Question

The average weight of a particular box of crackers is 26.0 ounces with a standard deviation of 0.5 ounce. The weights of the

Answer 1

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,

$\mu=26.0$ ounce

The standard deviation of weight,

$\sigma=0.5$ ounce.

Let $z$ be the standard normal variable,

$z=\frac{x-\mu}{\sigma}$

And, the probability of the boxes having weight x ounces is

$P(z)=\frac{1}{2\pi}e^{-\frac{z^2}{2}}$

For $x=24.5$,

$z=\frac{x-\mu}{\sigma}=\frac{24.5-26}{0.5}=-3$

(a) For the boxes having weight more than 24.5 ounces:$z>-3$

So, the probability of boxes for $z>-3$ is

$P(z>-3)=\int_{-3}^{\infty}\left(\frac{1}{2\pi}e^{-\frac{z^2}{2}}\right)dx$

=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

$P(z-3}=1-0.99865$

=0.00135

So, the percent of the boxes weigh more than 24.5 ounces is 0.135%.