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MATHS 3 Marks Question

Statistics · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 35 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
$1$st 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
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 $x_i$ $f_i$ $f_ix_i$ $|x_i - 358|$ $f_i|x_i - 358|$
$0 - 100$ $50$ $4$ $200$ $308$ $1232$
$100 - 200$ $150$ $8$ $1200$ $208$ $1664$
$200 - 300$ $250$ $9$ $2250$ $108$ $972$
$300 - 400$ $350$ $10$ $3500$ $8$ $80$
$400 - 500$ $450$ $7$ $3150$ $92$ $644$
$500 - 600$ $550$ $5$ $2750$ $192$ $960$
$600 - 700$ $650$ $4$ $2600$ $292$ $1168$
$700 - 800$ $750$ $3$ $2250$ $392$ $1176$
    $50$ $17900$   $7896$
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:
$x_i$ $15$ $21$ $27$ $30$ $35$
$fi$ $3$ $5$ $6$ $7$ $8$
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$
$x_i$ $|x_i - m|$
$36$ 1$1.5$
$42$ $5.5$
$45$ $2.5$
$46$ $1.5$
$46$ $1.5$
$49$ $1.5$
$51$ $3.5$
$53$ $5.5$
$60$ $12.5$
$72$ $24.5$
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.
$x_i$ $4$ $8$ $11$ $17$ $20$ $24$ $32$
$f_i$ $3$ $5$ $9$ $5$ $4$ $3$ $1$
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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.
$x_i$ $3$ $6$ $9$ $12$ $13$ $15$ $21$ $22$
$f_i$ $3$ $4$ $5$ $2$ $4$ $5$ $4$ $3$
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Question 233 Marks
Find the mean deviation about the mean for the following data.
$x_i$ $2$ $5$ $6$ $8$ $10$ $12$
$f_i$ $2$ $8$ $10$ $7$ $8$ $5$
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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 $x_1, x_2,..., x_n$​​​​​​​ is increased by a, where a is a negative or positive number, then show that the variance remains unchanged.
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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
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 313 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 323 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 333 Marks
Find the standard deviation for the following data:
$x_i$ $3$ $8$ $13$ $18$ $23$
$f_i$ $7$ $10$ $15$ $10$ $6$
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Question 343 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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