Suppose the average client charge per hour for out-of-court work by...

Question

Suppose the average client charge per hour for out-of-court work by lawyers in the state of Iowa is \$125. Suppose further that a random telephone sample of 32 lawyers in Iowa is taken and that the sample average charge per hour for out-of-court work is \$110. If the population variance is \$525, what is the probability of getting a sample mean of \$110 or larger? What is the probability of getting a sample mean larger than \$135 per hour? What is the probability of getting a sample mean of between \$120 and \$130 per hour?

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Solution:
Here we have, Population Mean (µ) = 125, Population Variance (σ²)= 525,
Population Standard Deviation (σ) = sqrt(σ²) = 22.9129, n = 32

Z score is defined as

Z = (µ-Sample Mean)/ (σ/sqrt(n)) or Z = (x-bar - mu)/(sigma/sqrt (n))

We need to find

1) what is the probability of getting a sample mean of \$110 or larger?...

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