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MATHS 3 Marks Question

Statistics · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 36 questions.

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Question 13 Marks
Find the mean and variance of the following data.
xi 92 93 97 98 102 104 109
fi 3 2 3 2 6 3 3
Answer
xi fi fixi (xi-100) (xi-100)2 fi(xi-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
xi 6 10 14 18 24 28 30
fi 2 4 7 12 8 4 3
Answer
1st of all we construct table. 

xi fi fixi $X_i - \bar X$ $(X_i - \bar X)^2$ $f_i (X_i - \bar X)^2$
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:

xi 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, 12 + 22 + 32 + .. + n2 = $\frac {n(n+1)(2n+1)}{6}$
$\therefore$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
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
$\therefore$ 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 xi fi fixi |xi - 358| fi|xi - 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:
xi 15 21 27 30 35
fi 3 5 6 7 8
Answer
xi fi 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 xi 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
xi 5 7 9 10 12 15
fi 8 6 2 2 2 6
Answer
xi fi c.f. |xi - 7| fi|xi - 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:
xi 10 30 50 70 90
fi 4 24 28 16 8
Answer
To calculate the mean deviation about mean we need to make the following table, 

xi

 fi xifi |di|=|xi-mean| fidi
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:
xi 5 10 15 20 25
fi 7 4 6 3 5
Answer
To find mean deviation about the mean we need to make the following table, 

xi fi xifi |di| = |xi - mean| fidi
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$

xi |xi - 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

xi |xi - 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 = $\frac{1}{n}\sum\limits_{i = 1}^n$ |xi - 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

xi $\left| {{x_i} - \bar x} \right|$
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 xi fi c.f. |xi - 38| fi|xi - 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 xi fi c.f. |xi - 27.86| fi|xi - 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 xi fi fixi |xi - 125.3| fi|xi - 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

xi $\left| {{x_1} - \bar x} \right|$
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.
xi 4 8 11 17 20 24 32
fi 3 5 9 5 4 3 1
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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
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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
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Question 223 Marks
Find the mean deviation about the median for the following data.
xi  3 6 9 12 13 15 21 22
fi  3 4 5 2 4 5 4 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.
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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
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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?
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Question 273 Marks
If each of the observation x1, x2,..., xn is increased by a, where a is a negative or positive number, then show that 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.
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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.
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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 cm2 23.1361 kg2

Find S.D and check which of them is more variable. 

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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?
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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?

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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
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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
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