Question

If 100 different random samples of 300 adult were obtained, one would expect ___ to result in more than 37% not owning a credit card.

Score: 2 of 4 pts 5 of 7 16 complete HW Score: 58.58%, 11. 8.2.18-T Question According to a wrvey in a country. 35% of adults do not own a credit card. Suppose a simple random sample of 300 adults is obtained. Complete parts (a) through id below. (a) Describe the sampling 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 distribution of p below. DA No normal because ns0.05N and rot-p) <10 a Approximately normal because no05N and 1- 10 C. Not normal because nSOOSN and np(1-210 D. Approximately normal becausen 0.05 and p1-) < 10 Determine the mean of the sampling distribution of 35 Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of 0 026 Round to three decimal places as needed.) () What is the probability that in rondom sample of 300 ts, more than 3% do not own a credit card? The probability is 2389 (Round to four decimal places as needed interpretis probably d , one would expect to rest in more nan 37 not owning a credit card 100 rentrando sapies of 300 dus wor Round to hear w as need) or your answer in the web and then Check A As FGHJK

Answer 1

solution = BUMB=Jº 300 0.028 a Here p= 35% =0.35, n=300 here np (1-P = 300(0.35 (1-0.35 = 68-25 > 10 so, we con approximate to nsumal YES, approximatly normal because nao.OSN, np (1-P) 2010. The sampling distubution of ħ is mean = up = 0.35 8.35 (1 =8-0275u 300 -0.028 - for $= 37% = 0.37, 0.37-0.35 0.02754 = 1.0893 P(more than 37%)=P(på 20.37 =P(2701-0893)=1-Plze 1.0893) =1-0-8620 = 0·1380 If coo different random samples of 300 adults were observed one would expect 0.1380 (300)= ul.u= ul to result in more than 37%. not earning a credit card