JEE Main Previous Year Statistics Questions and Solutions

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]

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

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2

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(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

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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]

0%

(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

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(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

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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

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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

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(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]

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

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(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

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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

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(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 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]

0%

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

Topic Question Set

Q 23 : The variance of the numbers 8, 21, 34, 47, ..., 320 is __________. [2025] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

0%

(4)

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Q 23 :

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