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

Answer:

$P(X>30000)=P(\frac{X-\mu}{\sigma}>\frac{30000-\mu}{\sigma})=P(Z<\frac{30000-40000}{5000})=P(z>-2)$

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

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

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

Using this formula we got:

$P(X>30000)=P(\frac{X-\mu}{\sigma}>\frac{30000-\mu}{\sigma})=P(Z<\frac{30000-40000}{5000})=P(z>-2)$

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

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