Question

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000.Refer to Exhibit 6-4. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000

Answer 1

`P(X>30000)=P((X-\mu)/(\sigma)>(30000-\mu)/(\sigma))=P(Z<(30000-40000)/(5000))=P(z>-2)`

And we can find this probability using the complement rule and the normal standard distribution

`P(z>-2)=1-P(z<-2)= 1-0.02275 = 0.97725`

Explanation:

Let X the random variable that represent the starting salaries of individuals with a MBA of a population, and for this case we know the distribution for X is given by:

`X \sim N(40000,5000)`

Where
`\mu=40000` and
`\sigma=5000`

We are interested on this probability

`P(X\geq 30000)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

Using this formula we got:

`P(X>30000)=P((X-\mu)/(\sigma)>(30000-\mu)/(\sigma))=P(Z<(30000-40000)/(5000))=P(z>-2)`

And we can find this probability using the complement rule and the normal standard distribution

`P(z>-2)=1-P(z<-2)= 1-0.02275 = 0.97725`