Step 1

The Z-score of a random variable X is defined as follows:

\(\displaystyle{Z}=\frac{{{X}-\mu}}{\sigma}\)

Here, \(\displaystyle\mu{\quad\text{and}\quad}\sigma\) are the mean and standard deviation of X, respectively.

Step 2

Consider a random variable X, that defines the variable of interest.

According to the given information, X follows normal distribution with mean \(\displaystyle\mu_{{{x}}}={136.4}\) and the standard deviation of \(\displaystyle\sigma_{{{x}}}={30.2}\).

The probability that a single randomly selected value is greater than 135 is,

\(\displaystyle{P}{\left({X}{>}{135}\right)}={1}-{P}{\left({\frac{{{X}-\mu}}{{\sigma}}}\leq{\frac{{{135}-{136.4}}}{{{30.2}}}}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}\leq-{0.046357615}\right)}\)

\(=1-0.4815 \begin{bmatrix}using\ the\ Excel\ formula\\ =NORM.S.DIST(-0.046357615,TRUE) \end{bmatrix} =0.5185\)

Therefore, the probability that a single randomly selected value is greater than 135 is 0.5185.