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Sagot :
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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