# Chapter 3 - Statistics for Describing, Exploring, and Comparing Data - 3-3 Measures of Variation - Beyond the Basics - Page 111: 45

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 the quantity of the distance of each point from the average squared over the number of points in the set. Hence: $σ^2=\frac{20.667}{3}\approx6.9$ b) We know the sample variance can be computed by: $s=\sqrt{\frac{n[Σ(f)⋅x^2]−[Σ(f⋅x)^2]}{n(n−1)}}.$ By plugging in the known values we get $s=6.9$. c) We know that the standard deviation is the quantity of the distance of each point from the average squared over the number of points in the set. Hence: mean=$\frac{0.25+0.25+6.25+6.25+9+9}{9}=3.4$ d) We can see that the $s^2$ is the best estimate and we can also see that it should be $n−1$, because for there are less than $9$ data points. e) We first find the standard deviations. Then we add them and divide by $9$ to find: $\overline{σ}=\frac{0.5+3+0.5+2.5+3+2.5}{9}=1.3$ $\overline{s}=\frac{3.5+4.2+0.7+3.5+4.2+0.7}{9}=1.9$

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