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Workers employed in a large service industry have an average wage of $9.00 per hour with a standard deviation of $0.50. The industry has 64 workers of a certain ethnic group. These workers have an average wage of $8.85 per hour. Calculate the probability of obtaining a sample mean less than or equal to $8.85 per hour. (Round your answer to four decimal places.)

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Answer:

The probability of obtaining a sample mean less than or equal to $8.85 per hour=0.0082

Step-by-step explanation:

We are given that

Average wage, [tex]\mu=[/tex]$9.00/hour

Standard deviation,[tex]\sigma=[/tex]$0.50

n=64

We have to find the  probability of obtaining a sample mean less than or equal to $8.85 per hour.

[tex]P(\bar{x} \leq 8.85)=P(Z\leq \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]

Using the values

[tex]P(\bar{x}\leq 8.85)=P(Z\leq \frac{8.85-9}{\frac{0.50}{\sqrt{64}}})[/tex]

[tex]P(\bar{x}\leq 8.85)=P(Z\leq \frac{-0.15}{\frac{0.50}{8}})[/tex]

[tex]P(\bar{x}\leq 8.85)=P(Z\leq -2.4)[/tex]

[tex]P(\bar{x}\leq 8.85)=0.0082[/tex]

Hence, the probability of obtaining a sample mean less than or equal to $8.85 per hour=0.0082

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