STATISTICS: A dentist is interested in obtaining information about delinquent (past due) accounts.

Question

Since the practice opened 7 years ago, delinquent accounts are normally distributed with an average of 44 days and a variance of 121.00 days. The dentist randomly selected a sample of 28 delinquent accounts with an average of 47 days.

A)The dentist wants to know what average number of days is less than 56% from the past 7 years.

B) Is the average days from the sample of 28 less than the 56% average from the past 7 years?

N/A

Yes

No

Answer 1

With the accounts being normally distributed we can use Z scores to answer part A)

Part A) is asking P(X<x) = 56%, where X = average number of days delinquent. Since we know the data comes from a normal distribution and that we can convert our random variable X to the scores on a Z distribution table, we first want to know P(Z<z) = 56%, so if we check a normal distribution table we will find that a Z-score of 0.15 has a probability of .56, i.e. P(Z<0.15) = 56%. Now we can figure out our value for x, because Z = (x-mean)/standard deviation = (x-44)/121 we use 44 and 121 because they are the parameters of our normal distribution.

So 0.15 = (x-44)/121

0.15*121 = x-44

(0.15*121)+44 = x = 62.15

P(X<62.15) = 65%, so 65% of delinquent accounts will have an average of 62.15 days or less

Part B)

We obtain a value of 62.15 days from Part A) and the sample mean form the 28 accounts is 47 days,

So since 62.15 days is larger than 47 days the answer would be Yes.