In: Statistics and Probability
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?
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