Answer
a) 6.9
b) 6.9
c) 3.4
d) Use $s^2$; use $n-1$
e) 1.3; 1.9
Work Step by Step
a) We know that the standard deviation is equal to the quantity of the distance each point is away from the average squared over the number of points in the set. Doing this, we find:
$\sigma^2 = \frac{20.667}{3}=6.9$
b) We know the following equation for sample variance:
$s = \sqrt{ \frac{n[\Sigma(f\cdot x^2]-[\Sigma (f\cdot x)^2]}{n(n-1)}}$
Plugging in the known values, it follows:
$s=6.9$
c) We know that the standard deviation is equal to the quantity of the distance each point is away from the average squared over the number of points in the set. Thus, we find:
$mean = \frac{.25+.25+6.25+6.25+9+9}{9}=3.4$
d) We see that the $s^2$ is the best estimate; we also see that it should be $n-1$, for there are less than 9 data points.
e) We first find the standard deviations. After ding this, we add them and divide by 9 to find:
$\bar{\sigma}=\frac{.5+3+.5+2.5+3+2.5}{9}=1.3$
$\bar{s}=\frac{3.5+4.2+.7+3.5+4.2+.7}{9}=\frac{16.8}{9}=1.9$
