Answer:
Probability that the mean outstanding bill is over $200 is 0.0526.
Step-by-step explanation:
We are given that the average outstanding bill for delinquent customer accounts for a national department store chain is $187.50 with standard deviation $54.50.
Also, a random sample of 50 delinquent accounts is selected.
<em>Firstly, Let </em><em> = mean outstanding bill</em>
The z score probability distribution for is given by;
Z = ~ N(0,1)
where, = average outstanding bill = $187.50
= standard deviation = $54.50
n = sample of delinquent accounts = 50
Probability that the mean outstanding bill is over $200 is given by = P( > $200)
P( > 200) = P(
>
) = P(Z > 1.62) = 1 - P(Z
1.62)
= 1 - 0.94738 = 0.0526
Therefore, probability that the mean outstanding bill is over $200 is 0.0526.