Question

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Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000 with a standard deviation of $40,000. Assume the population distribution is normal. A random sample of 100 households is taken.

1. What is the probability that sample mean will be more than $120,000?

2. What is the probability that sample mean will be between $100,000 and $120,000?

Answer 1

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Given- u=110000 6=40000 n=100 PCX 120000) = NORM DIST(120.000, 110000 400 false) Oooooo PC 10o00o < X < 120000) =NORNDIST (120000, 110000, 4000, Tue) NORM DIST (100 000, 110000, 400 E1-0.0000 = 1 + True

Sorry by mistake now I am using standard deviation 4000

P(xbar>120000)=NORMDIST(120000,110000,4000, FALSE )= 0.0062

P(xbar>120000)=NORMDIST(120000,110000,4000, FALSE ) - NORMDIST(100000,110000,4000, FALSE )= 0.9938 - 0.0062 = 0.9876