2) According to creditcard.com, 29% of adults Americans do not own a credit card

Question

2) According to creditcard.com, 29% of adults Americans do not own a credit card. (14 points) a) If we surveyed 500 adult Americans, how many would we expect to not own a credit if creditcard.com is correct? (2 points) b) If we surveyed 500 adult Americans, what would we expect the sample proportion's standard deviation to be? (2 points) c) If we surveyed 500 adult Americans, what is the sample proportion if 140 said that they did not own a credit card? (2 point) d) Assuming that creditcard.com is correct, what is the probability that, in a random survey of 500 adult Americans, more than 30% do not own a credit card? (2 points) e) Assuming creditcard.com is correct, what is the probability that, in a random sample of 500 adult Americans, between 25% and 30% do not own a credit card? (2 points) f) Assuming creditcard.com is correct, would it be unusual for a random sample of 500 adult Americans to result in 125 or fewer that do not own a credit card? WHY??? (4 points)

Explanation / Answer

2.
29% of adults americans do not have own credit card

a.
we suryed 500 adults americans, we would expect do not have own credit card is
500*0.29 = 145

b.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.29
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.29*0.71/500)
=0.0203
sample proportion standard deviation = 0.0203

c.
we suryed 500 adults americans,140 members did not have own credit card
sample proportion = 140/500 =0.28

d.
probabilty that a randomly sample of 500 adults americans more than 30% do not have own credit card
P(X > 0.3) = (0.3-0.29)/0.0203
= 0.01/0.0203 = 0.4926
= P ( Z >0.493) From Standard Normal Table
= 0.3111

e.
probabilty that a randomly sample of 500 adults americans between 25% and 30% do not have own credit card
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.25) = (0.25-0.29)/0.0203
= -0.04/0.0203 = -1.9704
= P ( Z <-1.9704) From Standard Normal Table
= 0.02439
P(X < 0.3) = (0.3-0.29)/0.0203
= 0.01/0.0203 = 0.4926
= P ( Z <0.4926) From Standard Normal Table
= 0.68886
P(0.25 < X < 0.3) = 0.68886-0.02439 = 0.6645

f.
random sample of 500 adults americans will results in 125 or fewer that do not have own credit card
proportion = 125/500 =0.25

P(X > 0.25) = (0.25-0.29)/0.0203
= -0.04/0.0203 = -1.9704
= P ( Z >-1.97) From Standard Normal Table
= 0.9756
P(X < = 0.25) = (1 - P(X > 0.25)
= 1 - 0.9756 = 0.0244
it is unusual because minimum 29% adults do not have own credit card,but less than 25%(125/500) is unusual

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