Question

According to a survey in a​ country, 2727​% of adults do not own a credit card. Suppose a simple random sample of 900900 adults is obtained. Complete parts​ (a) through​ (d) below. ​(a) Describe the sampling distribution of ModifyingAbove p with caretp​, 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 ModifyingAbove p with caretp below. A. Approximately normal because n less than or equals 0.05 Upper Nn≤0.05N and np left parenthesis 1 minus p right parenthesis greater than or equals 10np(1−p)≥10 B. Approximately normal because n less than or equals 0.05 Upper Nn≤0.05N and np left parenthesis 1 minus p right parenthesis less than 10np(1−p)<10 C. Not normal because n less than or equals 0.05 Upper Nn≤0.05N and np left parenthesis 1 minus p right parenthesis greater than or equals 10np(1−p)≥10 D. Not normal because n less than or equals 0.05 Upper Nn≤0.05N and np left parenthesis 1 minus p right parenthesis less than 10np(1−p)<10 Determine the mean of the sampling distribution of ModifyingAbove p with caretp. mu Subscript ModifyingAbove p with caret Baseline equalsμp=nothing ​(Round to two decimal places as​ needed.) Determine the standard deviation of the sampling distribution of ModifyingAbove p with caretp. sigma Subscript ModifyingAbove p with caretσpequals=nothing ​(Round to three decimal places as​ needed.) ​(b) What is the probability that in a random sample of 900900 ​adults, more than 3131​% do not own a credit​ card? The probability is nothing. ​(Round to four decimal places as​ needed.) Interpret this probability. If 100 different random samples of 900900 adults were​ obtained, one would expect nothing to result in more than 3131​% not owning a credit card. ​(Round to the nearest integer as​ needed.) ​(c) What is the probability that in a random sample of 900900 ​adults, between 2626​% and 3131​% do not own a credit​ card? The probability is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

A. Approximately normal because n≤0.05N and np(1−p)≥10

μp=0.27

std error of proportion=σp=√(p*(1-p)/n)=

0.015

b)

probability =P(X>0.31)=P(Z>(0.31-0.27)/0.015)=P(Z>2.67)=1-P(Z<2.67)=1-0.9962=0.0038

If 100 different random samples of 900900 adults were​ obtained, one would expect 0  to result in more than 31%

c)

probability =P(0.26<X<0.31)=P((0.26-0.27)/0.015)<Z<(0.31-0.27)/0.015)=P(-0.67<Z<2.67)=0.9962-0.2514=0.7448

.....one would expect 74     to result in......