Question

The mean salary offered to students who are graduating from Coastal State University this year is $ 24,260 , with a standard deviation of $ 3643 . A random sample of 80 Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for these 80 students is $ 24,500 or less? Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Answer 1

Let X be the salary offered to students who are graduating from coastal state university.

Given:

$\mu=24260$

$\sigma=3643$

n=80

The probability that the mean salary offer for these 80 students is $24500 or less given by

$P(\bar{x} \leq 24500)=P(\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \leq \frac{24500-\mu}{\frac{\sigma}{\sqrt{n}}})$

$P(\bar{x} \leq 24500)=P(z \leq \frac{24500-24260}{\frac{3643}{\sqrt{80}}})$

$P(\bar{x} \leq 24500)=P(z \leq \frac{240}{\frac{3643}{8.9443}})$

$P(\bar{x} \leq 24500)=P(z \leq \frac{240}{407.2985})$

$P(\bar{x} \leq 24500)=P(z \leq 0.5892)$

$P(\bar{x} \leq 24500)=P(z \leq 0.59)$

$P(\bar{x} \leq 24500)=0.7224$

### by using z table

Hence required probability is:

$P(\bar{x} \leq 24500)=0.7224$