the ages of jamess five cousins are listed below.\n9 12 18 11 8\n1) calculate the mean.\nμ = \n2) fill in…

Explanation:

Step1: Calculate the mean

The formula for the mean $\mu$ of a set of data $x_1,x_2,\cdots,x_n$ is $\mu=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here, $n = 5$, $x_1=9,x_2 = 12,x_3=18,x_4=11,x_5 = 8$. So, $\mu=\frac{9 + 12+18+11+8}{5}=\frac{58}{5}=11.6$.

Step2: Calculate differences and squared - differences

For $x = 9$: $x-\mu=9 - 11.6=-2.6$, $(x - \mu)^2=(-2.6)^2 = 6.76$. For $x = 12$: $x-\mu=12 - 11.6 = 0.4$, $(x - \mu)^2=(0.4)^2=0.16$. For $x = 18$: $x-\mu=18 - 11.6 = 6.4$, $(x - \mu)^2=(6.4)^2 = 40.96$. For $x = 11$: $x-\mu=11 - 11.6=-0.6$, $(x - \mu)^2=(-0.6)^2 = 0.36$. For $x = 8$: $x-\mu=8 - 11.6=-3.6$, $(x - \mu)^2=(-3.6)^2 = 12.96$. And $\sum(x - \mu)^2=6.76+0.16 + 40.96+0.36+12.96=61.2$.

Step3: Calculate the population standard deviation

The formula for the population standard deviation $\sigma=\sqrt{\frac{\sum(x - \mu)^2}{N}}$, where $N = 5$ and $\sum(x - \mu)^2=61.2$. So, $\sigma=\sqrt{\frac{61.2}{5}}=\sqrt{12.24}\approx3.50$.

Answer:

1. $\mu = 11.6$

$x$	$x-\mu$	$(x - \mu)^2$
9	$-2.6$	$6.76$
12	$0.4$	$0.16$
18	$6.4$	$40.96$
11	$-0.6$	$0.36$
8	$-3.6$	$12.96$
	$\sum(x - \mu)^2=61.2$

3. $\sigma\approx3.50$