91Ó°ÊÓ

To find the mean and standard deviation of all possible sample mean spans of binge drinking episodes for sample size $40$.

The mean $\left(\mathrm{μ}\right)$span of binge drinking episodes is $4.9$hours. The mean $\left({\mathrm{μ}}_X\right)$for sample size $40$:
${\mathrm{μ}}_x=\mathrm{μ}$
${\mathrm{μ}}_x=4.9$
For sample size 40, the standard deviation is $\left({\mathrm{μ}}_x\right)$.
The standard deviation $\left(\mathrm{σ}\right)$is assumed to be $1.1$ hours.
$\begin{array}{r}{\mathrm{σ}}_X=\frac{\mathrm{σ}}{\sqrt{n}}\end{array}$

$=\frac{1.1}{\sqrt{40}}$

As a result, the mean and standard deviation of all possible sample spans of binge drinking episodes for size $40$ binge drinkers are $4.9$hours and $0.17$ hours, respectively.

To find the mean and standard deviation of all possible sample mean spans of binge drinking episodes for sample size $n=120$.

Let, the Mean is $4.9$hours and the Standard deviation is$1.1$hours.

Determine the sample size $120$for the mean $\left({\mathrm{μ}}_X\right)$ as:
${\mathrm{μ}}_x=\mathrm{μ}$
${\mathrm{μ}}_x=44.8$
Then the standard deviation $\left({\mathrm{σ}}_X\right)$can be determine as:

$\begin{array}{r}{\mathrm{σ}}_X=\frac{\mathrm{σ}}{\sqrt{n}}\end{array}$
$=\frac{1.1}{\sqrt{120}}$
$=\frac{1.1}{6.3246}$
$=0.1004$

As a result, the mean and standard deviation of all potential sample spans of binge drinking episodes for samples of size $120$binge drinkers are $4.9$hours and $0.1004$hours, respectively.