JEE Main Previous Year Statistics Questions and Solutions

Q 1 :

Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

16/3
13/3
11/3
14/3

1

2

3

4

(2)

$Mean = \frac{- 3 + 4 + 7 + (- 6) + α + β}{6} = 2$

$\Rightarrow α + β = 10$ ...(i)

$Variance = \frac{\sum x_{i}^{2}}{n} - {(\frac{\sum x_{i}}{n})}^{2} = 23$

$\Rightarrow \sum x_{i}^{2} = 27 \times 6$ $(∵ \bar{x} = \frac{\sum_{x_{i}}}{n} = 2)$

$\Rightarrow 9 + 16 + 49 + 36 + α^{2} + β^{2} = 162$

$\Rightarrow α^{2} + β^{2} = 52$ ...(ii)

Solving (i) and (ii), we get $α = 4$ and $β = 6$

So, mean deviation about mean $= \frac{| - 3 - 2 | + | 4 - 2 | + | 7 - 2 | + | - 6 - 2 | + | 4 - 2 | + | 6 - 2 |}{6}$

$= \frac{5 + 2 + 5 + 8 + 2 + 4}{6}$ $= \frac{13}{3}$

Q 2 :

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken as 8 instead of 12. The correct standard deviation is [2024]

1.8
1.94
$\sqrt{3.86}$
$\sqrt{3.96}$

1

2

3

4

(4)

Incorrect mean, $\bar{x} = 10$

$\Rightarrow Incorrect \frac{\sum x_{i}}{20} = 10$

Incorrect sum, $\sum x_{i} = 200$

$Correct sum = 200 - 8 + 12 = 204$

$Correct mean = \frac{204}{20} = 10.2$

Incorrect S.D. = 2

Incorrect variance = 4

$\Rightarrow \frac{\sum x_{i}^{2}}{20} - {(\frac{\sum x_{i}}{20})}^{2} = 4 \Rightarrow \frac{\sum x_{i}^{2}}{20} - 100 = 4$

$\Rightarrow \sum x_{i}^{2} = 2080$

Incorrect $\sum x_{i}^{2} = 2080$

Correct $\sum x_{i}^{2} = 2080 - 8^{2} + 12^{2} = 2160$

Correct Variance $= \frac{2160}{20} - {(10.2)}^{2} = 108 - 104.04 = 3.96$

Correct S.D. = $\sqrt{3.96}$

Q 3 :

The frequency distribution of the age of students in a class of 40 students is given below. [2024]

Age 15 16 17 18 19 20

No. of students 5 8 5 12 $x$ $y$

If the mean deviation about the median is 1.25, then 4x + 5y is equal to:

43
44
46
47

1

2

3

4

(2)

Age	No. of Students	C.F.
15	5	5
16	8	13
17	5	18
18	12	30
19	$x$	$30 + x$
20	$y$	$30 + x + y$

$30 + x + y = 40$ (Given)

$\Rightarrow x + y = 10$ ...(i)

Median of the given data = 18

$Mean deviation about median = \frac{\sum f_{i} | x_{i} - 18 |}{\sum f_{i}}$

$\sum f_{i} | x_{i} - 18 | = 5 \times 3 + 8 \times 2 + 5 \times 1 + 12 \times 0 + x \times 1 + y \times 2$

$= 15 + 16 + 5 + x + 2 y = 36 + x + 2 y$

$\Rightarrow \frac{36 + x + 2 y}{40} = 1.25$ (Given)

$\Rightarrow x + 2 y = 14$ ...(ii)

On solving (i) and (ii), we get

$x = 6, y = 4$

Now,

$4 x + 5 y = 4 \times 6 + 5 \times 4 = 24 + 20 = 44$

Q 4 :

If the variance of the frequency distribution

$x$ $c$ $2 c$ $3 c$ $4 c$ $5 c$ $6 c$

$y$ 2 1 1 1 1 1

is 160, then the value of $c \in N$ is [2024]

6
8
5
7

1

2

3

4

(4)

We have, $\bar{x} = \frac{2 c + 2 c + 3 c + 4 c + 5 c + 6 c}{2 + 1 + 1 + 1 + 1 + 1}$ $= \frac{22 c}{7}$

$Var (X) = \frac{c^{2} (2 + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2})}{7} - {(\frac{22 c}{7})}^{2}$

$= \frac{92 c^{2}}{7} - \frac{484 c^{2}}{49} = \frac{(644 - 484) c^{2}}{49} = \frac{160 c^{2}}{49}$

$\Rightarrow 160 = \frac{160 c^{2}}{49}$

$\Rightarrow c^{2} = 49$

$\Rightarrow c = 7$ [ $∵ c \in N$ ]

Q 5 :

Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

32
30
28
31

1

2

3

4

(2)

The ascending order of 7 observations are 125, a, b, 170, 190, 210, 230

$M.D. (Med) = \frac{1}{7} \sum_{i = 1}^{7} | x_{i} - Med | = \frac{205}{7}$

$\Rightarrow | 125 - 170 | + | a - 170 | + | b - 170 | + | 170 - 170 | + | 190 - 170 | + | 210 - 170 | + | 230 - 170 | = 205$

$[∵ Median = 170]$

$\Rightarrow a + b = 300$ $. . . (i)$

$∴ Mean = \frac{125 + 300 + 170 + 190 + 210 + 230}{7} = \frac{1225}{7} = 175$ (By (i))

$∴ M.D. (Mean) = \frac{1}{7} \sum_{i = 1}^{7} | x_{i} - Mean |$ $= \frac{1}{7} (50 + 350 - 300 + 5 + 15 + 35 + 55)$

$= \frac{210}{7} = 30$

Q 6 :

Consider 10 observations $x_{1}, x_{2}, . . ., x_{10}$ such that $\sum_{i = 1}^{10} (x_{i} - α) = 2$ and $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 40$ , where $α, β$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{β}{α}$ is equal to : [2024]

2
1
$\frac{5}{2}$
$\frac{3}{2}$

1

2

3

4

(1)

We have, $\frac{\sum_{i = 1}^{10} x_{i}}{n} = \frac{6}{5} \Rightarrow \sum_{i = 1}^{10} x_{i} = 12$ $[∵ n = 10]$

Also, $\sum_{i = 1}^{10} (x_{i} - α) = 2 \Rightarrow \sum_{i = 1}^{10} x_{i} - 10 α = 2 \Rightarrow α = 1$

Now, $\frac{\sum_{i = 1}^{10} x_{i}^{2}}{n} - {(\bar{x})}^{2} = \frac{84}{25} \Rightarrow \frac{\sum_{i = 1}^{10} x_{i}^{2}}{10} = \frac{84}{25} + \frac{36}{25} = \frac{120}{25}$

$\Rightarrow \sum_{i = 1}^{10} x_{i}^{2} = 48$

Also, $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 40 \Rightarrow \sum_{i = 1}^{10} x_{i}^{2} + 10 β^{2} - 2 β \sum_{i = 1}^{10} x_{i} = 40$

$\Rightarrow 48 + 10 β^{2} - 24 β = 40 \Rightarrow 5 β^{2} - 12 β + 4 = 0$

$\Rightarrow (5 β - 2) \Rightarrow (β - 2) = 0 \Rightarrow β = \frac{2}{5} or β = 2$

If $β = 2$ , then $\frac{β}{α} = 2$

If $β = \frac{2}{5}$ , then $\frac{β}{α} = \frac{2}{5}$

Q 7 :

Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100 .$ Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to [2024]

5
$\sqrt{115}$
$\sqrt{5}$
10

1

2

3

4

(3)

Given, $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100$

$σ^{2} = \frac{\sum a_{k}^{2}}{10} -$ $({Mean)}^{2}$ $= \frac{\sum a_{k}^{2}}{10} - {(\frac{\sum a_{k}}{10})}^{2} = \frac{\sum a_{k}^{2}}{10} - {(\frac{50}{10})}^{2}$

$= \frac{\sum a_{k}^{2}}{10} - 25$ ...(i)

Also, ${(\sum a_{k})}^{2} = \sum a_{k}^{2} + 2 \sum_{\forall k < j} a_{k} a_{j}$

$\Rightarrow 2500 = \sum a_{k}^{2} + 2 (1100) \Rightarrow \sum a_{k}^{2} = 2500 - 2200 = 300$

From (i), we get

$σ^{2} = \frac{300}{10} - 25 = 30 - 25 = 5 \Rightarrow σ = \sqrt{5}$

Q 8 :

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to: [2024]

$\frac{77}{12}$
$\frac{105}{4}$
$\frac{5}{4}$
$\frac{4}{5}$

1

2

3

4

(3)

Let $a, b, c, d$ and $e$ be the five observations.

$∴$ $Mean of 5 observations, \bar{x} = \frac{24}{5}$

$\Rightarrow a + b + c + d + e = 24$ ...(i)

$Mean of four observations, {\bar{x}}_{1} = \frac{7}{2}$

$\Rightarrow a + b + c + d = 14$ ...(ii)

From (i) and (ii), we have $e = 10$

Now, variance of 5 observations $= \frac{194}{25}$

$\Rightarrow \frac{\sum x_{i}^{2}}{5} - {(\bar{x})}^{2} = \frac{194}{25}$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 5 (\frac{194}{25} + \frac{576}{25}) = 154$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 154 - 100 (∵ e = 10)$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 54$

$∴ Variance of 4 observations = \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4} - {({\bar{x}}_{1})}^{2}$

$= \frac{54}{4} - \frac{49}{4} = \frac{5}{4}$

Q 9 :

Let the mean and the variance of 6 observations $a, b$ , 68, 44, 48, 60 be 55 and 194, respectively. If a > b, then a + 3b is: [2024]

210
190
200
180

1

2

3

4

(4)

Given, mean $(\bar{x}) = 55$

$∴$ $\frac{a + b + 68 + 44 + 48 + 60}{6} = 55 \Rightarrow a + b = 110$ ...(i)

Given, variance $(σ^{2}) = 194$

$∴$ $\frac{a^{2} + b^{2} + {(68)}^{2} + {(44)}^{2} + {(48)}^{2} + {(60)}^{2}}{6} - {(55)}^{2} = 194$

$\Rightarrow a^{2} + b^{2} = 6850$ ...(ii)

From (i) and (ii), we get

$a = 75$ and $b = 35$

$∴$ $a + 3 b = 75 + 3 (35) = 180$

Q 10 :

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α,$ $β,$ 60 where $α > β$ , are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to _____ . [2024]

(6344)

Mean = 56

$Variance = \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - {(\bar{x})}^{2}$

$\Rightarrow 66.2 = \frac{1}{10} [{(65)}^{2} + {(68)}^{2} + {(58)}^{2} + {(44)}^{2} + {(48)}^{2} + {(45)}^{2} + (60)$ ²+α²+β²+(60)²]-(56)²

$\Rightarrow 3202.2 \times 10 = 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + α^{2} + β^{2} + 3600$

$\Rightarrow α^{2} + β^{2} = 6344$

Q 11 :

Let $a, b, c \in N$ and $a < b < c .$ Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, $a, b, c$ be 18, 4, and $\frac{136}{5}$ , respectively. Then $2 a + b - c$ is equal to _______ . [2024]

(33)

$a, b, c \in N$ and $a < b < c$

Since, mean = 18 $\Rightarrow \frac{9 + 25 + a + b + c}{5} = 18$

$\Rightarrow 34 + a + b + c = 90 \Rightarrow a + b + c = 56$

Mean deviation about the mean is 4

$\Rightarrow \frac{| 9 - 18 | + | 25 - 18 | + | a - 18 | + | b - 18 | + | c - 18 |}{5} = 4$

$\Rightarrow 9 + 7 + | a - 18 | + | b - 18 | + | c - 18 | = 20$

$\Rightarrow | a - 18 | + | b - 18 | + | c - 18 | = 4$

Variance $= \frac{136}{5}$

$\Rightarrow \frac{\sum | x_{i} - \bar{x} |^{2}}{n} = \frac{136}{5}$

$\Rightarrow \frac{81 + 49 + | a - 18 |^{2} + | b - 18 |^{2} + | c - 18 |^{2}}{5} = \frac{136}{5}$

$\Rightarrow | a - 18 |^{2} + | b - 18 |^{2} + | c - 18 |^{2} = 136 - 130 = 6$

Possible values are $| a - 18 |^{2} = 1, | b - 18 |^{2} = 1$ and $| c - 18 |^{2} = 4$

$∵$ $a < b < c$

$∴$ $18 - a = 1, b - 18 = 1$ and $c - 18 = 2$

$\Rightarrow a = 17, b = 19$ and $c = 20$

$Hence, 2 a + b - c = 34 + 19 - 20 = 33$

Q 12 :

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking, it was found that an observation was read as 10 in place of 12. If $μ$ and $σ^{2}$ denote the mean and variance of the correct observations respectively, then $15 (μ + μ^{2} + σ^{2})$ is equal to _________ . [2024]

(2521)

We have, mean = 12, standard deviation = 3

Let $x_{1}, x_{2}, \dots, x_{15}$ be 15 observations

$∴$ $n = 15$

$\frac{x_{1} + x_{2} + \dots + x_{14} + 10}{15} = 12$ $\Rightarrow x_{1} + x_{2} + \dots + x_{14} = 170$

Now, correct mean = $μ$

$\frac{x_{1} + x_{2} + \dots + x_{14} + 12}{15} = μ \Rightarrow \frac{170 + 12}{15} = μ \Rightarrow μ = \frac{182}{15}$

Standard deviation $= \frac{1}{15} \sqrt{15 \sum_{i = 1}^{15} x_{i}^{2} - {(\sum_{i = 1}^{15} x_{i})}^{2}}$

$\Rightarrow 3 = \frac{1}{15} \sqrt{15 \times \sum_{i = 1}^{15} x_{i}^{2} - {(180)}^{2}}$ $\Rightarrow 45 = \sqrt{15 \times \sum_{i = 1}^{15} x_{i}^{2} - 32400}$

$\Rightarrow 2025 = 15 \sum_{i = 1}^{15} x_{i}^{2} - 32400$

$\Rightarrow 2025 + 32400 = 15 (x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 100)$

$\Rightarrow 2295 = x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 100$

$\Rightarrow x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} = 2195$

Correct Variance, $σ^{2} = \frac{1}{n^{2}} (15 \times \sum_{i = 1}^{15} x_{i}^{2} - {(\sum_{i = 1}^{n} x_{i})}^{2})$

$= \frac{1}{{(15)}^{2}} (15 \times (x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 144) - {(182)}^{2})$

$= \frac{1}{225} (15 \times (2339) - 33124)$ $= \frac{1}{225} (35085 - 33124)$

$σ^{2} = \frac{1961}{225}$

$∴$ $15 (μ + μ^{2} + σ^{2}) = 15 (\frac{182}{15} + {(\frac{182}{15})}^{2} + \frac{1961}{225})$

$= 15 (\frac{2730 + 33124 + 1961}{225})$ $= \frac{37815}{15} = 2521$

Q 13 :

The variance $σ^{2}$ of the data

$x_{i}$ 0 1 5 6 10 12 17

$f_{i}$ 3 2 3 2 6 3 3

is ________ . [2024]

(29)

Mean $(\bar{x})$ $= \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{0 \times 3 + 1 \times 2 + 5 \times 3 + 6 \times 2 + 10 \times 6 + 12 \times 3 + 17 \times 3}{3 + 2 + 3 + 2 + 6 + 3 + 3}$

$= \frac{176}{22} = 8$

$∴$ Variance $σ^{2}$ $= \frac{\sum f_{i} {(x_{i} - \bar{x})}^{2}}{\sum f_{i}}$

$= \frac{3 {(0 - 8)}^{2} + 2 {(1 - 8)}^{2} + 3 {(5 - 8)}^{2} + 2 {(6 - 8)}^{2} + 6 {(10 - 8)}^{2} + 3 {(12 - 8)}^{2} + 3 {(17 - 8)}^{2}}{22}$

$= \frac{3 \times 64 + 2 \times 49 + 3 \times 9 + 2 \times 4 + 6 \times 4 + 3 \times 16 + 3 \times 81}{22}$

$= \frac{640}{22} = 29.09 \approx 29$

Q 14 :

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

106
103
100
105

1

2

3

4

(2)

Mean, $\bar{x} = 6 + 4 + a + 8 + b + 12 + 10 + 13$

$\Rightarrow 9 = \frac{53 + a + b}{8} \Rightarrow a + b = 72 - 53 = 19$

$∴$ a + b = 19 ... (i)

Variance, $σ^{2} = \frac{{(6 - 9)}^{2} + {(4 - 9)}^{2} + {(a - 9)}^{2} + {(8 - 9)}^{2} + {(b - 9)}^{2} + {(12 - 9)}^{2} + {(10 - 9)}^{2} + {(13 - 9)}^{2}}{8}$

$\Rightarrow 9.25 \times 8 = 9 + 25 + 1 + 9 + 1 + 16 + {(a - 9)}^{2} + {(b - 9)}^{2}$

$\Rightarrow 74 - 61 = {(a - 9)}^{2} + {(19 - a - 9)}^{2}$ [Using (i)]

$\Rightarrow 13 = {(a - 9)}^{2} + {(10 - a)}^{2}$

$\Rightarrow a^{2} + 81 - 18 a + a^{2} + 100 - 20 a = 13$

$\Rightarrow 2 a^{2} - 38 a = 13 - 181$

$\Rightarrow a^{2} - 19 a + 84 = 0$

$\Rightarrow (a - 12) (a - 7) = 0$

$\Rightarrow a = 12 or 7 \Rightarrow b = 7 or 12$

$∴$ a + b + ab = 12 + 7 + 84 = 103.

Q 15 :

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

17
16.4
16
17.4

1

2

3

4

(3)

We have, $\bar{x} = 5 and σ^{2} = 10$

Now, $\bar{x} = \frac{\sum_{}^{} x_{i}^{}}{n} = \frac{1 + 3 + a + 7 + b}{5} \Rightarrow \frac{11 + a + b}{5} = 5$

$\Rightarrow a + b = 14$

Also, $σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\bar{x})}^{2}$

$\Rightarrow 10 = \frac{1^{2} + 3^{2} + a^{2} + 7^{2} + b^{2}}{5} - 25$

$\Rightarrow a^{2} + b^{2} = 116$

Since, a > b $\Rightarrow$ a = 10 and b = 4

Now, $n + x_{n}$ : 2, 5, 13, 11, 9

$∴ σ^{2} = \frac{2^{2} + 5^{2} + 13^{2} + 11^{2} + 9^{2}}{5} - {(\frac{2 + 5 + 13 + 11 + 9}{5})}^{2} = 80 - 64 = 16$ .

Q 16 :

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

$\frac{3}{5}$
$\frac{2}{15}$
$\frac{28}{75}$
$\frac{11}{15}$

1

2

3

4

(3)

X	X=0	X=1	x=2
P(x)	$\frac{{}^{7}C_{2} \times {}^{3}C_{0}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{7}C_{1} \times {}^{3}C_{1}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{7}C_{0} \times {}^{3}C_{2}}{{}^{10}C_{2}} = \frac{1}{15}$

Mean, $E (X) = \sum_{}^{} X_{i} P (X_{i}) = 0 + \frac{7}{15} + \frac{2}{15} = \frac{3}{5}$

Now, variance $(X) = E (X^{2}) - (E {(X))}^{2} = \frac{28}{75}$

$= (0 + \frac{7}{15} + \frac{4}{15}) - {(\frac{3}{5})}^{2} = \frac{11}{15} - \frac{9}{25} = \frac{28}{75}$

Q 17 :

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is : [2025]

1
$\frac{3}{4}$
2
$\frac{1}{2}$

1

2

3

4

(1)

Given observations are 2, 3, 3, 4, 5, 7, a, b.

Also, mean = 4 and S.D. = $\sqrt{2}$

$\Rightarrow \frac{24 + a + b}{8} = 4 \Rightarrow a + b = 8$

${(\sqrt{2})}^{2} = \frac{2^{2} + 3^{2} + 3^{2} + 4^{2} + 5^{2} + 7^{2} + a^{2} + b^{2}}{8} - 16$

$\Rightarrow 112 + a^{2} + b^{2} = 18 \times 8$

$\Rightarrow a^{2} + b^{2} = 32 \Rightarrow a = b = 4$

New numbers are 2, 3, 3, 4, 5, 7, 4, 4

Arrange the numbers in a ascending order

2, 3, 3, 4, 4, 4, 5, 7

So, mode = 4.

$∴$ Mean deviation about mode

$= \frac{2 + 1 + 1 + 0 + 1 + 3 + 0 + 0}{8} = 1$ .

Q 18 :

The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to [2025]

447
445
451
449

1

2

3

4

(4)

We have, Mean = $(\bar{x}) = 40$

Mean $(\bar{x}) = \frac{\sum_{}^{} x_{i}^{}}{100}$

$\Rightarrow 40 \times 100 = \sum_{}^{} x_{i}^{} \Rightarrow \sum_{}^{} x_{i}^{} = 4000$

Correct mean, $μ = \frac{100 (40) - 50 + 40}{100}$

$= 40 - \frac{1}{10} = 39.9$

Also, S.D.( $σ$ ) = 5.1

$\Rightarrow {(5.1)}^{2} = \frac{\sum_{}^{} x_{i}^{2}}{100} - {(\bar{x})}^{2}$

$\Rightarrow \sum_{}^{} x_{i}^{2} = 100 \times {(40)}^{2} + 100 {(5.1)}^{2}$

$= 16 \times 10^{4} + {(5.1)}^{2} \times 100 = 162601$

Correct $σ^{2} = \frac{\sum_{}^{} x_{i}^{2} - 50^{2} + 40^{2}}{100} - {(μ)}^{2}$

$= 1617.01 - 1592.01 = 25 \Rightarrow σ = 5$

$∴ 10 (μ + σ) = 10 (39.9 + 5) = 10 \times 44.9 = 449$ .

Q 19 :

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is : [2025]

64
48
51
32

1

2

3

4

(2)

When a coin is tossed three times, outcomes are:

{HHH, HHT, HTH, THH, THT, TTH, HTT, and TTT}

Outcomes when tails does not follows a head (only once):

{HHH, THH, TTT, TTH}

Outcomes when tail follows a head:

{HHT, HTT, HTH, THT}

$∴$ Probability distribution table is given by

$X_{i}$	0	1
$P (X_{i})$	1/2	1/2

Mean $(μ) = \sum_{}^{} X_{i}^{} P_{i} = \frac{1}{2}$

and variance $(σ^{2}) = \sum_{}^{} X_{i}^{2} P_{i} - μ^{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

$∴ 64 (μ + σ^{2}) = 64 (\frac{1}{2} + \frac{1}{4}) = 64 \times \frac{3}{4} = 48$ .

Q 20 :

For a statistical data $x_{1}, x_{2}, . . ., x_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i = 1}^{10} x_{i}^{2} = 371$ . He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is [2025]

7
5
9
4

1

2

3

4

(1)

$\sum_{i = 1}^{10} x_{i}^{} = 5.5 \times 10 = 55$

Correct mean $= \frac{\sum_{i = 1}^{10} x_{i}^{} - 4 - 5 + 6 + 8}{10} = \frac{55 - 9 + 14}{10} = 6$

Corrected $\sum_{i = 1}^{10} x_{i}^{2} = 371 - 4^{2} - 5^{2} + 6^{2} + 8^{2} = 430$

Corrected variance $= \frac{Corrected \sum_{i = 1}^{10} x_{i}^{2}}{10} - {(Corrected \bar{x})}^{2}$

$= \frac{430}{10} - {(6)}^{2} = 43 - 36 = 7$

Q 21 :

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiable between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is [2025]

$\frac{18}{25}$
$\frac{14}{25}$
$\frac{28}{75}$
$\frac{26}{75}$

1

2

3

4

(3)

Given, x denote the number of defective oranges, out of two. So, the probability distribution is given below as:

$x_{i}$	x = 0	x = 1	x = 2
$P (x = x_{i})$	$\frac{{}^{7}C_{2}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{7}C_{1} \times {}^{3}C_{1}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{3}C_{2}}{{}^{10}C_{2}} = \frac{1}{15}$

Mean $(μ) = \sum_{i = 0}^{2} x_{i} p_{i} = \frac{9}{15} = \frac{3}{5}$

and $σ^{2} = \sum_{i = 0}^{2} p_{i} x_{i}^{2} - μ^{2} = \frac{11}{15} - {(\frac{3}{5})}^{2} = \frac{28}{75}$

Q 22 :

Let $x_{1}, x_{2}, . . ., x_{10}$ be ten observations such that $\sum_{i = 1}^{10} (x_{i} - 2) = 30$ , $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 98$ , $β > 2$ , and their variance is $\frac{4}{5}$ . If $μ$ and $σ^{2}$ are respectively the mean and the variance of $2 (x_{1} - 1) + 4 β, 2 (x_{2} - 1) + 4 β, . . ., 2 (x_{10} - 1) + 4 β$ , then $\frac{β μ}{σ^{2}}$ is equal to : [2025]

100
110
120
90

1

2

3

4

(1)

We have, $\sum_{i = 1}^{10} (x_{i} - 2) = 30 \Rightarrow \sum_{i = 1}^{10} x_{i} - 2 \times 10 = 30$

$\Rightarrow \sum_{i = 1}^{10} x_{i}^{} = 50$

Now, variance $= \frac{4}{5} \Rightarrow \frac{4}{5} = \frac{1}{10} \sum_{}^{} x_{i}^{2} - {(\frac{\sum_{}^{} x_{i}}{10})}^{2}$

$\Rightarrow \frac{4}{5} = \frac{1}{10} \sum_{}^{} x_{i}^{2} - 25 \Rightarrow \sum_{}^{} x_{i}^{2} = 258$

Now, $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 98 \Rightarrow \sum_{i = 1}^{10} (x_{i}^{2} - 2 β x_{i} + β^{2}) = 98$

$\Rightarrow 258 - 2 β \cdot 50 + 10 β^{2} = 98$

$\Rightarrow 10 β^{2} - 100 β + 160 = 0$

$\Rightarrow (β - 8) (β - 2) = 0 \Rightarrow β = 8$ [ $∵ β > 2$ ]

Now, $μ = \frac{1}{10} \sum_{i = 1}^{10} [2 (x_{i} - 1) + 4 β]$

$= \frac{1}{10} [2 \sum_{i = 1}^{10} x_{i} - 20 + 40 β]$

$= \frac{1}{10} [2 \times 50 - 20 + 320] = 40$

$σ^{2} = 2^{2} (\frac{4}{5}) = \frac{16}{5}$

$∴ \frac{β μ}{σ^{2}} = \frac{8 \times 40 \times 5}{6} = 100$ .

Q 23 :

The variance of the numbers 8, 21, 34, 47, ..., 320 is __________. [2025]

(8788)

Since given numbers are in A.P.

$∴ 8 + (n - 1) 13 = 320 \Rightarrow 13 n = 325 \Rightarrow n = 25$

Number of terms = 25

Now, Mean $= \frac{\sum_{}^{} x_{i}^{}}{n} = \frac{8 + 21 + . . . + 320}{25} = \frac{\frac{25}{2} (8 + 320)}{25} = 164$

and variance $(σ^{2}) = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(mean)}^{2}$

$= \frac{8^{2} + 21^{2} + . . . + 320^{2}}{25} - {(164)}^{2} = \frac{\sum_{n = 1}^{25} {(13 n - 5)}^{2}}{25} - 26896$

$= \frac{\sum_{n = 1}^{25} (169 n^{2} + 25 - 130 n)}{25} - 26896$

$= \frac{\frac{169 \times 25 \times 26 \times 51}{6} + 625 - \frac{130 \times 25 \times 26}{2}}{25} - 26896$

= 35684 –26896 = 8788.

Q 24 :

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $σ^{2}$ is equal to [2023]

10
11
12
9

1

2

3

4

(1)

Combine variance

$= \frac{n_{1} σ_{1}^{2} + n_{2} σ_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {(m_{1} - m_{2})}^{2}}{{(n_{1} + n_{2})}^{2}}$

$\Rightarrow 13 = \frac{15 \cdot 14 + 15 \cdot σ^{2}}{30} + \frac{15 \cdot 15 {(12 - 14)}^{2}}{30 \times 30}$

$\Rightarrow 13 = \frac{14 + σ^{2}}{2} + \frac{4}{4} \Rightarrow σ^{2} = 10$

Q 25 :

Let the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\frac{m}{n}$ ?, where $m$ and $n$ are coprime, then $m + n$ is equal to [2023]

314
315
317
316

1

2

3

4

(3)

Incorrect mean = $\frac{9}{2}$

$\Rightarrow Incorrect \sum x_{i} = \frac{9}{2} \times 12 = 54$

$So, correct mean = \frac{54 + 7 + 14 - 9 - 10}{12} = \frac{56}{12} = \frac{14}{3}$

$And 4 = Incorrect \frac{\sum x_{i}^{2}}{12} - {(\frac{9}{2})}^{2} \Rightarrow Incorrect \sum x_{i}^{2} = 291$

$Now, correct \sum (x_{i}^{2}) = 291 + 7^{2} + 14^{2} - 9^{2} - 10^{2} = 355$

$So, correct variance = \frac{355}{12} - {(\frac{14}{3})}^{2} = \frac{281}{36} = \frac{m}{n}$

$∴$ $m + n = 281 + 36 = 317$

Q 26 :

Let $μ$ be the mean and $σ$ be the standard deviation of the distribution

$x_{i}$ 0 1 2 3 4 5

$f_{i}$ $k + 2$ $2 k$ $k^{2} - 1$ $k^{2} - 1$ $k^{2} + 1$ $k - 3$

where $\sum f_{i} = 62$ . If $[x]$ denotes the greatest integer $\leq x$ , then $[μ^{2} + σ^{2}]$ is equal to [2023]

8
6
9
7

1

2

3

4

(1)

We have, $\sum f_{i} = 62$

$\Rightarrow$ $3 k^{2} + 4 k - 2 = 62 \Rightarrow 3 k^{2} + 16 k - 12 k - 64 = 0$

$\Rightarrow k (3 k + 16) - 4 (3 k + 16) = 0$

$\Rightarrow k = 4, \frac{- 16}{3} (not possible)$

$∴$ $k = 4, σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}} - μ^{2}$

$\Rightarrow σ^{2} = \frac{8 \times 1^{2} + 15 \times 2^{2} + 15 \times 3^{2} + 17 \times 4^{2} + 5^{2}}{62} - μ^{2}$

$\Rightarrow σ^{2} + μ^{2} = \frac{500}{62} \Rightarrow [σ^{2} + μ^{2}] = 8$

Q 27 :

Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is ______. [2023]

32
38
40
36

1

2

3

4

(2)

$A = {a_{1}, a_{2}, a_{3}, a_{4}, a_{5}},$    $B = {b_{1}, b_{2}, b_{3}, b_{4}, b_{5}}$

Given, $\sum_{i = 1}^{5} a_{i} = 25,$    $\sum_{i = 1}^{5} b_{i} = 40$

$\frac{\sum_{i = 1}^{5} a_{i}^{2}}{5} - {(\frac{\sum_{i = 1}^{5} a_{i}}{5})}^{2} = 12,$ $\frac{\sum_{i = 1}^{5} b_{i}^{2}}{5} - {(\frac{\sum_{i = 1}^{5} b_{i}}{5})}^{2} = 20$

$\Rightarrow \sum_{i = 1}^{5} a_{i}^{2} = 185,$ $\sum_{i = 1}^{5} b_{i}^{2} = 420$

Now, $C = {C_{1}, C_{2}, \dots, C_{10}}$

Such that $C_{i} =$ ${\begin{matrix} a_{i} - 3, & i = 1, 2, 3, 4, 5 \\ b_{i} + 2, & i = 6, 7, 8, 9, 10 \end{matrix}$

$∴$ $Mean of C, \bar{C} = \frac{(\sum a_{i} - 15) + (\sum b_{i} + 10)}{10} = \frac{10 + 50}{10} = 6$

$∴$ $σ^{2} = \frac{\sum_{i = 1}^{10} C_{i}^{2}}{10} - {(\bar{C})}^{2} = \frac{\sum {(a_{i} - 3)}^{2} + \sum {(b_{i} + 2)}^{2}}{10} - 6^{2}$

$= \frac{\sum a_{i}^{2} + \sum b_{i}^{2} - 6 \sum a_{i} + 4 \sum b_{i} + 65}{10} - 36 = 32$

$∴$ $Mean + Variance = \bar{C} + σ^{2} = 6 + 32 = 38$

Q 28 :

Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to [2023]

$\frac{7}{3}$
$\frac{8}{3}$
$\frac{10}{3}$
$3$

1

2

3

4

(2)

$Mean (\bar{x}) = \frac{1 + 2 + 4 + 5 + x + y}{6} = 5$

$\Rightarrow x + y = 18 (i)$

$Variance = 10$

$\Rightarrow \frac{1^{2} + 2^{2} + 4^{2} + 5^{2} + x^{2} + y^{2}}{6} - 25 = 10$

$\Rightarrow x^{2} + y^{2} = 164 (ii)$

$Solving (i) and (ii), we get:$

$x = 10, y = 8$

$Now, M . D (\bar{x}) = \frac{\sum | x_{i} - \bar{x} |}{6} = \frac{4 + 3 + 1 + 0 + 5 + 3}{6} = \frac{16}{6} = \frac{8}{3}$

Q 29 :

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is [2023]

11
12
14
13

1

2

3

4

(4)

Q 30 :

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is [2023]

1216
1456
1072
1792

1

2

3

4

(3)

Let the missing observations be $x$ and $y$ .

$∴$ $Mean = \frac{1 + 3 + 5 + x + y}{5}$

$\Rightarrow 5 = \frac{9 + x + y}{5} \Rightarrow x + y = 16$ ...(i)

and variance $= \frac{x^{2} + y^{2} + 35}{5} - 25 \Rightarrow 8 = \frac{x^{2} + y^{2} + 35 - 125}{5}$

$\Rightarrow x^{2} + y^{2} = 40 + 90 = 130$ ..(ii)

Consider ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$256 = 130 + 2 x y \Rightarrow 2 x y = 126$

Now, consider ${(x - y)}^{2}$

$\Rightarrow x^{2} + y^{2} - 2 x y = 130 - 126 \Rightarrow {(x - y)}^{2} = 4$

$\Rightarrow x - y = 2$ ...(iii) or $x - y = - 2$ ...(iv)

Using (i) and (iii), we have $x = 9$ and $y = 7$

Using (i) and (iv), we have $x = 7$ and $y = 9$

$∴$ $Sum of cubes of x and y = 7^{3} + 9^{3} = 343 + 729 = 1072$

Topic Question Set

Q 1 :

Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

Q 2 :

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken as 8 instead of 12. The correct standard deviation is [2024]

Q 3 :

The frequency distribution of the age of students in a class of 40 students is given below. [2024]

Age 15 16 17 18 19 20

No. of students 5 8 5 12 $x$ $y$

If the mean deviation about the median is 1.25, then 4x + 5y is equal to:

Q 4 :

If the variance of the frequency distribution

$x$ $c$ $2 c$ $3 c$ $4 c$ $5 c$ $6 c$

$y$ 2 1 1 1 1 1

is 160, then the value of $c \in N$ is [2024]

Q 5 :

Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

Q 7 :

Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100 .$ Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to [2024]

Q 8 :

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to: [2024]

Q 9 :

Let the mean and the variance of 6 observations $a, b$ , 68, 44, 48, 60 be 55 and 194, respectively. If a > b, then a + 3b is: [2024]

Q 10 :

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α,$ $β,$ 60 where $α > β$ , are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to _____ . [2024]

Q 11 :

Let $a, b, c \in N$ and $a < b < c .$ Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, $a, b, c$ be 18, 4, and $\frac{136}{5}$ , respectively. Then $2 a + b - c$ is equal to _______ . [2024]

Q 13 :

The variance $σ^{2}$ of the data

$x_{i}$ 0 1 5 6 10 12 17

$f_{i}$ 3 2 3 2 6 3 3

is ________ . [2024]

Q 14 :

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

Q 15 :

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

Q 16 :

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

Q 17 :

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is : [2025]

Q 18 :

The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to [2025]

Q 19 :

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is : [2025]

Q 21 :

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiable between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is [2025]

Q 23 :

The variance of the numbers 8, 21, 34, 47, ..., 320 is __________. [2025]

Q 24 :

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $σ^{2}$ is equal to [2023]

Q 26 :

Let $μ$ be the mean and $σ$ be the standard deviation of the distribution

$x_{i}$ 0 1 2 3 4 5

$f_{i}$ $k + 2$ $2 k$ $k^{2} - 1$ $k^{2} - 1$ $k^{2} + 1$ $k - 3$

where $\sum f_{i} = 62$ . If $[x]$ denotes the greatest integer $\leq x$ , then $[μ^{2} + σ^{2}]$ is equal to [2023]

Q 28 :

Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to [2023]

Q 29 :

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is [2023]

(4)

Q 30 :

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is [2023]

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$x_{i}$	0	1	2	3	4	5
$f_{i}$	$k + 2$	$2 k$	$k^{2} - 1$	$k^{2} - 1$	$k^{2} + 1$	$k - 3$

$x_{i}$	0	1	5	6	10	12	17
$f_{i}$	3	2	3	2	6	3	3

Topic Question Set

Q 1 : Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

Q 2 : The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken as 8 instead of 12. The correct standard deviation is [2024]

Q 3 : The frequency distribution of the age of students in a class of 40 students is given below. [2024] Age 15 16 17 18 19 20 No. of students 5 8 5 12 x y If the mean deviation about the median is 1.25, then 4x + 5y is equal to:

Q 4 : If the variance of the frequency distribution x c 2c 3c 4c 5c 6c y 2 1 1 1 1 1 is 160, then the value of c∈N is [2024]

Q 5 : Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and 2057 respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

Q 6 : Consider 10 observations x1,x2, ..., x10 such that ∑i=110(xi-α)=2 and ∑i=110(xi-β)2=40, where α,β are positive integers. Let the mean and the variance of the observations be 65 and 8425 respectively. Then βα is equal to : [2024]

Q 7 : Let a1,a2, …, a10 be 10 observations such that ∑k=110ak=50 and ∑∀k<jak·aj=1100. Then the standard deviation of a1,a2, …, a10 is equal to [2024]

Q 8 : If the mean and variance of five observations are 245 and 19425 respectively and the mean of the first four observations is 72, then the variance of the first four observations is equal to: [2024]

Q 9 : Let the mean and the variance of 6 observations a,b, 68, 44, 48, 60 be 55 and 194, respectively. If a > b, then a + 3b is: [2024]

Q 10 : If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, α,β, 60 where α>β, are 56 and 66.2 respectively, then α2+β2 is equal to _____ . [2024]

Q 11 : Let a,b,c∈N and a<b<c. Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, a,b,c be 18, 4, and 1365, respectively. Then 2a+b-c is equal to _______ . [2024]

Q 13 : The variance σ2 of the data xi 0 1 5 6 10 12 17 fi 3 2 3 2 6 3 3 is ________ . [2024]

Q 14 : If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

Q 15 : Let the Mean and Variance of five observations x1=1,x2=3,x3=a,x4=7 and x5=b,a>b be 5 and 10 respectively. Then the Variance of the observations n+xn,n=1,2,...,5 is [2025]

Q 16 : A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

Q 17 : Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and 2 respectively. Then the mean deviation about the mode of these observations is : [2025]

Q 18 : The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are μ and σ respectively, then 10(μ+σ) is equal to [2025]

Q 19 : A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X, then the value of 64(μ+σ2) is : [2025]

Q 20 : For a statistical data x1,x2,...,x10 of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371. He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is [2025]

Q 21 : Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiable between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is [2025]

Q 22 : Let x1,x2,...,x10 be ten observations such that ∑i=110(xi–2)=30, ∑i=110(xi–β)2=98, β>2, and their variance is 45. If μ and σ2 are respectively the mean and the variance of 2(x1–1)+4β,2(x2–1)+4β,...,2(x10–1)+4β, then βμσ2 is equal to : [2025]

Q 23 : The variance of the numbers 8, 21, 34, 47, ..., 320 is __________. [2025]

Q 24 : The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and σ2 respectively. If the variance of all the 30 numbers in the two sets is 13, then σ2 is equal to [2023]

Q 25 : Let the mean and variance of 12 observations be 92 and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is mn?, where m and n are coprime, then m+n is equal to [2023]

Q 26 : Let μ be the mean and σ be the standard deviation of the distribution xi 0 1 2 3 4 5 fi k+2 2k k2-1 k2-1 k2+1 k-3 where ∑fi=62. If [x] denotes the greatest integer ≤x, then [μ2+σ2] is equal to [2023]

Q 28 : Let the mean of 6 observations 1, 2, 4, 5, x and y be 5 and their variance be 10. Then their mean deviation about the mean is equal to [2023]

Q 29 : The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is [2023]

(4)

Q 30 : The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is [2023]

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Q 1 :

Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

Q 2 :

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken as 8 instead of 12. The correct standard deviation is [2024]

Q 3 :

The frequency distribution of the age of students in a class of 40 students is given below. [2024]

Age 15 16 17 18 19 20

No. of students 5 8 5 12 $x$ $y$

If the mean deviation about the median is 1.25, then 4x + 5y is equal to:

Q 4 :

If the variance of the frequency distribution

$x$ $c$ $2 c$ $3 c$ $4 c$ $5 c$ $6 c$

$y$ 2 1 1 1 1 1

is 160, then the value of $c \in N$ is [2024]

Q 5 :

Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

Q 7 :

Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100 .$ Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to [2024]

Q 8 :

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to: [2024]

Q 9 :

Let the mean and the variance of 6 observations $a, b$ , 68, 44, 48, 60 be 55 and 194, respectively. If a > b, then a + 3b is: [2024]

Q 10 :

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α,$ $β,$ 60 where $α > β$ , are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to _____ . [2024]

Q 11 :

Let $a, b, c \in N$ and $a < b < c .$ Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, $a, b, c$ be 18, 4, and $\frac{136}{5}$ , respectively. Then $2 a + b - c$ is equal to _______ . [2024]

Q 13 :

The variance $σ^{2}$ of the data

$x_{i}$ 0 1 5 6 10 12 17

$f_{i}$ 3 2 3 2 6 3 3

is ________ . [2024]

Q 14 :

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

Q 15 :

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

Q 16 :

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

Q 17 :

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is : [2025]

Q 18 :

The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to [2025]

Q 19 :

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is : [2025]

Q 21 :

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiable between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is [2025]

Q 23 :

The variance of the numbers 8, 21, 34, 47, ..., 320 is __________. [2025]

Q 24 :

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $σ^{2}$ is equal to [2023]

Q 26 :

Let $μ$ be the mean and $σ$ be the standard deviation of the distribution

$x_{i}$ 0 1 2 3 4 5

$f_{i}$ $k + 2$ $2 k$ $k^{2} - 1$ $k^{2} - 1$ $k^{2} + 1$ $k - 3$

where $\sum f_{i} = 62$ . If $[x]$ denotes the greatest integer $\leq x$ , then $[μ^{2} + σ^{2}]$ is equal to [2023]

Q 28 :

Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to [2023]

Q 29 :

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is [2023]

Q 30 :

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is [2023]