Answer: 0.2643
Step-by-step explanation:
Let p be the population proportion and [tex]\hat{p}[/tex] be the sample proportion .
As per given question , we have
[tex]p=0.14\\\\ \hat{p}=0.13\\\\ n=478[/tex]
z-score : [tex]\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z=\dfrac{0.13-0.14}{\sqrt{\dfrac{0.14(1-0.14)}{478}}}\\\\=-0.630087269176\approx-0.63[/tex]
The required probability ( using z-table ):-
[tex]P(z<-0.63)=1-P(z<0.63)\\\\=1- 0.7356527=0.2643473\approx0.2643[/tex]
Hence, the probability that the proportion of books checked out in a sample of 478 books would be less than 13% = 0.2643
