Question

Suppose that weekly income of migrant workers doing agricultural labor in Florida has a distribution with a mean of $520 and a standard deviation of $90. A researcher randomly selected a sample of 100 migrant workers. What is the probability that sample mean is less than $500

Answer 1

`z = (500-520)/((90)/(√(100)))= -2.22`

And we can find this probability using the normal standard distribution and we got:

`P(z<-2.22) =0.0132`

Explanation:

For this case we have the foolowing parameters given:

`\mu = 520` represent the mean

`\sigma =90` represent the standard deviation

`n = 100` the sample size selected

And for this case since the sample size is large enough (n>30) we can apply the central limit theorem and the distribution for the sample mean would be given by:

`\bar X \sim N(\mu , (\sigma)/(√(n)))`

And we want to find this probability:

`P(\bar X <500)`

We can use the z score formula given by:

`z = (500-520)/((90)/(√(100)))= -2.22`

And we can find this probability using the normal standard distribution and we got:

`P(z<-2.22) =0.0132`