# Confidence Interval

Pages: 3 (1075 words) Published: May 29, 2011
34. Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is \$110,000. This distribution follows the normal distribution with a standard deviation of \$40,000. a. If we select a random sample of 50 households, what is the standard error of the mean? b. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least \$112,000? d. What is the likelihood of selecting a sample with a mean of more than \$100,000? e. Find the likelihood of selecting a sample with a mean of more than \$100,000 but less than \$112,000. a) The standard error of the mean, S.E. = /Sqrt(n) = 40,000/Sqrt(50) = 5656.8542 b) It will be normally distributed with mean \$110,000 and standard deviation \$5656.85. Note that Z = (Xbar – 110,000)/5656.85 follows a Standard Normal distribution. c) The likelihood of selecting a sample with a mean of at least \$112,000 is given by,

P[Xbar 112,000] = P[(Xbar–110,000)/5656.85 (112,000-110,000)/5656.85]
= P[Z 0.3536] = 1- P[Z < 0.3536]
= 1 – 0.6382 = 0.3618
d) The likelihood of selecting a sample with a mean of more than \$100,000 is given by,
P[Xbar > 100,000] = P[(Xbar–110,000)/5656.85 > (100,000-110,000)/5656.85]
= P[Z > -1.7678] = 1- P[Z < 0.3536]
= 1 – 0.0385 = 0.9615
e) The likelihood of selecting a sample with a mean of more than \$100,000 but less than \$112,000 is given by, P[100,000 < Xbar < 112,000]
= P[(100,000-110,000)/5656.85 < (Xbar–110,000)/5656.85 < (112,000- 110,000)/5656.85]
= P[-1.7678 < Z < 0.3536]
= P[Z < 0.3536] - P[Z<-1.7678]
= 0.6382 - 0.0385 = 0.5997
Exercise 32. A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled “3 pounds.” Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight...

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