Question

According to a study conducted in one city, 29% of adults in the city have credit card debts of more than $2000. A simple random sample of n=100 adults is obtained from the city. Describe the sampling distribution of p , the sample proportion of adults who have credit card debts of more than $2000.

Select one:

a. Approximately normal; μ_p = 0.29, σ_p = 0.0021

b. Exactly normal; μ_p = 0.29, σ_p = 0.045

c. Binomial; μ_p = 29, σ_p = 4.54

d. Approximately normal; μ_p = 0.29, σ_p = 0.045

Answer 1

Mean of sampling distribution of is

= p = 0.29

Standard deviation of sampling distribution of is

= sqrt( p (1 - p) / n)

= sqrt ( 0.29 * 0.71 / 100)

= 0.045

So,

Sampling distribution of is

Approximately normal , = 0.29 , = 0.045