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This window shows what is correct and incorrect for the work you have completed so far Even iall of the work you have done so far is correct, you may not have completed everything 4. 2.00 points Information from the American Institute of Insurance indicates the mean amount of ide insurance per household in the United States is $102,000. This distributon follows the normal distribution with a standand deviation of $30,000 (a) H we select a random sample of 40 households, what is the standard error of the mean? (Round your answer to the nearest whole number.) Standard error of the mean 4743 0 (b) What is the expected shape of the distribution of the sample mean? ONot normal, the standard deviation is unknown. OONormal (c)What is 1kolhood of selecting a sample wth a mean of at least $108,000? (Round z value to 2 decimal places and final answer to 4 decimal places.) (d) What is the ikelihood of selecting a sample with a mean of more than $96,000? (Round z value to 2 decimal places and final answer to 4 decimal places Probability8970 (e) Find the lkelhood of selecting a sample with a mean of more than $06,000 but less than $108,000 (Round z value to 2 decimal places and final answer to 4 decimal places) 3 4 6 WE

Explanation / Answer

mean = 102000

standard deviation = 30000

c)For x = 108000, z = (108000 - 102000) /30000 = 0.2

Hence P(x > 108000) = P(z >0. 2) = [total area] - [area to the left of0. 2]

= 1 - 0.5793 = 0.4207

d)

For x = 96000, z = (96000 - 102000) / 30000 = -0.2

Hence P(x >96000) = P(z > -0.2) = [total area] - [area to the left of -0.2]

= 1 - 0.4207 = 0.5793

e)

For x = 108000 , z = (108000 - 102000) /30000 = 0.2 and for x = 96000, z = (96000 - 102000) / 30000 = -0.2

Hence P(96000 < x < 108000) = P(-0.2 < z < 0.2) = [area to the left of z = 0.2] - [area to the left of -0.2]

= 0.5793 - 0.4207 = 0.1586

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