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

boxes are normally distributed

a. What percent of the boxes weigh more than 24.5 ounces

b. What percent of the boxes weigh less than 25.0 ounces?

Answer 1

(a) 99.865%

(b) 0.135%

Step-by-step 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=(x-\mu)/(\sigma)`

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

`P(z)=(1)/(2\pi)e^{-(z^2)/(2)}`

For
`x=24.5`,

`z=(x-\mu)/(\sigma)=(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((1)/(2\pi)e^{-(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-P(z>-3}=1-0.99865`

=0.00135

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