ime261a6s - IME 2610 Fall 2010 Assignment 6 Material...

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IME 2610 - Fall 2010 Assignment 6 Material Covered: Chapter 5 Directions: Please complete the three problems listed below. SHOW YOUR WORK! From your text: Problem 5.16 (pg. 198-199) Binomial probability of correct answer = ¼ mean = np = 100*¼ = 25 standard deviation = np(1-p) = 100*.25(1-.25) = 4.33 a) P(X>30) 30-25 4.33 = 1.15 P(Z>1.15) = 0.5 P(0 < Z < 1.15) = 0.5 0.3749 = 0.1251 b) P(X<20) 20-25 4.33 = -1.15 P(Z<-1.15) = 0.5 P(-1.15 < Z < 0) = 0.5 - .3749 = 0.1251 c) P(15<X<35) 15-25 4.33 = -2.31 15-25 4.33 = 2.31 P(-2.31 < Z < 2.31) = P(-2.31< Z<0) + P(0<Z<2.31) = 0.4896 + 0.4895 = 0.9792 d) P(X > 60) 60-25 4.33 = 8.08 P(Z > 8.08) = 0.5 P(0 < Z < 8.08) = 0.5 0.5 = 0
_____________________________________________________________________________________ From your text: Problem 5.26 (pg. 206) (Construct the Normal Probability Plot only) ____________________________________________________________________________________ From your text: Problem 5.44 (pg. 218) Mean of sample averages = 40 Standard deviation of sample averages = 5 2 100 = 5 100 = 0.5 a) P( X > 38.5) 38.5-40 0.5 = -3 P(Z >-3) = 0.5 + P(-3< Z <0) = 0.5 + 0.49865 = .99865 b) P( X > x ) = 0.95 0.95 - 0.5 = 0.45 Look up this number in table Using the table the z-value = -1.645 -1.645 = x - 40 0.5 -1.645(0.5) + 40 = 39.18 c) We are assuming the sample means are normally distributed. This assumption is met because the sample is large enough. 1.600 1.595 1.590 1.585 1.580 1.575 1.570 99 95 90 80 70 60 50 40 30 20 10 5 1 Voltage Percent Mean 1.582 StDev 0.004949 N 24 AD 0.311 P-Value 0.528 Probability Plot of Voltage Normal - 95% CI

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