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help with statistics homework

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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 with a standard deviation of $37.000

Find the likelihood of selecting a sample with a mean of more than $106,000 but less than $120,ooo( Round z value to 2 decimal places and final answer to 4 decimal places

Oct 23rd, 2017

Z(120,000) = 120000-115000 / 37000 = 0.14

P(0.14) = 0.5557

Z(106,000) = 106000-115000 / 37000 = - 0.24

P(-0.24) = 0.4052

Therefore,

106,000 < P < 120,000 = 0.5557 - 0.4052 = 0.1505

Apr 11th, 2015

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