Question

(a) Weekly demand at a grocery store for a breakfast cereal is normally distributed with a mean of 800 boxes and a standard deviation of 75 boxes. (i) What is the probability that the weekly demand is less than 650 boxes or greater than 950 boxes?

Answer 1

Answer: 0.0455

Explanation:

Given : Weekly demand at a grocery store for a breakfast cereal is normally distributed .

Population mean :
`\mu=800`

Standard deviation :
`\sigma=75`

To find : Probability that the weekly demand is less than 650 boxes or greater than 950 boxes.

We first find z-score corresponds to 650 and 950.

Since
`z=(x-\mu)/(\sigma)`

Then , for x= 650

`z=(650-800)/(75)=-2`

x=950

`z=(950-800)/(75)=2`

Then , the probability that the weekly demand is less than 650 boxes or greater than 950 boxes is given :-

`P(z<-2)+P(z>2)=P(z<-2)+1-P(z<2)\\\\=0.0227501+1- 0.9772498=0.0455003\approx0.0455`

Hence, the probability that the weekly demand is less than 650 boxes or greater than 950 boxes = 0.0455