Respuesta :
Answer:
The probability that the proportion of books checked out in a sample of 774 books would be less than 10% is 0.0066
Step-by-step explanation:
We are given that The director of the library believes that 13% of the library's collection is checked out.
[tex]\mu = p = 0.13[/tex]
Total no. of books = 774
Standard deviation =[tex]\sqrt{\frac{p(1-p)}{n}}[/tex]
Standard deviation = [tex]\sqrt{\frac{0.13(1-0.13)}{774}}=0.012[/tex]
We are supposed to find the probability that the proportion of books checked out in a sample of 774 books would be less than 10%
[tex]P(\hat{p}<0.1)\\Z=\frac{x-\mu}{\sigma}\\Z=\frac{0.1-0.13}{0.012}\\Z=-2.48\\P(Z<-2.48)=0.0066[/tex]
So,[tex]P(\hat{p}<0.1)=0.0066[/tex]
Hence the probability that the proportion of books checked out in a sample of 774 books would be less than 10% is 0.0066
