Let the mean and the variance of 5 observations x_1,x_2,x_3,x_4,x_5 be 245 and 19425 respe…

Question

Let the mean and the variance of 5 observations $x_1,x_2,x_3,x_4,x_5$ be $\dfrac{24}{5}$ and $\dfrac{194}{25}$ respectively. If the mean and variance of the first 4 observations are $\dfrac72$ and a respectively, then $(4a+x_5)$ is equal to:

Accepted Answer

$\sum_{i=1}^5 x_i = 24$; first four sum to $4\times\dfrac72=14 \implies x_5=10$

$\sum x_i^2 = 5\left(\dfrac{194}{25}+\dfrac{576}{25}\right)=154$

Sum of squares of first four $=154-100=54$

$$a = \frac{54}{4}-\left(\frac72\right)^2 = \frac{54}{4}-\frac{49}{4}=\frac54$$

$$4a+x_5 = 4\times\frac54+10 = 5+10=15$$

Solutions · 1