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:

Question

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:

TestHub · Accepted Answer

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

Mathematics - Statistics Question with Solution | TestHub

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Doubts & Discussion

$x :$	$x_{1} = 2$	$x_{2} = 6$	$x_{3} = 8$	$x_{4} = 9$	Total
$f :$	$4$	$4$	$α$	$β$	$8 + α + β$
$f x :$	$8$	$24$	$8 α$	$9 β$	$32 + 8 α + 9 β$
$f {(x - x)}^{2} :$	$64$	$0$	$4 α$	$9 β$	$64 + 4 α + 9 β$