# Confidence Interval

**Topics:**Normal distribution, Standard deviation, Sample size

**Pages:**3 (1075 words)

**Published:**May 29, 2011

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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