Answer:13.05 %
Step-by-step explanation:
Given
mean [tex]\mu =9.4 lb[/tex]
standard deviation [tex]\sigma =4.2 lb[/tex]
Let X be the amount of paper discarded from households per week normally distributed
we need to find [tex]P(X> 8)[/tex]
suppose [tex]z=\frac{X-9.4}{4.2}[/tex]
[tex]P(X>8)=P(\frac{X-9.4}{4.2}>\frac{8-9.4}{4.2})[/tex]
[tex]=P(z> -0.333)[/tex]
Since standard normal is perfectly symmetric about the mean
therefore [tex]P(z > -0.33)=P(z < 0.33) [/tex]
For Normal distribution
[tex]P(a\leq z\leq b)=P\left ( a\leq z\leq b\right )=\frac{1}{\sqrt{2\pi }}\int_{a}^{b}e^{\frac{z^2}{2}}dz[/tex]
[tex]P(z<0.33)=\frac{1}{\sqrt{2\pi }}\int_{0}^{3}e^{\frac{z^2}{2}}dz[/tex]
[tex]P(z<0.33)=\frac{0.1636\sqrt{2}}{\sqrt{\pi }}[/tex]
[tex]P(z<0.33)=0.13054[/tex]
Thus [tex]P(z> -0.33)=0.13054[/tex]
Therefore 13.05 % of households throw out at least 8 lb of paper a week
