Answer:
54%
Explanation:
We are given that
S.D=4.2 lb
Mean=[tex]\mu=9.4 lb[/tex]
We have to find the percentage of household throw out at least 9 lb of paper a week.
Normal distribution formula :
[tex]P(X\geq a)=P(z\geq \frac{a-\mu}{\sigma})[/tex]
We have a=9
[tex]P(X\geq 9)=P(z\geq \frac{9-9.4}{4.2})=P(z\geq -0.1)[/tex]
[tex]P(X\geq 9)=1-P(z<-0.1)[/tex]
[tex]P(X\geq 9)=1-0.4602=0.5398\times 100=54[/tex]%
Hence, the percentage of household throw out at least of paper a week=54%