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emmainna [20.7K]
3 years ago
13

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
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
6 0

Answer:

P(X>30000)=P(\frac{X-\mu}{\sigma}>\frac{30000-\mu}{\sigma})=P(Z-2)

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

P(z>-2)=1-P(z

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-2)

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

P(z>-2)=1-P(z

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