Question 7 of 8) 12.50 points Exercise 7-42 The mean amount of life insurance pe

Question

Question 7 of 8) 12.50 points Exercise 7-42 The mean amount of life insurance per household is $128,000. This distribution is positively skewed. The standard deviation of the population is $46,000. Use Appendix B.1 for the values. a. A random sample of 55 households revealed a mean ofs131.000. 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 131,000 samples of households. What is the expected shape of the distribution of the sample mean? Sample mean (Click to select) c. What is the likelihood of selecting a sample with a mean of at least S131.000? ound the final answer to 4 decimal places.) Sample mean d. a sample with a mean of more than S113000? (Round the final answer to 4 decimal places) Sample mean e. Find the likelihood of selecting a sample with a mean of more than $113,000 but less than $131.000. (Round the final answer to 4 decimal places.) Sample mean

Explanation / Answer

Result:

a).

standard error = sd/sqrt(n) = 46000/sqrt(55) =6202.6387

6202.64 ( two decimals)

b).

Approximately normal.

c).

z value for 131000, z =(131000-128000)/ 6202.64 =0.48

P( x 131000) = P( z >0.48)

=0.3156

d).

z value for 113000, z =(113000-128000)/ 6202.64 =-2.42

P( x 113000) = P( z >-2.42)

=0.9922

e).

P( 113000<x<131000) = P( -2.42<x<0.48)

P( z <0.48) – P( z< -2.42)

=0.6844-0.0078

=0.6766

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