Question

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.

Answer 1

Answer:0.1059

Explanation:

Please see attached a solved copy of the question

Answer 2

Answer:

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=$\sigma^2$=196

                                  $\sigma=\sqrt{196}=14$

Sample size n=20

Sample mean $\bar{x}$=47

Part a:

Using values from the reference question found online.

Given α=43%

α=0.43

At 1-α=0.57

$z_{\alpha}$ from the table yields 0.18

Now the value is given as

$\bar{x}=\mu+z_{\alpha}\sqrt{\frac{\sigma}{n}}\\\bar{x}=43+0.18\sqrt{\frac{14}{20}}\\\bar{x}=43.15 days$

Part b:

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