(b) 99%; n = 15; σ is unknown; population appears to be normally distributed.

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1 MTH 345 Exam 3 Fall 2013 Jutify all anwer with neat and organized work. Clearly indicate your anwer. 100 point poible. 1. (12 pt.) Women height are normally ditributed with mean 63.6 in. and tandard deviation 2.5 in. A Tall Club ha a memberhip requirement that women mut be at leat 68.5 in. tall. What percentage of women meet that requirement? 2. (12 pt.) You are the operation manager for American Airline and you are conidering a higher fare level for paenger in aile eat. You want to etimate the percentage of paenger who now prefer aile eat. How many randomly elected air paenger mut you urvey? Aume that you want to be 80% confident that the ample percentage i within 1.5 percentage point of the true population percentage. Aume that nothing i known about the percentage of paenger who prefer aile eat. 1

2 3. (12 pt.) In a KRC Reearch poll of 1002 ubject, 531 of them reported that they felt vulnerable to identity theft. Find a 99% confidence interval etimate of the proportion who felt vulnerable to identity theft. 4. (12 pt.) In each problem, aume we want to contruct a confidence interval uing the given confidence level. Chooe and do only one of thee three choice, whichever i appropriate. If the normal (z) ditribution hould be ued, then find the critical value z α/2. If the t ditribution hould be ued, then find the critical value t α/2. If neither the normal (z) nor the t ditribution applie, then tate thi. (a) 95%; n = 20; σ i unknown; population appear to be normally ditributed. (b) 99%; n = 15; σ i unknown; population appear to be normally ditributed. 2

3 5. (12 pt.) Standing eye height of men are normally ditributed with a mean of 1634 mm and a tandard deviation of 66 mm (baed on anthropometric urvey data from Gordon, Churchill, et al.). A window i poitioned to be comfortable for the lowet 89% of eye height of men. What tanding eye height of men eparate the lowet 89% of tanding eye height from the highet 11%? 6. (12 pt.) Aume that weight of male are normally ditributed with a mean of lb and a tandard deviation of 40.8 lb. If an elevator i loaded with 16 male paenger, find the probabilty that it i overloaded becaue they have a mean weight greater than lb. 3

4 7. (12 pt.) The lited value are waiting time (in minute) of cutomer at the Jefferon Valley Bank, where cutomer enter a ingle waiting line that feed three teller window. (Aume that a imple random ample i elected from a normally ditributed population.) Contruct a 90% confidence interval for the population tandard deviation σ (8 pt.) Baed on a Bellowe urvey of adult, there i a 0.48 probability that a randomly elected adult ue a tax preparer to file taxe. Find the probability that among 16 randomly elected adult, exactly 9 ue tax preparer to file their taxe. 4

5 9. (8 pt.) The accompanying table decribe reult from group of 10 birth from 10 different et of parent. The random variable x repreent the number of girl among 10 children. x P (x) (a) Find the probability of getting exactly 8 girl in 10 birth. (b) Find the probability of getting 8 or more girl in 10 birth. (c) Which probability i relevant for determining whether 8 i an unuually high number of girl in 10 birth: the reult from part (a) or part (b)? 5

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7 Formula Ch 3: Decriptive Statitic x = Σx n Mean x = Σf x Mean (frequency table) Σf Σ(x x) = 2 Sample tandard deviation n 1 n(σx = 2 ) (Σx) 2 Sample tandard deviation (hortcut) n(n 1) n[σ(f x = 2 )] [Σ(f x)] 2 Sample tandard deviation (frequency table) n(n 1) Sample variance = 2 z = x x or x µ σ x = x + z or x = µ + zσ Ch 4: Probability Standard core P (A or B) = P (A) + P (B) if A, B are mutually excluive P (A or B) = P (A) + P (B) P (A and B) if A, B are not mutually excluive P (A and B) = P (A) P (B) if A, B are independent P (A and B) = P (A) P (B A) if A, B are dependent P (A) = 1 P (A) Rule of complement np r = n! (n r)! Permutation (no element alike) n! n 1! n 2! n k! Permutation (n 1 alike,... ) nc r = n! (n r)! r! Combination Ch 5: Probability Ditribution µ = Σ[x P (x)] Mean (prob. dit.) σ = p [Σ(x 2 P (x) ) ] µ 2 Standard deviation (prob. dit.) P (x) = n! (n x)! x! px q n x = n C x p x q n x Binomial probability µ = n p Mean (binomial) σ 2 = n p q Variance (binomial) σ = n p q Standard deviation (binomial) P (x) = µx e µ x! Poion Ditribution where e

8 Ch 6: Normal Ditribution z = x x or x µ σ Standard core x = µ + z σ µ x = µ Central limit theorem σ x = σ n Central limit theorem (Standard error) z = x µ (σ/ n) Central limit theorem Ch 7: Confidence Interval (one population) ˆp E < p < ˆp + E Proportion where E = z α/2 r ˆpˆq n x E < µ < x + E Mean where E = t α/2 (σ unknown, ee flowchart) n σ or E = z α/2 n (σ known, ee flowchart) (n 1) 2 χ 2 R (n 1) 2 χ 2 R (n 1)2 χ 2 Variance L (n 1) < σ < 2 Standard Deviation < σ 2 < χ 2 L Ch. 7: Sample Size Determination n = [z α/2] E 2 Proportion (ˆp and ˆq are unknown) n = [z α/2] 2 ˆpˆq E 2 Proportion (ˆp and ˆq are known) h zα/2 σi 2 n = Mean E 8

NEGATIVE z Scores. TABLE A-2 Standard Normal (z) Distribution: Cumulative Area from the LEFT. (continued)

NEGATIVE z Scores. TABLE A-2 Standard Normal (z) Distribution: Cumulative Area from the LEFT. (continued) NEGATIVE z Score z 0 TALE A- Standard Normal (z) Ditribution: Cumulative Area from the LEFT z.00.01.0.03.04.05.06.07.08.09-3.50 and lower.0001-3.4.0003.0003.0003.0003.0003.0003.0003.0003.0003.000-3.3.0005.0005.0005.0004.0004.0004.0004.0004.0004.0003-3..0007.0007.0006.0006.0006.0006.0006.0005.0005.0005-3.1.0010.0009.0009.0009.0008.0008.0008.0008.0007.0007-3.0.0013.0013.0013.001.001.0011.0011.0011.0010.0010