According to survey in a country 26% of adults do not have any credit cards. Sup

Question

According to survey in a country 26% of adults do not have any credit cards. Suppose a simple random sample of 500 adults is obtained. Describe the sampling distribution of p the sample proportion of adults who do not have a credit card. A) Approximately normal because n<0.05N and np(1-p)<10 B) Approximately normal because n<0.05 and np(1-p)>10 C)Not normal because n<0.05N and np(1-p)<10 D)Not normal n<0.05Nand np(1-p)>10 , Determine the mean of the samplingdistribution up=, Determine the standard deviation 0p= b) In a random sample of 500 adults what is the probability that less than 24% have no credit cards? c) Would it be unusual if a random sample of 500 adults results in 155 or more having no credit cards

Explanation / Answer

Now, given n=500,p=0.26.

Thus, np=500*0.26=130 and np(1-p)=96.2

Since, both np and np(1-p) are both greater than 10 thus the sampling distribution of p the sample proportion of adults who do not have a credit card is normally distributed with mean p and standard deviation p(1-p)/n

The mean is 0.26

The standard deviation is 0.26*(1-0.26)/500

=0.01

Find P(X<0.24)

Z=(x-)/=(0.24-0.26)/0.01=-2

Thus P(z<-2)=0.0228

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