Information from the American Institute of Insurance indicates the mean amount o
ID: 3127812 • Letter: I
Question
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States a $102.000. The distribution follows the normal distribution with a standard deviation of $30.000 If we select a random sample of 40 households, whet is the standard error of the moan? (Round your answer to the nearest whole number.) What is the expected shape of the distibution of the sample mean? What is the likelihood of selecting a sample with a mean of at least $108.000? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability What is the likelihood of selecting a sample with s moan of more then $96.000? (Round z value to 2 decimal places and final answer to 4 decimal places.)Explanation / Answer
Given that, mean = $102,000
Standard deviation (sd) = $30,000
1) If we select a random sample of 40 households, what is the standard error of mean?
n = 40
standara error = sd / sqrt(n) = 30000 / sqrt(40) = 4743
2) The expected shape of the distribution of the sample mean is normal.
3) That is here we have to find P(Xbar >=108000).
First convert Xbar into z-score.
z = (x - mean) / se = (108000 - 102000) / 4743 = 1.31
Now we have to find P(Z >= 1.31).
This probability we can find by using EXCEL.
syntax,
=1 - NORMSDIST(z) (because EXCEL always gives left tail probability)
where z = 1.31
P(Z >= 1.31) = 0.0956
4) Here we have to find P(Xbar > 96000).
z = (96000 - 102000) / 4743 = -1.26
Now we have to find P(Z > -1.26).
By using EXCEL,
P(Z > -1.26) = 0.8970
5) Here we have to find P(96000 < Xbar < 108000).
Convert 108000 and 96000 into z-score,
z-score for x = 108000,
z1 = 1.31
z-score for x = 96000,
z2 = -1.26
Now we have to find P(-1.26 < Z < 1.31).
P(-1.26 < Z < 1.31) = P(Z <=1.31) - P(Z <=-1.26)
These probabilities we can find by using EXCEL.
syntax is,
=NORMSDIST(z)
where z is test statistic value.
P(-1.26 < Z < 1.31) = 0.9044 - 0.1038 = 0.8006