91Ó°ÊÓ

Consider the provided details, let X be the time taken to complete one review. Here, X is normally distributed with mean =$4$hours and standard deviation =$1.2$hours. The sample size is $16$.

Consider X be the time taken to complete one review.

Here, X remains normally distributed

The mean =$4$hours

The standard deviation =$1.2$hours. The sample size is $16$.

Let $\stackrel{-}{x}$be the mean time to complete $16$reviews

The random variable localid="1648622508636" $\stackrel{¯}{X}−N\left({\mathrm{μ}}_X,\frac{{\mathrm{σ}}_X}{\sqrt{n}}\right)$

Therefore,

$\stackrel{¯}{X}~N(4,\frac{1.2}{\sqrt{16}})$

$\stackrel{¯}{X}−N(4,0.30)$

a. Graph with the shaded region for the probability that mean time to complete $16$reviews will take between $3.5$to $4.25$hours is given as below:

P(________________) = _______

Use the Ti-83 calculator to compute the value of $P(3.5<\stackrel{¯}{X}<4.25)$

To find this, click on $2$^nd, then DISTR, and then scroll down to the normal CDF option and enter the supplied details.

After this let's click on ENTER button on the calculator to have the chosen result.

The result of the screenshot is given as below:

Therefore the probability is $P(3.5<\stackrel{¯}{X}<4.25)=0.7499$