Probability question?

Question

An auditor knows from past history that the average accounts receivable for ABC Company is $348.23 with a standard deviation of $462.44. If the auditor takes a sample of 100 accounts, what is the probability that the mean of the sample will be within $100 of the population mean?

Answer 1

The auditors best estimate of the sd for a 100 account sample is sqrt(462.44^2/99) or 44 (appx). The z-score for the sample mean+ or - 100 is 100/44 or about 2.3. So the probability will correspond to a z-score of 4.6 (very high)

Answer 2

You need to use the Z table since the sample size is rather large. The z score you are looking up is z <= 100/462.44 = .216 and z >= -.216 which is the z score equivalent of between +- $100 on the actual accounts scale.

Go to the standard normal probability calculator found at the cited source. [See source.] Calculate P(z > .216) as specified by the calculator. The probability answer represents the area under the normal curve between the z score and the RHS tail of the curve. Since the curve is symmetric, it also represents the are of the curve between minus the z score and the LHS tail of the curve.

Since the total area under a standard normal curve is 1.00, the area between the two z scores (which represent +- $100 from the average ($348.23) is 1.00 - 2P(z > .216); and that's the probability that the mean of the sample will be within the average. You can use the table and do the math. (Hint: it'll be a low probability because your z score is quite close to the standard normal mean.)

PS: The given standard deviation is greater than the given mean. This is unlikely if the distribution is normal. Roughly, the width of a normal curve is six standard deviations (thus the six sigma models you may have heard about). Three sigmas less than the $348 average would put the value into a negative realm...not a very likely oucome in accounts receivable. In any case, a $100 variation around the mean ($348) would hover close to the mean; certainly less than plus and minus one sigma. That would carve out a low area (low probability); not a high probability as suggested elsewhere.