Question Help Generation Y has been defined as those individuals who were born b

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Question Help Generation Y has been defined as those individuals who were born between 1981 and 1991 A 2010 survey by a credit counseling foundation found that 54% of the g adults in Generation Y pay their monthly bills on tine) Suppose we take a random sample of 200 people from Generation Y Complete parts a through e below a. Calculate the standard error of the proportion. ?p-O (Round to four decimal places as needed ) b. What is the probability that 125 or fewer will pay their monthly bills on time? P(125 or fewer Generation Y individuals will pay their monthly bills n time)- Round to four decimal places as needed) c. What is the probability that 100 or fewer will pay their monthly bills on time? P(100 or fewer Generation Y individuals will pay their monthly bills on tine-L? Round to four decimal places as needed) d. What is the probability that 115 or more will pay their monthly bills on time? P(115 or more Generation Y individuals will pay their monthly blls ontime)- Round to four decimal places as needed) e. What is the probability that between 106 and 110 of them will pay their monthly bills on time? P(Between 106 and 110 ofthem wil pay their monthly bills on time)-? Round to four decimal places as needed)

Explanation / Answer

Solution:- Given that p = 0.54, n = 200, q = 1-p = 1-0.54 = 0.46

mean = n*p = 200*0.54 = 108

standard deviation = sqrt(n*p*q) = sqrt(200*0.54*0.46) = 7.0484

a. standard error = sqrt(pq/n) = sqrt(0.54*0.46/200) = 0.0352

b. P(X < 125) = P(Z < (125-108)/7.0484)
= P(Z < 2.4119)
= 0.9920

c. P(X < 100) = P(Z < (100-108)/7.0484)
= P(Z < -1.1350)
= 0.1271

d. P(X > 115) = P(Z > (115-108)/7.0484)
= P(Z > 0.9931)
= 0.1611

e. P(106 < X < 110) = P((106-108)/7.0484 < Z < (110-108)/7.0484)
= P(-0.2838 < Z < 0.2838)
= 0.2206

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