Download App

MATHS (Each question 4 marks)

STATISTICS · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 18 questions.

Questions

(Each question 4 marks)

Take a timed test

18 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 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 $\Sigma x_i$ = 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.
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
View full question & answer→
Question 24 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.
View full question & answer→
Question 34 Marks
The mean and standard deviation of 20 observation 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 cases of it is replaced by 12 .
Answer
Here we are given that, n = 20, $\bar x = 10$ and $\sigma = 2$
$\therefore \;\bar x = \frac{1}{n}\Sigma {x_i} \Rightarrow n \times \bar x = \Sigma {x_i }\\ \Rightarrow\Sigma {x_i}=20\times10=200$
Therefore Incorrect $\Sigma {x_i} = 200$
Now $\frac{1}{n}\Sigma x_i^2 - {(\bar x)^2} =\sigma^2$
$\Rightarrow \frac{1}{{20}}\Sigma x_i^2 - {(10)^2} = 4 \Rightarrow \Sigma x_i^2 = 2080$
If it is replaced by 12,
When wrong item 8 is replaced by 12
Therefore, Correct $\Sigma {x_i}$ = Incorrect $\Sigma {x_i}$ - 8 + 12
= 200 - 8 + 12 = 204
$\therefore$ Correct mean = $\frac{{204}}{{20}}$ = 10.2
Also correct $\Sigma x_i^2$ = Incorrect $\Sigma x_i^2 - (8)^2 + (12)^2$
= 2080 - 64 + 144 = 2160
$\therefore$ Correct variance $= \frac{1}{{20}}(correct\;\Sigma x_1^2)$ - (correct mean)$^2$
$ = \frac{{2160}}{{20}} - {\left( {\frac{{204}}{{20}}} \right)^2}$
$= \frac{{2160}}{{20}} - \frac{{41616}}{{400}} = \frac{{43200 - 41616}}{{400}} = \frac{{1584}}{{400}}$
Correct S.D. $= \sqrt {\frac{{1584}}{{400}}} = \sqrt {3.96}$ = 1.989
View full question & answer→
Question 44 Marks
The mean and standard deviation of 20 observation is found to be 10 and 2 respectively. On rechecking, it was found that observation 8 was incorrect. Calculate the correct mean and standard deviation in cases of the wrong items is omitted.
Answer
Here n = 20, $\bar x = 10$ and $\sigma = 2$
$\therefore \;\bar x = \frac{1}{n}\Sigma {x_i} \Rightarrow n \times \bar x = \Sigma {x_i }\\ \Rightarrow\Sigma {x_i}=20\times10=200$
​​Therefore Incorrect $\Sigma {x_i} = 200$
Now $\frac{1}{n}\Sigma x_i^2 - {(\bar x)^2} =\sigma^2$
$\Rightarrow \frac{1}{{20}}\Sigma x_i^2 - {(10)^2} = 4 \Rightarrow \Sigma x_i^2 = 2080$If wrong item is omitted.
When wrong item 8 is omitted from the data then we have 19 observations.
​​Therefore Correct $\Sigma {x_i}$ = Incorrect $\Sigma {x_i} - 8$
Correct $\Sigma {x_i} = 200 - 8 = 192$
​​Therefore Correct mean = $\frac{{192}}{{19}}$ = 10.1
Also correct $ \Sigma x_i^2$ = Incorrect $ \Sigma x_i^2 - {(8)^2}$
$\Rightarrow$ Correct $\Sigma x_i^2 = 2080 - 64$ = 2016
​​Hence Correct variance $ = \frac{1}{{19}}\left( {correct\;\Sigma x_i^2} \right)$ - (correct mean)$^2$
$= \frac{1}{{19}} \times 2016 - {\left( {\frac{{192}}{{19}}} \right)^2}$
$= \frac{{2016}}{{19}} - \frac{{36864}}{{361}} = \frac{{38304 - 36884}}{{361}} = \frac{{1440}}{{361}}$
Correct S.D. $ = \sqrt {\frac{{1440}}{{361}}} = \sqrt {3.99}$ = 1.997
View full question & answer→
Question 54 Marks
Given that $\bar x $ is the mean and ${\sigma ^2}$ is the variance of n observations $x_1, x_2, ..... x_n$ Prove that the mean and variance of the observation $ax_1, ax_2, .... ax_n$ are $a$$\bar x$ and $a^2{\sigma ^2}$ 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]$
$= {a^2}{\sigma ^2}$
View full question & answer→
Question 64 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
View full question & answer→
Question 74 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
$\frac{{2 + 4 + 10 + 12 + 14 + x + y}}{7} = 8$
$\therefore 42+x+y=56 \Rightarrow x+y=14$
$\text { Also } \frac{1}{7}\left(2^2+4^2+10^2+12^2+14^2+x^2+y^2\right)-(8)^2=16$
$\Rightarrow \frac{1}{7}\left(4+16+100+144+196+x^2+y^2\right)-64=16$
$\Rightarrow 460+x^2+y^2=560 \Rightarrow x^2+y^2=100 \ldots \text {..(ii) }$
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 $y=8$
View full question & answer→
Question 84 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
$\therefore 60+x+y=72 \Rightarrow x+y=12 \ldots \text { (i) }$
$\text { Also } \frac{1}{8}\left(6^2+7^2+10^2+12^2+12^2+13^2+x^2+y^2\right)-(9)^2=9.25$
$\Rightarrow \frac{1}{8}\left(36+49+100+144+144+169+x^2+y^2\right)-81=9.25$
$\Rightarrow 642+x^2+y^2=722$
$\Rightarrow x^2+y^2=80 \ldots \text { (ii) }$
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 $y=8$
View full question & answer→
Question 94 Marks
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
$\sum\limits_{i = 1}^{50} {{x_i}}$ = 212, $\sum\limits_{i = 1}^{50} {x_i^2}$ = 902.8, $\sum\limits_{i = 1}^{50} {{y_i}}$ = 261, $\sum\limits_{i = 1}^{50} {y_i^2}$ = 1457.6
Which is more varying, the length or weight?
Answer
Here $\sum\limits_{i = 1}^{50} {{x_i}}$ = 212, $\sum\limits_{i = 1}^{50} {x_i^2}$ = 902.8, $\sum\limits_{i = 1}^{50} {{y_i}}$ = 261, $\sum\limits_{i = 1}^{50} {y_i^2}$ = 1457.6
Now, $\bar x = \frac{{212}}{{50}}$ = 4.24
$\sigma _x^2 = \frac{1}{{50}} \times 902.8 - {\left( {\frac{{212}}{{50}}} \right)^2}$ = 18.056 - 17.978 = 0.078
${\sigma _x} = \sqrt {0.078}$ = 0.28
Also $\bar y = \frac{{261}}{{50}}$ = 5.22
$\sigma _y^2 = \frac{1}{{50}} \times 1457.6 - {\left( {\frac{{261}}{{50}}} \right)^2}$ = 29.152 - 27.248 = 1.904
${\sigma _y} = \sqrt {1.904}$ = 1.38
C.V. of length = $ \frac{{0.28}}{{4.24}} \times$ 100 = 6.6
C.V. of weight = $\frac{{1.38}}{{5.22}} \times$ 100 = 26.45
C.V. of weight > C.V. of length
Thus weights have more variability than lengths.
View full question & answer→
Question 104 Marks
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 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?
Answer
Team A
x f $f_ix_i$ $f_ix_i{}^2$
0 1 0 0
1 9 9 9
2 7 14 28
3 5 15 45
4 3 12 48
  25 50 130
Mean $({\bar x}) = \frac{1}{N}\Sigma fx = \frac{1}{{25}} \times$ 50 = 2
S.D. $({\sigma _A}) = \sqrt {\frac{{\Sigma f{x^2}}}{N} - {{\left( {\frac{{\Sigma fx}}{N}} \right)}^2}}$
= $\sqrt {\frac{{130}}{{25}}-{{\left( {\frac{{50}}{{25}}} \right)}^2}} = \sqrt {5.2 - 4} = \sqrt {1.2}$ = 1.09
C.V. of team A = $\frac{{1.09}}{2} \times$ 100 = 54.5
C.V. of team B = $\frac{{1.25}}{2} \times$ 100 = 62.5
Since C.V. of team A < C.V. of team B
Thus team A is more consistent.
View full question & answer→
Question 114 Marks
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 wage earners 586 648
Mean of monthly wages ₹ 5253 ₹ 5253
Variance of the distribution of wages. 100 121
  1. which firm A or B pays larger amount as monthly wages?
  2. which firm A or B shows greater variability in individual wages?
Answer
  1. Firm A:
    Number of wage earners ($n_1$) = 586
    Mean of monthly wages $({\bar x_1})$ = ₹ 5253
    $\therefore$ Total monthly wages = 5253 $\times$ 586 = 3078258
    Firm B:
    Number of wage earners ($n_2$) = 648
    Mean of monthly wages $({\bar x_2})$ = ₹ 5253
    $\therefore$ Total monthly wages = 5253 $\times$ 648 = ₹ 3403944
  2. Since both the firms have same mean of monthly wages, so the firm with greater variance will have more variability in individual wages. Thus firm B will have more variability in individual wages.
View full question & answer→
Question 124 Marks
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
Answer
X Y $(X - \bar X)$ $(Y - \bar Y)$ ${(X - \bar X)^2}$ ${(Y - \bar Y)^2}$
35 108 - 16 3 256 9
54 107 3 2 9 4
52 105 1 0 1 0
53 105 2 0 4 0
56 106 5 1 25 1
58 107 7 2 49 4
52 104 1 - 1 1 1
50 103 - 1 - 2 1 4
51 104 0 - 1 0 1
49 101 - 2 - 4 4 16
510 1050 350 40

$\bar x = \frac{{510}}{{10}}$ = 5; $\bar y = \frac{{1050}}{{10}}$ = 105
${\sigma _x} = \sqrt {\frac{{\Sigma {{(x - \bar x)}^2}}}{n}} = \sqrt {\frac{{350}}{{10}}} $ = 5.92
${\sigma _y} = \sqrt {\frac{{\Sigma {{(y - \bar y)}^2}}}{n}} = \sqrt {\frac{{40}}{{10}}}$ = 2
C.V. of x = $\frac{{5.92}}{{51}} \times $ 100 = 11.61
C.V. of y = $\frac{2}{{105}} \times$ 100 = 1.9
C.V. of Y < C.V. of X
Thus prices of share Y are more stable.

View full question & answer→
Question 134 Marks
From the following data, 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
Answer
For group A,
Let assumed mean a = 45
Class interval Mid-point $(x_i)$ $u_i$ = $\frac{x_i - 45}{10}$ $u_i^2$ $f_i$ $f_iu_i$ $f_iu_i^2$
10 - 20 15 -3 9 9 -27 81
20 - 30 25 -2 4 17 -34 68
30 - 40 35 -1 1 32 -32 32
40 - 50 45 0 0 33 0 0
50 - 60 55 1 1 40 40 40
60 - 70 65 2 4 10 20 40
70 - 80 75 3 9 9 27 81
Total       150 -6 342
Here,
$\sum f_i$ = N = 150, $\sum f_{i} u_{i}$ = -6, and $\sum f_{i} u_{i}^{2}$ = 342, b = 10
$\therefore$ $\overline x$ = a + $\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times$ b = 45 + $\frac{(-6)}{150}$ $\times$ 10 = 45 - 0.4 = 44.6
Variance, $\sigma_{A}=\frac{b^{2}}{N^{2}}\left[N \sum f_{i} u_{i}^{2}-\left(\sum f_{i} u_{i}\right)^{2}\right]$
= $\frac{100}{22500}$[150 $\times 342 - (-6)^2$]
= $\frac{1}{225}$ (51300 - 36) = $\frac{51264}{225}$ = 227.84
$\therefore$ Standard deviation, $\sigma _A$ = $\sqrt{227.84}$ = 15.09
Coefficient of variation (CV) = $\frac{\sigma_{{A}}}{\overline{x}} \times$ 100 = $\frac{15.09}{44.6}$ $\times$ 100 = 33.83
For group B,
Let the assumed mean a = 45
Class interval Mid-point $(x_i)$ $u_i$ = $\frac{x_i - 45}{10}$ $f_i$ $u_i^2$ $f_iu_i$ $f_iu_i^2$
10 - 20 15 -3 10 9 -30 90
20 - 30 25 -2 20 4 -40 80
30 - 40 35 -1 30 1 -30 30
40 - 50 45 0 25 0 0 0
50 - 60 55 1 43 1 43 43
60 - 70 65 2 15 4 30 60
70 - 80 75 3 7 9 21 63
Total     150   -6 366
$\sum F_i$ = N = 150,
$\sum f_{i} u_{i}$ = -6 and $\sum f_{i} u_{i}^{2}$ = 366
$\therefore$ $\overline{x}=a+\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times b$
= 45 + $\frac{(-6)}{150}$ $\times$ 10 = 45 - 0.4 = 44.6
Variance, $\sigma_{B}^{2}=\frac{b^{2}}{N^{2}}\left[N \sum f_{i} u_{i}^{2}-\left(\sum f_{i} u_{i}\right)^{2}\right]$
= $\frac{100}{22500}$ [150 $\times 366 - (-6)^2$]
= $\frac{1}{225}$(54900 - 36) = $\frac{1}{225}$ $\times$ 54864 = 243.84
$\therefore$ Standard deviation, $\sigma _ B$ = $\sqrt{243.84}$ = 15.61
$\therefore$ Coefficient of variation (CV) = $\frac{\sigma_{B}}{\overline{x}} \times$ 100 = $\frac{15.61}{44.6}$ $\times$ 100 = 35
Since, CV(B) > CV(A)
So, group B is more variable than group A.
View full question & answer→
Question 144 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 $(\sigma ) = \sqrt {105.58} = 10.27$
View full question & answer→
Question 154 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$
View full question & answer→
Question 164 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
View full question & answer→
Question 174 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}} $
$= \frac{1}{{100}}\sqrt {28600} = \frac{1}{{100}} \times 169.1 = 1.69.$
View full question & answer→
Question 184 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
View full question & answer→
Generate Paper FreeAll STATISTICS topics