Question

In: Statistics and Probability

A CPA knows from past history that the average accounts receivable for a company is $521.72...

A CPA knows from past history that the average accounts receivable for a company is $521.72 with a standard deviation of $584.64. If the auditor takes a simple random sample of 100 accounts, what is the probability that the mean of the sample is within $120 of the population mean?

Solutions

Expert Solution

Solution :

Given that ,

mean = = 521.72

standard deviation = = 584.64

n = 100

=   =521.72

= / n = 584.64 / 100 = 58.464

within 120 = 521.72 ± 521.72 = 401.72, 641.72

P(401.72 < < 641.72)  

= P[(401.72 - 521.72) / 58.464 < ( - ) / < (641.72 - 521.72) / 58.464)]

= P(-2.05 < Z < 2.05)

= P(Z < 2.05) - P(Z < -2.05)

Using z table,  

= 0.9798 - 0.0202

= 0.9596

Probability =0.9596


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