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

Answer:

$z = \frac{500-520}{\frac{90}{\sqrt{100}}}= -2.22$

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

$P(z$

Step-by-step 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:

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

And we want to find this probability:

$P(\bar X$

We can use the z score formula given by:

$z = \frac{500-520}{\frac{90}{\sqrt{100}}}= -2.22$

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

$P(z$