We will employ the z score table in this problem.
The shaded part is our area of interest.We will be required to seek the value of the z score at 0.11.
when we ahave a probability of 0.11 or 11%
This corresponds with a z-score of -1.225.
We then substitute into our z-score equation:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
where
[tex]\begin{gathered} z=z\text{ score} \\ x=\text{cut off point} \\ \mu=\operatorname{mean} \\ \sigma=\text{standard deviation} \end{gathered}[/tex]
Therefore, we have
[tex]\begin{gathered} x=\sigma z+\mu \\ x=52(-1.225)+470 \\ x=406.3 \end{gathered}[/tex]
