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Suppose the mean and variance of the frequency distribution are

X $x_1=2$ $x_2=6$ $x_3=8$ $x_4=9$

F $4$ $4$ $\alpha$ $\beta$

$6.8$ respectively. If $x^{3}$ changes from $8\ to\ 7$ , the average of the new data will be:

Option: 1
$15.6$

Option: 2
$20$

Option: 3
$\frac{17}{3}$

Option: 4
$\frac{89}{25}$

Answers (1)

best_answer

Given that:

$32+8\alpha+9\beta=\left(8+\alpha+\beta\right)\times6$

Or, $2\alpha+3\beta=16\$ ------------------ (1)

Now, also we have:

$4\times16+4\times\alpha+9\beta=\left(8+\alpha+\beta\right)\times6.8$

Or, $640+40\alpha+90\beta=544+68\alpha+68\beta$

Or, $14\alpha-11\beta=48$ ------------------ (2)

From (1) and (2), we get:

$\alpha=5$ and $\beta=2$

Therefore,

The new mean = $\frac{32+35+18}{15}$

$=\frac{85}{15}=\frac{17}{3}$

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