Respuesta :
Answer:
The probability that the sample proportion would differ from the population proportion by greater than 3% is 0.008.
Step-by-step explanation:
The complete question is:
A statistician calculates that 7% of Americans own a Rolls Royce. If the statistician is right, what is the probability that the proportion of Rolls Royce owners in a sample of 508 Americans would differ from the population proportion by greater than 3%? Round your answer to four decimal places.
Solution:
Let the random variable X represent the number of Americans owning a Rolls Royce.
The proportion of Americans who own a Rolls Royce is, p = 0.07.
A random sample of n = 508 Americans were selected.
As the sample size is quite large, i.e. n = 508 > 30, according to the central limit theorem the sampling distribution of sample proportion would follow a normal distribution approximately.
The mean and standard deviation are as follows:
[tex]\mu_{\hat p}=p=0.07[/tex]
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.07(1-0.07)}{508}}=0.01132[/tex]
Compute the probability that the sample proportion would differ from the population proportion by greater than 3% as follows:
[tex]P(|\hat p-p|>0.03)=P(\hat p<-0.03\ \cup\ \hat p>0.03)[/tex]
[tex]=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}}<\frac{-0.03}{0.01132}\ \cup\ \frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}}>\frac{0.03}{0.01132})\\\\=P(Z<-2.65\ \cup\ Z>2.65)\\\\=P(Z<-2.65)+P(Z>2.65)\\\\=P(Z<-2.65)+1-P(Z<2.65)\\\\=0.00402+1-0.99598\\\\=0.00804\\\\\approx 0.008[/tex]
*Use the z-table.
Thus, the probability that the sample proportion would differ from the population proportion by greater than 3% is 0.008.
