu$	$fu^2$
0-30	15	2	- 3	- 6	18
30-60	45	3	- 2	- 6	12
60-90	75	5	- 1	- 5	5
90-120	105	10	0	0	0
120-150	135	3	1	3	3
150-180	165	5	2	10	20
180-210	195	2	3	6	18
		30		2	76

Mean $(\bar x) = A + \frac{{\sum {fu} }}{N} \times h = 105 + \frac{2}{{30}} \times 30 = 107$
Variance $({\sigma ^2}) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\Sigma f{u^2} - {{(\Sigma fu)}^2}} \right]$
=900/900[30(76)-4]=2276

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Question 105 Marks

Find the mean and standard deviation using short cut method.

$x_i$	60	61	62	63	64	65	66	67	68
$f_i$	2	1	12	29	25	12	10	4	5

Answer

$x_i$	$f_i$	$u = x - 64$	$fu$	$fu^2$
60	2	- 4	- 8	32
61	1	- 3	- 3	9
62	12	- 2	- 24	48
63	29	- 1	- 29	29
64	25	0	0	0
65	12	1	12	12
66	10	2	20	40
67	4	3	12	36
68	5	4	20	80
	100		0	286

Mean $(\bar x) = A + \frac{{\sum {fu} }}{N} = 64 + \frac{0}{{100}} = 64$
S.D. $(\sigma ) = \frac{1}{{100}}\sqrt {N\sum {f{u^2}} - {{\left( {\sum {fu} } \right)}^2}}$$ = \frac{1}{{100}}\sqrt {100 \times 286 - {{(0)}^2}} $
$= \frac{1}{{100}}\sqrt {28600} = \frac{1}{{100}} \times 169.1 = 1.69.$

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Question 115 Marks

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.]

Answer

Diameters	Classes	Mid vales $x_i$	$f_i$	$u = \frac{{x - 42.5}}{4}$	$fu$	$fu^2$
33-36	32.5-36.5	34.5	15	- 2	- 30	60
37-40	36.5-40.5	38.5	17	- 1	- 17	17
41-44	40.5-44.5	42.5	21	0	0	0
45-48	44.5-52.5	46.5	22	1	22	22
49-52	48.5-52.5	50.5	25	2	50	100
			100		25	199

Mean $(\bar x) = A + \frac{{\Sigma fu}}{N} \times h$
$ = 42.5 + \frac{{25}}{{100}} \times 4$
= 42.5 + 1 = 43.5 mm
Standard deviation $(\sigma ) = \frac{h}{N}\sqrt {N\Sigma f{u^2} - {{(\Sigma fu)}^2}}$
$ = \frac{4}{{100}}\sqrt {100 \times 199 - {{(25)}^2}} = \frac{1}{{25}}\sqrt {19275}$
$ = \frac{1}{{25}} \times 138.83$ = 5.55

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