Step 1

Solution:

Let X be the value.

From the given information, X follows normal distribution with mean \(\displaystyle\mu={182.5}\) and a standard deviation \(\displaystyle\sigma={49.4}\). The sample size is 15.

Step 2

The probability that a single randomly selected value is greater than 169.7 is

\(\displaystyle{P}{\left({X}{>}{169.7}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{>}{\frac{{{169.7}-{182.5}}}{{{49.4}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{0.259}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}{<}-{0.259}\right)}\)

\(\displaystyle={1}-{0.3978}={0.6022}\begin{bmatrix}Using\ the\ excel\ function \\=NORM.DIST(-0.259,0,1,TRUE) \end{bmatrix}\)