Question

A dentist is interested in obtaining information about delinquent (past due) accounts. Since the practice opened 12 years ago, delinquent accounts are normally distributed with an average of 50 days and a variance of 81.00 days. The dentist randomly selected a sample of 36 delinquent accounts with an average of 47 days.

a) The dentist wants to know what average number of days is less than 45% from the past 12 years.

b) Is the average days from the sample of 36 more than the 45% average from the past 12 years? (yes) (no) (NA)

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	50
std deviation   =σ=	9.000

sample size       =n=	36
std error=σx̅=σ/√n=	1.5000

a)

for 45th percentile critical value of z=		-0.13
therefore corresponding value=mean+z*std deviation=50-0.13*1.5 =			49.81 days

b)

since 47 is less than 45% average from the past 12 years ,

therefore No