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MATHS 5 Marks Questions

Statistics · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 11 questions.

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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
$\therefore$ Incorrect $\Sigma$xi = 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.
Correct $\Sigma {x_i}$ = Incorrect $\Sigma {x_i}$ - 21 - 21 - 18
= 2000 - 21 - 21 - 18 = 1940
$\therefore$ Correct mean $= \frac{{1940}}{{97}}$= 20
Also correct $\Sigma x_i^2$ = Incorrect $\Sigma x_i^2$ - (21)2 - (21)2 - (18)2
= 40900 - 441 - 441 - 324 = 39694
$\therefore$ Correct variance =$\frac{1}{{97}}$ (correct $\Sigma x_i^2$) - (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 x1, x2, ..... xn Prove that the mean and variance of the observation ax1, ax2, .... axn are $a$$\bar x$ and a2${\sigma ^2}$ respectively $(a \ne 0)$
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 x1, x2, x3, x4, x5, x6 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 $\frac{1}{7}$ (22 + 42 + 102 + 122 + 142 + x2 + y2) - (8)2 = 16
$\Rightarrow \frac{1}{7}$(4 + 16 + 100 + 144 + 196 + x2 + y2) - 64 = 16
$\Rightarrow$ 460 + x2 + y2 = 560 $\Rightarrow$ x2 + y2 = 100 .... (ii)
Now (x + y)2 + (x - y)2 = 2(x2 + y2)
$\Rightarrow$  (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 ...(i)
Also $\frac{1}{8}$(62 + 72 + 102 + 122 + 122 + 132 + x2 + y2) - (9)2 = 9.25
$\Rightarrow \frac{1}{8}$(36 + 49 + 100 + 144 + 144 + 169 + x2 + y2) - 81 = 9.25
$\Rightarrow$ 642 + x2 + y2 = 722
$\Rightarrow$ x2 + y2 = 80. . . (ii)
Now (x + y)2 + (x - y)2 = 2(x2 + y2)
$\Rightarrow$ (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 xi fi $u = \frac{{x - 92.5}}{5}$ fu fu2
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 xi fi $u = \frac{{x - 25}}{{10}}$ fu fu2
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 xi fi $u = \frac{{x - 105}}{{30}}$ fu fu2
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.
xi 60 61 62 63 64 65 66 67 68
fi 2 1 12 29 25 12 10 4 5
Answer
xi fi u = x - 64 fu fu2
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 xi fi $u = \frac{{x - 42.5}}{4}$ fu fu2
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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