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]

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

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]

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.

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%

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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 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]

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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]