Question

The mean amount of life insurance per household is $111,000. This distribution is positively skewed. The standard deviation of the population is $39,000. Use Appendix B.1 for the z-values.

a. A random sample of 40 households revealed a mean of $115,000. What is the standard error of the mean? (Round the final answer to 2 decimal places.)

Standard error of the mean            

b. Suppose that you selected 115,000 samples of households. What is the expected shape of the distribution of the sample mean?

Shape             (Click to select)  Unknown  Normal  Not normal, the standard deviation is unknown  Uniform

c. What is the likelihood of selecting a sample with a mean of at least $115,000? (Round the final answer to 4 decimal places.)

Probability            

d. What is the likelihood of selecting a sample with a mean of more than $96,000? (Round the final answer to 4 decimal places.)

Probability            

e. Find the likelihood of selecting a sample with a mean of more than $96,000 but less than $115,000. (Round the final answer to 4 decimal places.)

Probability            

Answer 1

a)

std error=σx̅=σ/√n=

6166.44

b) Normal

c)

probability =P(X>115000)=P(Z>(115000-111000)/6166.44)=P(Z>0.65)=1-P(Z<0.65)=1-0.7422=0.2578

d)

probability =P(X>96000)=P(Z>(96000-111000)/6166.441)=P(Z>-2.43)=1-P(Z<-2.43)=1-0.0075=0.9925

e)

probability =P(96000<X<115000)=P((96000-111000)/6166.441)<Z<(115000-111000)/6166.441)=P(-2.43<Z<0.65)=0.7422-0.0075=0.7347