Question

An auditor knows from past history that the average accounts receivable for ABC Company is $441.27 with a standard deviation of $555.48. Suppose the auditor takes a sample of 100 accounts.

a. What is the probability that the mean of the sample will be within $100 of the population mean?

b. What is the probability that the mean of the sample will more than $580?

Answer 1

(a)

= 441.27

= 555.48

n = 100

SE = /

= 555.48/ = 55.548

To find P(341.27 < < 541.27):

Case 1: For from 341.27 to mid value:

Z = (341.27 - 441.27)/55.548 = - 1.80

Table of Area Under Standard Normal Curve gives area = 0.4641

Case 2: For from mid value to 541.27:

Z = (541.27 - 441.27)/55.548 = 1.80

Table of Area Under Standard Normal Curve gives area = 0.4641

So,

P(Sample within $100) = 0.4641X 2 = 0.9282

So,

Answer is:

0.9282

(b)

To find P(>580):

Z = (580 - 441.27)/55.548 = 2.50

Table of Area Under Standard Normal Curve gives area = 0.4938

So,

P(>580) = 0.5 - 0.4938 = 0.0062

So,

Answer is:

0.0062