Question

A recent study by the Island Resort Taxi Drivers Association showed that the mean fare charged for service from the beach to the airport is $29 and the standard deviation is $4.5. We select a sample of 18 fares. Use Appendix B.1 for the z-values.

a. What is the likelihood that the sample mean is between $26 and $31? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)

Probability            

b. What must you assume to make the above calculation?

(Click to select)  Assume the population is normally distributed.  Assume the population is uniformly distributed.  Assume the population is not normally distributed.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	29
std deviation   =σ=	4.500
sample size       =n=	18
std error=σx̅=σ/√n=	1.06066

likelihood that the sample mean is between $26 and $31:

probability =P(26<X<31)=P((26-29)/1.061)<Z<(31-29)/1.061)=P(-2.83<Z<1.89)=0.9706-0.0023=0.9683

b)

  Assume the population is normally distributed.