Answer
a) 4.7
b)
Variance, probability:
0,3/9
.5,2/9
8,2/9
12.5, 2/9
c) 4.7
d) Yes
Work Step by Step
a. We know that the average is: 6. Thus, we find:
$\sigma^2 = \frac{\Sigma(x-\bar{x})^2}{n}=4.7$
b. We know:
$s^2 = \frac{n[\Sigma(f\cdot x^2]-[\Sigma (f\cdot x)^2]}{n(n-1)}$
Thus, we obtain:
Variance, probability:
0,3/9
.5,2/9
8,2/9
12.5, 2/9
c. Using the equation for mean, we see that the mean of the above table is 4.7.
d. Yes, it is. After all, we see that it has the same value as $\sigma^2$, so it follows that it is an unbiased predictor.
