A dentist is interested in obtaining information about delinquent (past due) accounts. Since the practice opened 8 years ago, delinquent accounts are normally distributed with an average of 43 days and a variance of 196.00 days. The dentist randomly selected a sample of 20 delinquent accounts with an average of 47 days.

The average number of days is more than 43% from the past 8 years is 43.15 days and it is less than Sample mean of 47 for 20 samples

Explanation:

From the data

Population mean = μ=43 days

Population variance==196

Sample size n=20

Sample mean =47

Part a:

Using values from the reference question found online.

Given α=43%

α=0.43

At 1-α=0.57

from the table yields 0.18

Now the value is given as

Part b:

Sample mean of 47 for 20 samples is more than 43% of the days from last 8 years.

0.1059

Explanation:

Please see attached a solved copy of the question

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