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

Answer:

$z=\frac{510-520}{\frac{90}{\sqrt{100}}}= -1.11$

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

$P(z<-1.11)= 0.1335$

Step-by-step 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:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

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

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score formula we got:

$z=\frac{510-520}{\frac{90}{\sqrt{100}}}= -1.11$

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

$P(z<-1.11)= 0.1335$