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##### help with statistics homework

label Statistics
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Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the U.S is \$115,000. This distribution follows the normal distribution of \$\$

c. What is the likelihood of selecting a sample with a mean of at least \$120,000 (Round z values to 2 decimal places and final answer to 4 decimal places)

d) What is the likelihood of selecting a sample with a mean of more than \$106,000? (Round z values to 2 decimal places and final answer to 4 decimal places)

Oct 20th, 2017

Assuming the same standard deviation as from the other problem:

c.  Z(120000) = 120000-115000 / 37000 = 0.14

P(Z=0.14) = 0.5557

But we want AT LEAST \$120,000 which is the right portion of the normal curve

So, the correct P value is 1 - 0.5557 = 0.4443

d.  Z(106000) = 106000 - 115000 /  37000 = - 0.24

P(Z= -0.24) = 0.4052

But we want MORE than \$106,000 so again we are looking at the right portion of the normal curve

So, P = 1- 0.4052 = 0.5948

Apr 11th, 2015

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Oct 20th, 2017
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Oct 20th, 2017
Oct 20th, 2017
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