Respuesta :
Answer:
The probability that the sample proportion is within 0.03 of the population proportion is 0.468.
Step-by-step explanation:
The complete question is:
A company makes auto batteries. They claim that 84% of their LL70 batteries are good for 70 months or longer. Assume that this claim is true. Let p^ be the proportion in a random sample of 60 such batteries that are good for 70 months or more. What is the probability that this sample proportion is within 0.03 of the population proportion? Round your answer to two decimal places.
Solution:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p\\[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
The information provided is:
[tex]p=0.84\\n=60[/tex]
As the sample size is large, i.e. n = 60 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample proportion of LL70 batteries that are good for 70 months or longer.
Compute the probability that the sample proportion is within 0.03 of the population proportion as follows:
[tex]P(-0.03<\hat p-p<0.03)=P(\frac{-0.03}{\sqrt{\frac{0.84(1-0.84)}{60}}}<\frac{\hat p-p}{\sigma_{\hat p}}<\frac{-0.03}{\sqrt{\frac{0.84(1-0.84)}{60}}})\\\\=P(-0.63<Z<0.63)\\\\=P(Z<0.63)-P(Z<-0.63)\\\\=0.73565-0.26763\\\\=0.46802\\\\\approx 0.468[/tex]
Thus, the probability that the sample proportion is within 0.03 of the population proportion is 0.468.
