Homeowrk 4 Solution

Homeowrk 4 Solution - HW4 Solution 5.4 We are concerned...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW4 Solution 5.4 We are concerned with the sampling distribution of p , which is governed by a binomial distribution. Letting success =responder, we have p=.2 and 1-p=.8. The number of trials is n=15. (a) The event p =.2 occurs if there are 3 successes in the 15 trials (because 3/15=.2). Thus, to find the probability that p =.2, we can use the binomial formula (1 ) j n j n j C p p-- with j=3, so n-j=12: 3 12 15 3 3 12 Pr( .2) (1 ) (455)(.2) (.8) .2501 p C p p = =- = = (b) The event p =0 occurs if there are 0 successes in the 15 trials (because 0/15=0). Thus, to find the probability that p =0, we can use the binomial formula with j=0, so n-j=15: 15 15 15 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) 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, =176 and =30. For y=186, 186 176 .33. 30 y z -- = = = From Table 3, the area below .33 is .6293. For y=166, 166 176 .33. 30 y z -- = = = - 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 = = the standard deviation is 30 10, 9 Y n = = = And the shape of the distribution is normal because the population distribution is normal as I mentioned in the lecture note 8....
View Full Document

Page1 / 5

Homeowrk 4 Solution - HW4 Solution 5.4 We are concerned...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online