33 OMIS 1000 C - Assignment 3 - Solution

33 OMIS 1000 C - Assignment 3 - Solution - OMIS 1000 C...

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OMIS 1000 C I NTRODUCTION TO S TATISTICS F ALL 2007 A LEXANDER S HOUMAROV A SSIGNMENT 3 S OLUTION Total of 23 marks Question 1 (6 marks: 2 for part a, 1 for the rest) p = 0.6 n = 20 a. E(x) = (0.6)(20) = 12 SD(x) = 1909 . 2 4.8 0.4) (20)(0.6)( npq = = = b. P(x=15) = 0.0746 c. P(x≤15) = 0.9490 d. P(10<x<20) = 0.7553 e. Can determine this as the probability of less than 15 ‘failures’ or more than 5 ‘successes’: P(x<15|p=0.4) = 0.9984 P(x>5|p=0.6) = 0.9984 You easily see why both solutions work by looking at the binomial table for n=20 and p=0.4 or 0.6. Both probabilities above are found by summing exactly the same values. Question 2 (3 marks) We’re looking for that point on the x-axis that has 1% of the data below it and 99% of the data above it, i.e. the 1 st percentile. This point will lie to the left of the mean. Z = -2.326 X = μ – (Z)(σ) = 75 – (2.326)(8) = 75 – 18.608 = 56.392 months ≈ 56 months and 12 days (it is not necessary to convert to days) Question 3 (total of 6 marks: 2 marks for each part) 6.22. a. P( x < 9.6) = P(z < (9.6 – 10)/(1.75/ 25 ) = P(z < -1.14) = 0.5 – 0.3729 = 0.1271
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b. P( x > 9.6) = P(z > (9.6 – 9)/(1.75/ 25 ) = P(z > 1.71) = 0.5 – 0.4564 = 0.0436 c. n x z σ μ - = ; -1.645 =
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