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The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $2000. What is the probability of randomly selecting one employee who earned less than or equal to $45,000

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Answer:

The probability of randomly selecting one employee who earned less than or equal to $45,000=0.00621

Step-by-step explanation:

We are given that

Mean,[tex]\mu=50000[/tex]

Standard deviation,[tex]\sigma=2000[/tex]

We have to find the probability of randomly selecting one employee who earned less than or equal to $45,000.

[tex]P(x\leq 45000)=P(\frac{x-\mu}{\sigma}\leq \frac{45000-50000}{2000})[/tex]

[tex]P(x\leq 45000)=P(Z\leq-\frac{5000}{2000})[/tex]

[tex]P(x\leq 45000)=P(Z\leq -2.5)[/tex]

[tex]P(x\leq 45000)=0.00621[/tex]

Hence,  the probability of randomly selecting one employee who earned less than or equal to $45,000=0.00621

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