Question

The population mean annual salary for a environmental compliance specialist is about 63,000A random sample of 30 specialist is drawn from the population.Assume sigma =6,300 What is the probability that the sample mean is less than 60,000

Answer 1

Given:

`\mu=\text{ \$63000 ; n=30 ; }\sigma\text{=6300}`
`P(\bar{x}<60000)=P(Z<\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}})`
`P(\bar{x}<60000)=P(Z<\frac{60000-63000}{\frac{6300}{\sqrt[]{30}}})`
`P(\bar{x}<60000)=P(Z<\frac{-3000}{\frac{6300}{\sqrt[]{30}}})`
`P(\bar{x}<60000)=P(Z<-2.608)`
`P(\bar{x}<60000)=P(Z<-2.61)`

From the standard normal table

`P(Z<-2.61)=0.00453`

Therefore, 0.0045 is the probability of the given sample.