Let the mean and variance of the frequency distribution x: x _ 1 = 2 x _ 2 = 6 x _ 3 = 8 x _ 4 = 9 f: 4 4 α β be 6 and 6 . 8 respectively. If x _ 3 is changed from 8 to 7, then the mean for the new data will be:

Correct answer: C. (17)/(3) Given x: x _ 1 = 2 x _ 2 = 6 x _ 3 = 8 x _ 4 = 9 Total f: 4 4 α β 8 + α + β fx: 8 24 8α 9β 32 + 8α + 9β f (x - x) ^ 2 : 64 0 4α 9β 64 + 4α + 9β x = ∑ _ i = 1 ^ 4 f _ i x _ i ∑ _ i = 1 ^ 4 f _ i 32 + 8α + 9β = (8 + α + β) × 6 ⇒ 2α + 3β = 16 . . . (1) Also, for the variance, σ ^ 2 = ∑ f _ i ( x _ i - x) ^ 2 _ i = 1 ^ 4 ∑ f _ i _ i = 1 ^ 4 64 + 4α + 9β = (8 + α + β) × 6 . 8 ⇒ 640 + 40α + 90β = 544 + 68α + 68β ⇒ 28α - 22β = 96 ⇒ 14α - 11β = 48 . . . (2) from (1) and (2) α = 5 and β = 2 so, new mean = (32 + 35 + 18)/(15) = (85)/(15) = (17)/(3)

JEE Main 202127 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Statistics

JEE Main 2021 · 27 July · Evening Shift · Mathematics · Statistics

Let the mean and variance of the frequency distribution x: x _ 1 =...

Let the mean and variance of the frequency distribution

$x:$	${x}_{1} = 2$	${x}_{2} = 6$	${x}_{3} = 8$	${x}_{4} = 9$
$f:$	$4$	$4$	$\alpha$	$\beta$

be $6$ and $6 . 8$ respectively. If ${x}_{3}$ is changed from $8$ to $7,$ then the mean for the new data will be:

Options

A $4$
B $5$
C $\dfrac{17}{3}$
D $\dfrac{16}{3}$

Correct answer

C. $\dfrac{17}{3}$

Step-by-step solution

Given

$x:$	${x}_{1} = 2$	${x}_{2} = 6$	${x}_{3} = 8$	${x}_{4} = 9$	Total
$f:$	$4$	$4$	$\alpha$	$\beta$	$8 + \alpha + \beta$
$fx:$	$8$	$24$	$8\alpha$	$9\beta$	$32 + 8\alpha + 9\beta$
$f{\left(x - x\right)}^{2}:$	$64$	$0$	$4\alpha$	$9\beta$	$64 + 4\alpha + 9\beta$

$x = \dfrac{\mathop{∑}_{i = 1}^{4}{f}_{i}{x}_{i}}{\mathop{∑}_{i = 1}^{4}{f}_{i}}$

$32 + 8\alpha + 9\beta = \left(8 + \alpha + \beta\right) \times 6$

$⇒ 2\alpha + 3\beta = 16 . . . \left(1\right)$

Also, for the variance, ${\sigma}^{2} = \mathop{∑ {f}_{i}{\left({x}_{i} - x\right)}^{2}}_{i = 1}^{4}\mathop{∑ {f}_{i}}_{i = 1}^{4}$

$64 + 4\alpha + 9\beta = \left(8 + \alpha + \beta\right) \times 6 . 8$

$⇒ 640 + 40\alpha + 90\beta = 544 + 68\alpha + 68\beta$

$⇒ 28\alpha - 22\beta = 96$

$⇒ 14\alpha - 11\beta = 48 . . . \left(2\right)$

from \left(1\right) \text{ and } \left(2\right) $</p><p>$ \alpha = 5 \text{ and } \beta = 2 $</p><p>so, new mean$ = \dfrac{32 + 35 + 18}{15} = \dfrac{85}{15} = \dfrac{17}{3}$

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