Question

According to a survey in a country, 10% of adults do not own a credit card. Suppose a simple random sample of 300 adults is obtained. Complete parts (a) through (d) below (a) Describe the samping distribution of the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling diatribution of below. O A. Approximately normal becausens OSN and noft-p) 10 OB. Not normal because ns0.05N and npit-p)< 10 OC. Not normal because ns0.05N and np(1-2) 10 OD. Approximately normal because 0.05 and not - p)<10 Determine the mean of the sampling distribution of - Round to two decimal places as needed) Determine the standard deviation of the samping distribution of . (Round to three decimal places as needed) 0. (b) What is the probability that in a random sample of 300 adults, more than 21% do not own a credit card? The probability is (Round to four decimal places as needed) Interpret this probability If 100 different random samples of 300 adults were obtained, one would expect to result in more than 21 not owning a credit card. (Round to the nearest Integer as needed.) (c) What is the probability that in a random sample of 300 adults, between 16% and 21% do not own a credit card? The probability is

Answer 1

ANSWER Here P = 0.18 h=300 Boo (0.18) (1-0.18) 44.28 Approximally normal because n 20.05N and he (1-1) 210 Mê - P=0.18 ve PQ 10.18 (1.0.18) 300 20-0 2218 20.022 We show that 2² -Me ६ 6 for all P=0.21 1.35856 0.21-0.18 0.02218 = 1.35 P(more than arr. 8 (2 > 0.21) P(Z 51.358) 1- P(241358) 1-P( 21.35 È \ -0.01149 =0.0885 when N=100 Expected number is loo (0.0885) = 8.85 - 9 CS Scanned with CamScanner

o for 167. ,8 = 0.16 0.16 0.18 -6.90171 6.02918 -0.gol P (Between 167.821.)=P(0.16 58 4 0.21) = P6.90158 2 1853) :P(22\-35) - (25-0.90) = 0.9115 - 01841 when N=100 co.7274 1 Exceelled humber is 100 XO.7274 = 72.74 13 0.16 0.18 0.16) = 2 ر d here 0.02218 = -0.902 P (less than 16+9=P(p <0.16) -P12K05636) =P(z<-0.90) - 0.1841 here this is not unusual because The Probability is more than 0.05. CS Scanned with CamScanner