Answer:
A. Only 1.01% of samples of 30 specialists will have a mean salary less than $61,500. This is an unusual event.
Step-by-step explanation:
• Population Mean = 64,000
,
• Sample = 30
,
• Raw Score = 61,500
,
• Standard Deviation = 5,900
First, we find the z-score using the formula below:
![z=\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}}](https://img.qammunity.org/2023/formulas/mathematics/college/ok06jndjxvlf8gu2t8xz0qtman16lp42pn.png)
Substitute the given values:
![\begin{gathered} z=\frac{61500-64000}{5900/\sqrt[]{30}} \\ z=-2.32 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/l21cgbojf8a0kkmjq277fgxv91jemlj0po.png)
Next, from a z-score table:

Therefore, the probability that the mean salary of the sample is less than $61,500 is 0.0102.
Next, we interpret the result.
• As a general rule, z-scores lower than -1.96 or higher than 1.96 are considered unusual. Since -2.32 is lower than -1.96, it is unusual.
Only 1.01% of samples of 30 specialists will have a mean salary less than $61,500. This is an unusual event.