Answer: 0.0455
Step-by-step explanation:
Given : Weekly demand at a grocery store for a breakfast cereal is normally distributed .
Population mean : [tex]\mu=800[/tex]
Standard deviation : [tex]\sigma=75[/tex]
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 [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Then , for x= 650
[tex]z=\dfrac{650-800}{75}=-2[/tex]
x=950
[tex]z=\dfrac{950-800}{75}=2[/tex]
Then , the probability that the weekly demand is less than 650 boxes or greater than 950 boxes is given :-
[tex]P(z<-2)+P(z>2)=P(z<-2)+1-P(z<2)\\\\=0.0227501+1- 0.9772498=0.0455003\approx0.0455[/tex]
Hence, the probability that the weekly demand is less than 650 boxes or greater than 950 boxes = 0.0455
