Let the mean and variance of the frequency distribution
$$\matrix{
{x:} & {{x_1} = 2} & {{x_2} = 6} & {{x_3} = 8} & {{x_4} = 9} \cr
{f:} & 4 & 4 & \alpha & \beta \cr
} $$
be 6 and 6.8 respectively. If x3 is changed from 8 to 7, then the mean for the new data will be :
Solution
Given 32 + 8$\alpha$ + 9$\beta$ = (8 + $\alpha$ + $\beta$) $\times$ 6<br><br>$\Rightarrow$ 2$\alpha$ + 3$\beta$ = 16 ..... (i)<br><br>Also, 4 $\times$ 16 + 4 $\times$ $\alpha$ + 9$\beta$ = (8 + $\alpha$ + $\beta$) $\times$ 6.8<br><br>$\Rightarrow$ 640 + 40$\alpha$ + 90$\beta$ = 544 + 68$\alpha$ + 68$\beta$<br><br>$\Rightarrow$ 28$\alpha$ $-$ 22$\beta$ = 96<br><br>$\Rightarrow$ 14$\alpha$ $-$ 11$\beta$ = 48 ..... (ii)<br><br>from (i) & (ii)<br><br>$\alpha$ = 5 & $\beta$ = 2<br><br>So, new mean = ${{32 + 35 + 18} \over {15}} = {{85} \over {15}} = {{17} \over 3}$
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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