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STATISTICS · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 36 questions.

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Question 13 Marks
Find the mean and variance of the following data.
$x_i$ 92 93 97 98 102 104 109
$f_i$ 3 2 3 2 6 3 3
Answer
$x_i$ $f_i$ $f_ix_i$ $(x_i-100)$ $(x_i-100)^2$ $f_i(x_i-100)^2$
92 3 276 - 8 64 192
93 2 186 - 7 49 98
97 3 291 - 3 9 27
98 2 196 - 2 4 8
102 6 612 2 4 24
104 3 312 4 16 48
109 3 327 9 81 243
  22 2200     640
Mean $(\bar x) = \frac{1}{N}\sum {{f_i}{x_i}} = \frac{1}{{22}} \times 2200 = 100$
Variance $({\sigma ^2}) = \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{{({x_i} - \bar x)}^2}} = \frac{1}{{22}} \times 640 = 29.09$
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Question 23 Marks
Find the mean and variance for each of the data
$x_i$ 6 10 14 18 24 28 30
$f_i$ 2 4 7 12 8 4 3
Answer
1st of all we construct table.
$x_i$ $f_i$ $f_ix_i$ $X_i - \bar X$ $(X_i - \bar X)^2$ $f_i (X_i - \bar X)^2$
6 2 12 -13 169 338
10 4 40 -9 81 324
14 7 98 -5 25 175
18 12 216 -1 1 12
24 8 192 5 25 200
28 4 112 9 81 324
30 3 90 11 121 363
  40 760     1736
Here, N = 40, $\sum\limits_{i=1}^{7} f_i x_i = 760$
$\therefore \bar x = \frac {\sum\limits_{i=1}^{7} f_i x_i}{N} = \frac {760}{40} = 19$
Variance = $ \frac 1N \sum\limits_{i=1}^{7} f_i(x_i - \bar x)^2 = \frac {1}{40} \times 1736 = 43.4$
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Question 33 Marks
Find the mean and variance for: First $10$ multiples of $3$
Answer
The First $10$ multiples of $3$ are given by,
$3, 6, 9, 12, 15, 18, 21, 24, 27, 30$
We know that Mean, $\overline{\mathrm{x}}=\frac{\sum_{i=1}^{\mathrm{a}} \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$
$\therefore$ $\overline{\mathrm{x}}=\frac{3+6+9+12+15+18+21+24+27+30}{10}$ = $\frac{165}{10}$ = 16.5
From the given data, we can form the table:
$x_i$ Deviation from mean (xi - $\overline{\mathbf{X}}$) (xi - $\overline{\mathbf{X}}$)$^2$
3 3 - 16.5 = 13.5 182.25
6 6 - 16.5 = 10.5 110.25
9 9 - 16.5 = 7.5 56.25
12 12 - 16.5 = -4.5 20.25
15 15 - 16.5 = -1.5 2.25
18 18 - 16.5 = 1.5 2.25
21 21 - 16.5 = 4.5 20.25
24 24 - 16.5 = 7.5 56.25
27 27 - 16.5 = 10.5 110.25
30 30 - 16.5 = 13.5 182.25
    $\sum_{i=1}^{10}\left(x_{i}-\bar{x}\right)^{2}$ = 742.5
We know that Variance, $\sigma^{2}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{a}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$
$\therefore$ $\sigma^{2}$ = (1/10) $\times$ 742.5 = 74.25
Mean = 16.5 and Variance = 74.25
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Question 43 Marks
Find the mean and variance for each of the data First n natural numbers.
Answer
We have to calculate mean and Variance of First n natural numbers = $1, 2, ..., n$.
Mean = $=\frac{1}{n}(1+2+3+\ldots+n)=\frac{n(n+1)}{2 n}=\frac{n+1}{2}$
variance ($\sigma^2$) = $\frac {\sum (x_i - \bar x)^2}{n}$
As the data is very large hence, $(x_i - \bar x)^2$ calculation id difficult.
Hence, we can use the formula: variance ($\sigma^2$) = $\frac {\sum(x_i)^2}{n} - (Mean)^2$
$= \frac {1^2+2^2+...n^2}{n} - \left(\frac {n+1}{2}\right)^2$
Since, $1^2 + 2^2 + 3^2 + .. + n^2$ = $\frac {n(n+1)(2n+1)}{6}$
$\therefore$variance ($\sigma^2$) = $\frac {n(n+1)(2n+1)}{6n} - \frac {(n+1)^2}{4}$
$\frac {(n+1)(2n+1)}{6} - \frac {(n+1)^2}{4}$
$\frac {(n+1)}{2} \left(\frac {2n+1}{3} - \frac {n+1}{2}\right)$
$\frac {n+1}{2} \left(\frac {4n+2 -3n - 3}{6} \right)$
$\frac {n+1}{2} \times \frac {n-1}{6}$ = $\frac {n^2 -1}{12}$
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Question 53 Marks
Find the mean and variance for each of the data
6, 7, 10, 12, 13, 4, 8, 12
Answer
$\text { Here } x=6,7,10,12,13,4,8,12$
$\therefore \sum x=6+7+10+12+13+4+8+12=72$
$n=8 \therefore \bar{x}=\frac{72}{8}=9$
$\sum x^2=(6)^2+(7)^2+(10)^2+(12)^2+(13)^2+(4)^2+(8)^2+(12)^2$
$=36+49+100+144+169+16+64+144=722$
$\therefore \text { Variance } \sigma^2=\frac{N \sum x^2-\left(\sum x\right)^2}{N^2}=\frac{8 \times 722-(72)^2}{(8)^2}$
$=\frac{5776-5184}{64}=\frac{592}{64}=9.25$
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Question 63 Marks
Find the mean deviation about the mean for the data
Income per day 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Number of persons 4 8 9 10 7 5 4 3
Answer
Income per day Mid values $x_i$ $f_i$ $f_ix_i$ $|x_i - 358|$ $f_i|x_i - 358|$
0 - 100 50 4 200 308 1232
100 - 200 150 8 1200 208 1664
200 - 300 250 9 2250 108 972
300 - 400 350 10 3500 8 80
400 - 500 450 7 3150 92 644
500 - 600 550 5 2750 192 960
600 - 700 650 4 2600 292 1168
700 - 800 750 3 2250 392 1176
    50 17900   7896
Mean $\bar x = \frac{1}{N}\sum {{f_i}{x_i}} = \frac{1}{{50}} \times 17900 = 358$
Mean deviation about mean $= \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}\left| {{x_i} - \bar x} \right|}$
$ = \frac{1}{{50}} \times 7896 = 157.92$
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Question 73 Marks
Find the mean deviation from the median for the following data:
$x_i$ 15 21 27 30 35
$f_i$ 3 5 6 7 8
Answer
$x_i$ $f_i$ Cum. Freq. $\left|d_{i}\right|=\left|x_{i}-30\right|$ $f_{i}\left|d_{i}\right|$
15 3 3 15 45
21 5 8 9 45
27 6 14 3 18
30 7 21 0 0
35 8 29 5 40
  29     Total = 148

$\frac{N}{2}= \frac{29} {2} =14.5$
To calculate median we will locate the above value in column of cumulative frequency and the corresponding value of $x_i​​​​​​​$ will be our median.
Median = 30
$\mathrm{MD}=\frac{148}{29} = 5.10$
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Question 83 Marks
Find the mean deviation about the median for the data
$x_i$ 5 7 9 10 12 15
$f_i$ 8 6 2 2 2 6
Answer
$x_i$ $f_i$ c.f. $|x_i - 7|$ $f_i|x_i - 7|$
5 8 8 2 16
7 6 14 0 0
9 2 16 2 4
10 2 18 3 6
12 2 20 5 10
15 6 26 8 48
  26     84
$\frac{N}{2} = \frac{{26}}{2} = 13$
The c.f. just greater than 13 is 14 and corresponding value of x is 7.
$\therefore$ Median = 7
$\therefore$ M.D. about median $ = \frac{1}{N}\sum {{f_i}} \left| {{x_i} - M} \right| = \frac{1}{{26}} \times 84 = 3.23$
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Question 93 Marks
Find the mean deviation from the mean for the data:
$x_i$ 10 30 50 70 90
$f_i$ 4 24 28 16 8
Answer
To calculate the mean deviation about mean we need to make the following table,
$x_i$ $f_i$ $x_if_i$ $|d_i|=|x_i-mean|$ $f_id_i$
10 4 40 40 160
30 24 720 20 480
50 28 1400 0 0
70 16 1120 20 320
90 8 720 40 320
  80 4000   1280
$Mean=\frac{1}{n} \sum f_{i} x_{i}=\frac{4000}{80}=50$
$\therefore \quad \mathrm{M.D.}=\frac{1}{n} \Sigma f_{i}\left|d_{i}\right|=\frac{1}{80}[1280]=16$
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Question 103 Marks
Find the mean deviation from the mean for the data:
$x_i$ 5 10 15 20 25
$f_i$ 7 4 6 3 5
Answer
To find mean deviation about the mean we need to make the following table,
$x_i$ $f_i$ $x_if_i$ $|d_i| = |x_i - mean|$ $f_id_i$
5 7 35 9 63
10 4 40 4 16
15 6 90 1 6
20 3 60 6 18
25 5 125 11 55
  25 350   158
Mean = $\frac{1}{n} \sum f_{i} x_{i}=\frac{350}{25}$ = 14
$\therefore$ M.D = $\frac{1}{n} \Sigma f_{i}\left|\mathrm{d}_{i}\right|=\frac{1}{25}$[158] = 6.32
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Question 113 Marks
Find the mean deviation about the median for the data in: $36, 72, 46, 60, 45, 53, 46, 51, 49,42$
Answer
Arrange the data in ascending order, we have
$36, 42, 45, 46, 49, 51, 53, 60, 72$
Here $n = 10$ (which is even)
So median is average of $5^{th}$ and $6^{th}$ observation
$\therefore $ Median $ = \frac{{46 + 49}}{2} = \frac{{92}}{2} = 47.5$
$x_i$ $|x_i - m|$
36 11.5
42 5.5
45 2.5
46 1.5
46 1.5
49 1.5
51 3.5
53 5.5
60 12.5
72 24.5
Total 70
M.D. about medican $ = \frac{1}{n}\sum\limits_{i - l}^n {\left| {{x_i} - M} \right|} $
$ = \frac{1}{{10}} \times 70 = 7$
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Question 123 Marks
Find the mean deviation about the median for the data in: $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.$
Answer
Arrange the data in ascending order, we have
$10, 11, 11,12, 13, 13, 14, 16, 16, 17, 17, 18$
Here $n = 12$ (which is even)
So median is average of $6^{th}$ and $7^{th}$ observations
$\therefore$ Median = $\frac{{13 + 14}}{2} = \frac{{27}}{2}$ = 13.5
$x_i$ $|x_i - M|$
10 3.5
11 2.5
11 2.5
12 1.5
13 0.5
13 0.5
14 0.5
16 2.5
16 2.5
17 3.5
17 3.5
18 4.5
Total 28
M.D. about median = $\frac{1}{n}\sum\limits_{i = 1}^n |x_i - M|$
= $\frac{1}{{12}} \times$ 28 = 2.33
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Question 133 Marks
Find the mean deviation about the mean for the data: $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$.
Answer
$\bar x = \frac{{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}}{{10}}$ = $\frac{{500}}{{10}}$ = 50
$x_i$ $\left| {{x_i} - \bar x} \right|$
38 12
70 20
48 2
40 10
42 8
55 5
63 13
46 4
54 4
44 6
Total 84
M.D. about mean = $\frac{1}{n}\sum\limits_{i = 1}^n {\left| {{x_i} - \bar x} \right|} $
= $\frac{1}{{10}} \times$ 84 = 8.4
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Question 143 Marks
Calculate the mean deviation about median age for the age distribution of $100$ persons gives 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
[Hint: Convert the given data into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval]
Answer
Age Exclusive class intervals Mid values $x_i$ $f_i$ c.f. $|x_i - 38|$ $f_i|x_i - 38|$
16-20 15.5-20.5 18 5 5 20 100
21-25 20.5-25.5 23 6 11 15 90
26-30 25.5-30.5 28 12 23 10 120
31-35 30.5-35.5 33 14 37 5 70
36-40 35.5-40.5 38 26 63 0 0
41-45 40.5-45.5 43 12 75 5 60
46-50 45.5-50.5 48 16 91 10 160
51-55 50.5-55.5 53 9 100 15 135
      100     735
$\frac{N}{2} = \frac{{100}}{2} $ = 50
$\therefore$ Median class is 35.5 - 40.5
Median = 35.5 + $\frac{{50 - 37}}{{26}} \times  5 = 35.5 + 2.5 = 38$
M.D. about median = $\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}\left| {{x_i} - M} \right|} = \frac{1}{{100}} \times $ 735 = 7.35
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Question 153 Marks
Find the mean deviation about median for the following data:
Marks 0-10 10-20 20-30 30-40 40-50 50-60
Number of Girls 6 8 14 16 4 2
Answer
Marks Mid values $x_i$ $f_i$ c.f. $|x_i - 27.86|$ $f_i|x_i - 27.86|$
0 - 10 5 6 6 22.86 137.16
10 - 20 15 8 14 12.86 102.88
20 - 30 25 14 28 2.86 40.04
30 - 40 35 16 44 7.14 114.24
40 - 50 45 4 48 17.14 68.56
50 - 60 55 2 50 27.14 54.28
    50     517.16
$\frac{N}{2} = \frac{{50}}{2} = 25$
$\therefore$ Median class is 20 - 30
Median $ = 20 + \frac{{25 - 14}}{{14}} \times 10 = 20 + 7.86 = 27.86$
M.D. about median $ = \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}\left| {{x_i} - M} \right|} = \frac{1}{{50}} \times 517.16 = 10.34$
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Question 163 Marks
Find the mean deviation about the mean for the data
Height in cm 95-105 105-115 115-125 125-135 135-145 145-155
Number of boys 9 13 26 30 12 10
Answer
Height in cms Mid values $x_i$ $f_i$ $f_ix_i$ $|x_i - 125.3|$ $f_i|x_i - 125.3|$
95 - 105 100 9 900 25.3 227.7
105 - 115 110 13 1430 15.3 198.9
115 - 125 120 26 3120 5.3 137.8
125 - 135 130 30 3900 4.7 141
135 - 145 140 12 1680 14.7 176.4
145 - 155 150 10 1500 24.7 247
    100 12530   1128.8
Mean $\bar x = \frac{1}{N}\sum {{f_i}{x_i} = \frac{1}{{100}} \times 12530 = 125.3}$
Mean deviation about mean $\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}} \left| {{x_i} - \bar x} \right| = \frac{1}{{100}} \times 1128.8 = 11.28$
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Question 173 Marks
Find the mean deviation about the mean for the data: $4, 7, 8, 9, 10, 12, 13, 17.$
Answer
Mean of the given data is
$\bar x = \frac{{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}}{8} = \frac{{80}}{8}$ = 10
$x_i$ $\left| {{x_1} - \bar x} \right|$
4 6
7 3
8 2
9 1
10 0
12 2
13 3
17 7
Total 24
M.D. about mean = $\frac{1}{n}\sum\limits_{i = 1}^n {\left| {{x_i} - \bar x} \right|} $
= $\frac{1}{8} \times$ 24 = 3
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Question 183 Marks
Find the variance and standard deviation of the following data.
$x_i$ 4 8 11 17 20 24 32
$f_i$ 3 5 9 5 4 3 1
Answer
By using formula,
${\sigma ^2} = \frac{1}{N}\left[ {\sum\limits_{i = 1}^n {{f_i}} {{\left( {{x_i} - \bar x} \right)}^2}} \right]$
$x_i$ $f_i$ $f_ix_i$ $x_i - \overline x$ $(x_i - \overline x)^2$ $f_i(x_i - \overline x)^2$
4 3 12 -10 100 $300$
8 5 40 -6 36 $180$
11 9 99 -3 9 $81$
17 5 85 3 9 $45$
20 4 80 6 36 $144$
24 3 72 10 100 $300$
32 1 32 18 324 $324$
Total 30 420     $1374$
Given, N = $\sum f_i$ = 30, $\sum f_ix_i$ = 420 and $\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}$ = 1374
$\therefore$ $\bar x = \frac{{\sum\limits_{i = 1}^7 {{f_i}} {x_i}}}{N}$ = $\frac{420}{30}$ = 14
Variance ($\sigma^2$) = $\frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}} {\left( {{x_i} - \bar x} \right)^2}$ = $\frac{1}{30}$ $\times$ $1374 = 45.8$
Standard deviation, $\sigma$ = $\sqrt{\sigma^2}$ = $\sqrt{45.8}$ = 6.77
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Question 193 Marks
Find the variance of the following data:
$6, 8, 10, 12, 14, 16, 18, 20, 22, 24$
Answer
The mean is calculated by the step-deviation method taking $14$ as assumed mean. The number of observations is $n = 10$
$x_i​​​​​​​$ $d_i = \frac {x_i - 14}{2}$ Deviations from mean ($x_i - \bar x$) ($x_i - \bar x$)
6 -4 -9 81
8 -3 -7 49
10 -2 -5 25
12 -1 -3 9
14 0 -1 1
16 1 1 1
18 2 3 9
20 3 5 25
22 4 7 49
24 5 9 81
  5   330
Therefore Mean $\bar x$ = assumed mean + $\frac {\sum\limits_{i=1}^{n} d_i}{n} \times h$ = $14 + \frac 5{10} \times 2 = 15$
and Variance ($\sigma$) = $\frac 1n {\sum\limits_{i=1}^{10}}{(x_i - \bar x)^2} = \frac 1{10} \times 330 = 33$
Thus Standard deviation ($\sigma$) = $\sqrt {33} = 5.74$
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Question 203 Marks
Calculate the mean deviation about median for the following data.
Class 0-10 10-20 20-30 30-40 40-50 50-60
frequency 6 7 15 16 4 2
Answer
We make the table from the given data.
Class $f_i$ cf Mid-point($x_i$) $|x_i - M|$ $f_i|x_i - M|$
0-10 6 6 5 23 138
10-20 7 13 15 13 91
20-30 15 28 25 3 45
30-40 16 44 35 7 112
40-50 4 48 45 17 68
50-60 2 50 55 27 54
  50       508
Here, $\frac{N}{2}$ = $\frac{50}{2}$ = 25
Here, 25th item lies in the class 20-30. Therefore, 20-30 is the median class.
Here, l = 20, cf = 13, f = 15, b = 10 and N = 50
$\because$ Median, M = $l+\frac{\frac{N}{2}-c f}{f} \times b$
$\Rightarrow$ M = 20 + $\frac{25-13}{15}$ $\times$ 10 = 20 + 8 = 28
Thus, mean deviation about median is given by
MD (M) = $\frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}} \left| {{x_i} - M} \right|$ = $\frac{1}{50}$ $\times$ 508 = 10.16
Hence, mean deviation about median is 10.16.
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Question 213 Marks
Find the mean deviation about the mean for the following data.
Marks obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Number of students 2 3 8 14 8 3 2
Answer
We make the table from the given data.
Marks obtained Number of students $(f_i)$ Mid-point $(x_i)$ $f_ix_i$ $|x_i - \overline x|$ $f_i|x_i - \overline x|$
10-20 2 15 30 30 60
20-30 3 25 75 20 60
30-40 8 35 280 10 80
40-50 14 45 630 0 0
50-60 8 55 440 10 80
60-70 3 65 195 20 60
70-80 2 75 150 30 60
Total 40   1800   400
Here, $N = \sum\limits_{i = 1}^7 {{f_i}} = 40,\sum\limits_{i = 1}^7 {{f_i}} {x_i} = 1800$
Therefore, Mean ($\overline x$) = $\frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}} x$ = $\frac{1800}{40}$ = 45
$\sum\limits_{i = 1}^7 {{f_i}} \left| {{x_i} - \bar x} \right|$ = 400
Now, mean deviation,
MD ($\overline x$) = $\frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}} \left| {{x_i} - \bar x} \right|$ = $\frac{1}{40}$ $\times$ 400 = 10
Hence, mean deviation from mean is $10$.
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Question 223 Marks
Find the mean deviation about the median for the following data.
$x_i$ 3 6 9 12 13 15 21 22
$f_i$ 3 4 5 2 4 5 4 3
Answer
$N = 30$, which is even.
So, median is the mean of the $15^{th}$ and $16^{th}$ observations. Both of these observations lie in the cumulative frequency $18$ for which the corresponding observation is $13$.
Therefore,
Median (M) = $\frac{15 \text { th observation }+16 \text { th observation }}{2}$ = $\frac{13+13}{2}$ = 13
$x_i$ $f_i$ cf $|x_i - M|$ $f_i|x_i - M|$
$3$ 3 $3$ $10$ $30$
$6$ 4 $7$ $7$ $28$
$9$ 5 $12$ $4$ $20$
$12$ 2 $14$ $1$ $2$
$13$ 4 $18$ $0$ $0$
$15$ 5 $23$ $2$ $10$
$21$ 4 $27$ $8$ $32$
22 3 $30$ 9 $27$
We have, $\sum f_{i}$ = 30 and $\sum f_{i}\left|x_{i}-M\right|$ = 149
$\therefore$ Mean deviation about median,
MD (M) = $\frac{1}{N} \sum f_{i}\left|x_{i}-M\right|$ = $\frac{1}{30}$ $\times$ 149 = 4.97
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Question 233 Marks
Find the mean deviation about the mean for the following data.
$x_i$ 2 5 6 8 10 12
$f_i$ 2 8 10 7 8 5
Answer
We make the table from the given data.
$x_i$ $f_i$ $f_ix_i$ $|x_i - \overline x|$ $f_i|x_i - \overline x|$
2 2 4 5.5 11
5 8 40 2.5 20
6 10 60 1.5 15
8 7 56 0.5 3.5
10 8 80 2.5 20
12 5 60 4.5 22.5
  $\sum f_i$=40 $\sum f_{i} x_i$=300   $\sum f_{i}\left|x_{i}-\overline{x}\right|$=92
Here, N = $\sum f_i$ = 40, $\sum f_{i} x_i$ = 300 and $\sum f_{i}\left|x_{i}-\overline{x}\right|$ = 92
Now, mean($\overline x$) = $\frac{1}{N} \sum f_{i} x_{i}$ = $\frac{1}{40}$ $\times$ 300 = 7.5
$\therefore$ Mean deviation about the mean,
MD($\overline x$) = $\frac{1}{N} \sum f_{i}\left|x_{i}-\overline{x}\right|$ = $\frac{1}{40}$ $\times$ 92 = 2.3
Hence, the mean deviation about mean is 2.3
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Question 243 Marks
Find the mean deviation about the median for the following data:
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Answer
Arranging the data into ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
Now Median = $\left( \frac {11+1}{2}\right)^{th} \;or \;6^{th} observation = 9$
The absolute values of the respective deviations from the median, i.e., |$x_i - M$| are
6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12
Therefore $\sum\limits_{i=1}^{11} |x_i - M| = 58$
and M.D. (M) = $\frac 1{11} \sum\limits_{i=1}^{11} |x_i - M| = \frac 1{11} \times 58 = 5.27$
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Question 253 Marks
Find the mean deviation about the mean for the following data:
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5
Answer
From above data we have,
$\bar x = \frac {1}{20} \sum\limits_{i=1}^{20} x_i = \frac {200}{20} = 10$
The respective absolute values of the deviations from mean, i.e., $|x_i - \bar x|$ are
2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5
Therefore $\sum\limits_{i=1}^{20} |x_i - \bar x| = 124$
and $M.D. (\bar x) = \frac {124}{20} = 6.2$
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Question 263 Marks
The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$, respectively by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?
Answer
We have, n = 100, $\overline x$ = 40 and $\sigma$ = 5.1
$\therefore \overline{x}=\frac{1}{n} \Sigma x_{i}$
$\Rightarrow \Sigma x_{i} =n \overline{x} = 100 \times 40 = 4000$
$\therefore$ Incorrect $\Sigma x_i = 4000$
and,
$\sigma$ = 5.1
$\Rightarrow \sigma^2 = 26.01$
$\Rightarrow \frac{1}{n} \Sigma x_i^2- (mean)^2 = 26.01$
$\Rightarrow \frac{1}{100} \Sigma x_i^2- 1600 = 26.01$
$\Rightarrow \Sigma x_i^2 = 1626.01 \times 100$
$\therefore$ Incorrect $\Sigma x_i^2= 162601$
To correct the $\sum x_i$ , we need to subtract the incorrect observation 50 and add correct observation is 40.
We have, incorrect $\Sigma x_i= 4000$
$\therefore$ Correct $\Sigma x_i= 4000 - 50 + 40 = 3990$
and,
Similarly, to obtain correct $\sum x_i^2$ we need to subtract $50^2$ and add $40^2$ to incorrect one.
Incorrect $\Sigma x_i^2 = 162601$
$\therefore$ Correct $\Sigma x_i^2= 162601 - 50^2 + 40^2 = 161701$
Now, Correct mean = $\frac{3990}{100}$ = 39.90
Correct variance = $\frac{1}{100}$ (Correct $\Sigma x_i^2$) - (Correct mean)$^2$
$\Rightarrow$ Correct variance = $\frac{161701}{100}-\left(\frac{3990}{100}\right)^{2}$
$\Rightarrow$ Correct variance = $\frac{161701 \times 100-(3990)^{2}}{(100)^{2}}$
$\Rightarrow$ Correct variance = $\frac{16170100-15920100}{10000}$ = 25
$\therefore$ Correct standard deviation = $\sqrt{25}$ = 5
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Question 273 Marks
If each of the observation $x_1, x_2,..., x_n$ is increased by a, where a is a negative or positive number, then show that the variance remains unchanged.
Answer
Suppose $\overline x$ be the mean of $ x_1, x_2, ...,x_n.$
The variance,
$\sigma_{1}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}$ ...(i)
If a is added to each observation,
$y_i = x_i + a$ ....(ii)
Suppose the mean of the new observation be $\overline y$.
$\overline{y}=\frac{1}{n} \sum_{i=1}^{n} y_{i}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}+a\right)$
= $\frac{1}{n}\left[\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} a\right]$ = $\frac{1}{n}\left[\sum_{i=1}^{n} x_{i}+n a\right]$ = $\overline x$ + a
i.e., $\overline y$ = $\overline x$ + a ...(iii)
The variance of new observation
$\sigma_{2}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(y_{i}-\overline{y}\right)^{2}$ = $\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}+a-\overline{x}-a\right)^{2}$ [using Eqs.(ii) and (iii)]
= $\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}$
$\sigma_{2}^{2}=\sigma_{1}^{2}$
The variance of new observation is same as that of original observation.
Hence, the variance remains unchanged.
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Question 283 Marks
The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three of the observations are $1, 2$ and $6$, find the other two observations.
Answer
Let the other two observations be be $x$ and $y$.
$1+2+6+x+y=5 \times 4.4$
$x+y=22-9$
$x+y=13 \ldots \text { (i) }$
$\text { Variance }=\frac{\left[(1-4.4)^2+(2-4.4)^2+(6-4.4)^2+(x-4.4)^2+(y-4.4)^2\right]}{5}=8.24$
$11.56+5.76+2.56+(x-4.4)^2+(y-4.4)^2=41.2$
$(x-4.4)^2+(y-4.4)^2=21.32$
$(x-4.4)^2+(13-x-4.4)^2=21.32[\text { using eq }(i)]$
$(x-4.4)^2+(8.6-x)^2=21.32$
$x^2-8.8 x+19.36+73.96-17.2 x+x^2=21.32$
$2 x^2-26 x+72=0$
$x^2-13 x+36=0$
$x^2-4 x-9 x+36=0$
$x(x-4)-9(x-4)=0$
$(x-4)(x-9)=0$
$x=4 \text { or } x=9$
If, $x=4$, then $y=9$
Thus, the other two observations are $4$ and $9$ .
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Question 293 Marks
The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.
Answer
We have, n = 20 and $\sigma^{2}=5$
Now each observation is multiplied by 2.
Suppose X = 2x be the new data.
$\therefore \overline{X}=\frac{1}{20} \Sigma 2 x_{i}=\frac{1}{20} \times 2 \Sigma x_{i}=2 \overline{x}$
$\Rightarrow \quad \sum X_{i}^{2}=4 \sum x_{i}^{2}$
Since, $\sigma^{2}=5$
$\Rightarrow \frac{1}{n} \Sigma x_{i}^{2}-(\overline{x})^{2}=5$
Now, for the new data:
$\sigma^{2}=\frac{1}{n} \sum X_{i}^{2}-(\overline{X})^{2}$ $=\frac{4}{n} \sum x_{i}^{2}-(2 \overline{x})^{2}=4\left(\frac{\sum x_{i}^{2}}{n}-(\overline{x})^{2}\right)=4 \times 5=20$
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Question 303 Marks
The following values are calculated in respect of heights and weights of the students of a section of Class XI:
  Height Weight
Mean $162.6 cm$ $52.36 kg$
Variance $127.69 cm^2$ $23.1361 kg^2$
Find S.D and check which of them is more variable.
Answer
To compare the variability, we have to calculate their coefficients of variation.For this we 1st find standard deviation.
Given Variance of height = $127.69cm^2$
Therefore Standard deviation of height = $\sqrt{127.69}cm = 11.3 cm$
Also Variance of weight = $23.1361 kg^2$
Therefore Standard deviation of weight = $\sqrt{23.1361}kg = 4.81 kg$
Now, the coefficient of variations (C.V.) are given by
(C.V.) in heights = $\frac {Standard \;Deviation }{ Mean} \times 100$
$\frac {11.3}{162.6} \times 100 = 6.95$
and (C.V.) in weights = $\frac {4.81}{52.36} \times 100 = 9.18$
Clearly C.V. in weights is greater than the C.V. in heights
Therefore, we can say that weights show more variability than heights.
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Question 313 Marks
Coefficient of variation of the two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively. What are their arithmetic means?
Answer
Coefficient of variation = $\frac{\sigma}{\overline{X}} \times$100
So, we have:
60 = $\frac{21}{\overline{X}} \times$ 100 $\Rightarrow$ $\overline X$ = $\frac{21}{60} \times$100 = 35
70 = $\frac{16}{\overline X} \times$ 100 $\Rightarrow$ $\overline X$ = $\frac{16}{70} \times$ 100 = 22.85
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Question 323 Marks
Two plants A and B of a factory show following results about the number of workers and the wages paid to them:

Plant A Plant B
No. of workers 5000 6000
Average monthly wages ₹ 2500 ₹ 2500
Variance of distribution of wages 81 100

In which plant A or B is there greater variability in individual wages?

Answer
We observe that the average monthly wages in both firms is the same i.e. ₹ 2500. Therefore the plant with a greater variance will have greater variability.
Variance of plant A = 81 and Variance of plant B = 100.
Thus, plant B has greater variability in individual wages.
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Question 333 Marks
Calculate the mean, variance and standard deviation for the following distribution:
Class 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Frequency 3 7 12 15 8 3 2
Answer
Here, we construct the following table:
Class Frequency $(f_i)$ Mid-point$(x_i)$ $f_ix_i$ $(x_i - \overline x)^2$ $f_i(x_i - \overline x)^2$
30-40 3 35 105 729 2187
40-50 7 45 315 289 2023
50-60 12 55 660 49 588
60-70 15 65 975 9 135
70-80 8 75 600 169 1352
80-90 3 85 255 529 1587
90-100 2 95 190 1089 2178
  N = 50   $\sum f_i x_i$= 3100   $ \sum f_{i}\left(x_{i}-\overline{x}\right)^{2}$= 10050
Thus, N = 50, $\sum f_i x_i$ = 3100
$\therefore$ Mean $\overline x$ = $\frac{1}{N} \sum f_{i} x_{i}$ = $\frac{3100}{50}$ = 62
Variance, $\sigma^2$ = $\frac{1}{N} \sum f_{i}\left(x_{i}-\overline{x}\right)^{2}$ = $\frac{1}{50}$ $\times$ 10050 = 201 and standard deviation, $\sigma$ = $\sqrt{201}$ = 14.18
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Question 343 Marks
Find the standard deviation for the following data:
$x_i$ 3 8 13 18 23
$f_i$ 7 10 15 10 6
Answer
We make the table from the given data.
$x_i$ $f_i$ $f_ix_i$ $x_i^2$ $f_ix_i^2$
3 7 21 9 63
8 10 80 64 640
13 15 195 169 2535
18 10 180 324 3240
23 6 138 529 3174
  $\sum f_i = N = 48$ $\sum f_ix_i=614$   $\sum f_i x_i^2$ = 9652
$\therefore$ $\sigma=\frac{1}{N} \sqrt{N \sum f_{i} x_{i}^{2}-\left(\sum f_{i} x_{i}\right)^{2}}$
=$\frac{1}{48} \sqrt{48 \times 9652-(614)^{2}}$ = $\frac{1}{48} \sqrt{463296-376996}$ = $\frac{\sqrt{86300}}{48}$
=$\frac{1}{48}$ $\times$ 293.77 = 6.12
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Question 353 Marks
Calculate the mean, variance and standard deviation for the following distribution:
Class 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Frequency 3 7 12 15 8 3 2
Answer
Here, we construct the following table:
Class Frequency $(f_i)$ Mid-point$(x_i)$ $f_ix_i$ $(x_i - \overline x)^2$ $f_i(x_i - \overline x)^2$
30-40 3 35 105 729 2187
40-50 7 45 315 289 2023
50-60 12 55 660 49 588
60-70 15 65 975 9 135
70-80 8 75 600 169 1352
80-90 3 85 255 529 1587
90-100 2 95 190 1089 2178
  N = 50   $\sum f_i x_i$= 3100   $ \sum f_{i}\left(x_{i}-\overline{x}\right)^{2}$= 10050
Thus, N = 50, $\sum f_i x_i$ = 3100
$\therefore$ Mean $\overline x$ = $\frac{1}{N} \sum f_{i} x_{i}$ = $\frac{3100}{50}$ = 62
Variance, $\sigma^2$ = $\frac{1}{N} \sum f_{i}\left(x_{i}-\overline{x}\right)^{2}$ = $\frac{1}{50}$ $\times$ 10050 = 201 and standard deviation, $\sigma$ = $\sqrt{201}$ = 14.18
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Question 363 Marks
Find the mean deviation about the mean for the following data:
$6, 7, 10, 12, 13, 4, 8, 12$
Answer
Step 1 Mean of the given data is $\bar x = \frac {6+7+10+12+13+4+8+12}{8} = \frac {72}{8} = 9$
Step 2 The deviations of the respective observations from the mean $\bar x$, i.e., $x_i$ – $\bar x$ are
$6 – 9, 7 – 9, 10 – 9, 12 – 9, 13 – 9, 4 – 9, 8 – 9, 12 – 9,$
or $–3, –2, 1, 3, 4, –5, –1, 3$
Step 3 The absolute values of the deviations, i.e., $|x_i - \bar x|$ are
$3, 2, 1, 3, 4, 5, 1, 3$
Step 4 The required mean deviation about the mean is
$M.D. (\bar x) = \sum\limits_{i=1}^8 |x_i - \bar x|$
$\frac {3+2+1+3+4+5+1+3}{8} =\frac {22}{8} = 2.75$
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