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 $510

Answer 1

`z=(510-520)/((90)/(√(100)))= -1.11`

And we can find the probability using the normal standard distribution table and with the complement rule we got:

`P(z<-1.11)= 0.1335`

Explanation:

For this problem we have the following parameters:

`\mu = 520, \sigma = 90`

We select a sample size of n =100 and we want to find this probability:

`P(\bar X <510)`

The distribution for the sample mean using the central limit theorem would be given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

And we can solve this problem with the z score formula given by:

`z=(\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score formula we got:

`z=(510-520)/((90)/(√(100)))= -1.11`

And we can find the probability using the normal standard distribution table and with the complement rule we got:

`P(z<-1.11)= 0.1335`