Question

According to a survey in a country, 17% of adults do not have any credit cards. Suppose a simple random sample of 300 adults is obtained. Determine the mean of the sampling distribution of p. HA (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of p. (Round to three decimal places as needed.) (b) In a random sample of 300 adults, what is the probability that less than 14% have no credit cards? The probability is(Round to four decimal places as needed.) (c) Would it be unusual if a random sample of 300 adults results in 66 or more having no credit cards? OA. The result is not unusual because the probability that p is greater than or equal to this sample proportion is greater than 5%. B. The result is not unusual because the probability that p is greater than or equal to this sample proportion is less than 5 %. O C. The result is unusual because the probability that p is greater than or equal to this sample proportion is less than 5% O D. The result is unusual because the probability that p is greater than or equal to this sample proportion is greater than 5%. Click to select your answer(s).

Answer 1

Here we have

n=300, p=0.17

The sampling distribution of sample proportion will be approximately normal distribtuion with mean

Mp=p=0.17

and standard deviation

p(1-P) Op = V 0.17(1 -0.83) - = 0.022 300 = V n

b) probability that less than 14 % have no credit cards is

67 by defination probability that less than 14-1. have no credit card is. P(P<0.14) = P(Å - up <0.14 - 0.17 l ê 0.022 = P(Z <-13636) P(Å0 14)= 0.0863 7 Prob that 66 or more adults a will not having bedit card of the sample 1-PC$2560) = 1-P ($20.22) =1-P(P-up <0.22-017) 0022 = 1-P(Z <2:2727 = 1 -0.9885 = 0.0115 the result is unusual because the probability, § is greater than or equal to this sample proportion is less than 5 lt. (0-05) Therefore correct option is Ic]