Respuesta :
Answer:
(a) 99.865%
(b) 0.135%
Explanation:
Given that the weight of the boxes are normally distributed.
The average weight of the particular box,
[tex]\mu=26.0[/tex] ounce
The standard deviation of weight,
[tex]\sigma=0.5[/tex] ounce.
Let [tex]z[/tex] be the standard normal variable,
[tex]z=\frac{x-\mu}{\sigma}[/tex]
And, the probability of the boxes having weight x ounces is
[tex]P(z)=\frac{1}{2\pi}e^{-\frac{z^2}{2}}[/tex]
For [tex]x=24.5[/tex],
[tex]z=\frac{x-\mu}{\sigma}=\frac{24.5-26}{0.5}=-3[/tex]
(a) For the boxes having weight more than 24.5 ounces:[tex]z>-3[/tex]
So, the probability of boxes for [tex]z>-3[/tex] is
[tex]P(z>-3)=\int_{-3}^{\infty}\left(\frac{1}{2\pi}e^{-\frac{z^2}{2}}\right)dx[/tex]
=0.99865
So, the percent of the boxes weigh more than 24.5 ounces is 99.865%.
(b) For the boxes having weight less than 24.5 ounces: z<-3
So, the probability of boxes for z<-3 is
[tex]P(z<-3}=1-P(z>-3}=1-0.99865[/tex]
=0.00135
So, the percent of the boxes weigh more than 24.5 ounces is 0.135%.
