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Unformatted text preview: HW4 Solution 5.4 We are concerned with the sampling distribution of ˆ p , which is governed by a binomial distribution. Letting “success” =”responder,” we have p=.2 and 1p=.8. The number of trials is n=15. (a) The event ˆ p =.2 occurs if there are 3 successes in the 15 trials (because 3/15=.2). Thus, to find the probability that ˆ p =.2, we can use the binomial formula (1 ) j n j n j C p p with j=3, so nj=12: 3 12 15 3 3 12 ˆ Pr( .2) (1 ) (455)(.2) (.8) .2501 p C p p = = = = (b) The event ˆ p =0 occurs if there are 0 successes in the 15 trials (because 0/15=0). Thus, to find the probability that ˆ p =0, we can use the binomial formula with j=0, so nj=15: 15 15 15 ˆ Pr( 0) (1 ) (1)(1)(.8) .0352 p C p p = = = = 5.7 (a) 5 5 (252)(.6 )(.4 ) .2007 = (b) 6 4 (210)(.6 )(.4 ) .2508 = (c) 7 3 (120)(.6 )(.4 ) .2150 = (d) .2007+.2508+.2150=.6665 (e) .6665 (from part(d)) 5.8 (a) ˆ p Probability .0 5 .7 .1681 = .2 4 (5)(.3)(.7 ) .3602 = .4 2 3 (10)(.3 )(.7 ) .3087 = .6 3 2 (10)(.3 )(.7 ) .1323 = .8 4 (5)(.3 )(.7) .0284 = 1 5 .3 .0024 = (b) 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 p hat 0.2 0.4 0.6 0.8 1 Compared with Figure 5.4, this distribution is more spread out (more dispersed) and is more skewed. 5.15 (a) In the population, μ =176 and σ =30. For y=186, 186 176 .33. 30 y z μ σ = = = From Table 3, the area below .33 is .6293. For y=166, 166 176 .33. 30 y z μ σ = = =  From Table 3, the area below .33 is .3707. Thus the percentage with 166 186 y j is .6293.3707=.2586, or 25.86%. (b) We are concerned with the sampling distribution of Y for n=9. From the proposition and the central limit theorem, the mean of the sampling distribution of Y is 176, Y μ μ = = the standard deviation is 30 10, 9 Y n σ σ = = = And the shape of the distribution is normal because the population distribution is normal as I mentioned in the lecture note 8....
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This note was uploaded on 03/28/2011 for the course AMS 110 taught by Professor N/a during the Spring '08 term at SUNY Stony Brook.
 Spring '08
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