If the mean and variance of the data 65, 68, | vedclass

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Question

(C) Given data: $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$. Total number of observations $n = 10$.Mean $\overline{x} = \frac{\sum x_i}{n} = 56$.$

Vedclass · Accepted Answer

$6344$

Vedclass · Answer

(C) Given data: $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$. Total number of observations $n = 10$.Mean $\overline{x} = \frac{\sum x_i}{n} = 56$.$\frac{65+68+58+44+48+45+60+60+\alpha+\beta}{10} = 56$$\frac{448+\alpha+\beta}{10} = 56$ $\Rightarrow 448+\alpha+\beta = 560$ $\Rightarrow \alpha+\beta = 112$.Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2 = 66.2$.$\frac{65^2+68^2+58^2+44^2+48^2+45^2+60^2+60^2+\alpha^2+\beta^2}{10} - (56)^2 = 66.2$.$\frac{4225+4624+3364+1936+2304+2025+3600+3600+\alpha^2+\beta^2}{10} - 3136 = 66.2$.$\frac{25678+\alpha^2+\beta^2}{10} = 3136 + 66.2 = 3202.2$.$25678+\alpha^2+\beta^2 = 32022$.$\alpha^2+\beta^2 = 32022 - 25678 = 6344$.

$X$	$35$	$54$	$52$	$53$	$56$	$58$	$52$	$50$	$51$	$49$
$Y$	$108$	$107$	$105$	$105$	$106$	$107$	$104$	$103$	$104$	$101$

If the mean and variance of the data $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$ where $\alpha > \beta$ are $56$ and $66.2$ respectively,then $\alpha^2 + \beta^2$ is equal to

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