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**Bushy** The head of a business school claims that the average MBA graduate is offered a starting salary of $55,000. The standard deviation of the offers is $4,600.

**a) What is the probability that in a random sample of 38 MBA graduates, the mean starting salary is less than $53,000?**

I am saying that

$\displaystyle P( \bar{X}< 53,000) = P\left(Z<\frac{53,000-55,000}{\frac{4,600}{\sqrt{38}}}\right)= \dots$

**b) What if only 16 students had been sampled?**

This is confusing, can I say

$\displaystyle P(\bar{X}<53,000) = P\left(Z<\frac{53,000-55,000}{\frac{4,600}{\sqrt{16}}}\right)= \dots$ ??

as $\displaystyle \sigma$ is known, can $\displaystyle Z$ be used for $\displaystyle n=16$ ?