Question

The salaries of Pet Detectives are normally distributed with a mean of $32,000 and a standard deviation of $3000. If 100 Pet Detectives are randomly selected, find the probability that their average salary is greater than $32,500. Select the single best answer choice

Answer 1

Answer:

$P(\bar X> 32500)$

And for this case we can use the z score formula given by:

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

And replacing the info given we got:

$z =\frac{32500-32000}{\frac{3000}{\sqrt{100}}}= 1.67$

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

$P(z> 1.67) = 1-P(z<1.67) = 1-0.953 = 0.047$

Step-by-step explanation:

Let X the random variable that represent the salaries of Pet Detectives of a population, and for this case we know the distribution for X is given by:

$X \sim N(32000,3000)$  

Where $\mu=32000$ and $\sigma=3000$

We select a sample size of n =100 Pet Detectives and we want to find the following probability:

$P(\bar X> 32500)$

And for this case we can use the z score formula given by:

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

And replacing the info given we got:

$z =\frac{32500-32000}{\frac{3000}{\sqrt{100}}}= 1.67$

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

$P(z> 1.67) = 1-P(z<1.67) = 1-0.953 = 0.047$