Let x_1, x_2, , x_10 be ten observations such that _i=1^10(x_i-2) = 30 and _i=1^10(x_i-)^2…

Question

Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\displaystyle\sum_{i=1}^{10}(x_i-2) = 30$ and $\displaystyle\sum_{i=1}^{10}(x_i-\beta)^2 = 98$, $\beta > 2$, and their variance is $\dfrac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1-1)+4\beta, 2(x_2-1)+4\beta, \ldots, 2(x_{10}-1)+4\beta$, then $\beta\mu$ is equal to:

Accepted Answer

From $\sum(x_i-2)=30$: $\sum x_i = 50$, so mean $=5$.

Variance $=\dfrac{4}{5}$: $\dfrac{4}{5} = \dfrac{\sum x_i^2}{10} - 25 \implies \sum x_i^2 = 258$

From $\sum(x_i-\beta)^2 = 98$:
$$\sum x_i^2 - 2\beta\sum x_i+10\beta^2 = 98 \implies 258-100\beta+10\beta^2=98 \implies (\beta-8)(\beta-2)=0$$

Since $\beta>2$: $\beta = 8$.

The transformed data $2(x_i-1)+4\beta = 2x_i+30$ (using $\beta=8$).

$$\mu = 2(5)+30 = 40, \qquad \sigma^2 = 2^2\left(\frac{4}{5}\right) = \frac{16}{5}$$

The required quantity $\dfrac{\beta\mu}{\sigma^2} = \dfrac{8\times40}{16/5} = 100$.

Solutions · 1