Question

evaluate $sigma=sqrt{\frac{sum (x - mu)^2}{n}}$ given $mu = 7.32$, $n = 6$, and the x - values in the table.

x

7.9

8.5

6.8

6.2

7.4

7.1

$sigmaapproxsquare$ (round to two decimal places as needed.)

Explanation:

Step1: Calculate $(x - \mu)^2$ for each $x$ - value

For $x = 7.9$: $(7.9 - 7.32)^2=(0.58)^2 = 0.3364$
For $x = 8.5$: $(8.5 - 7.32)^2=(1.18)^2 = 1.3924$
For $x = 6.8$: $(6.8 - 7.32)^2=(- 0.52)^2 = 0.2704$
For $x = 6.2$: $(6.2 - 7.32)^2=(-1.12)^2 = 1.2544$
For $x = 7.4$: $(7.4 - 7.32)^2=(0.08)^2 = 0.0064$
For $x = 7.1$: $(7.1 - 7.32)^2=(-0.22)^2 = 0.0484$

Step2: Calculate $\sum(x - \mu)^2$

$\sum(x - \mu)^2=0.3364 + 1.3924+0.2704 + 1.2544+0.0064 + 0.0484=3.3084$

Step3: Calculate $\frac{\sum(x - \mu)^2}{N}$

Given $N = 6$, $\frac{\sum(x - \mu)^2}{N}=\frac{3.3084}{6}=0.5514$

Step4: Calculate $\sigma$

$\sigma=\sqrt{0.5514}\approx0.74$

Answer:

$0.74$