Question

(3 points + 1 point BONuS) Many people grab a granola bar for breakfast or for a snack to make it through the afternoon slump at work. A Kashi GoLean Crisp Chocolate Caramel bar weights 45 grams. The mean amount of protein in each bar is 7.8 grams. Suppose the distribution of protein in a bar is normally distributed with a standard deviation of 0.2 grams and a random Kashi bar is selected. (0.5 pts.) a) What is the probability that the amount of protein is between 7.65 and 8.2 grams?

Answer 1

Answer: 0.7506

Explanation:

Given :Mean :
`\mu=\text{ 7.8 grams}`

Standard deviation :
`\sigma =\text{ 0.2 grams}`

The formula for z -score :

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

For x= 7.65 ,

`z=(7.65-7.8)/(0.2)=-0.75`

For x= 8.2 ,

`z=(8.2-7.8)/(0.2)=2`

The p-value =
`P(-0.75<z<2)=P(z<2)-P(z<-0.75)`

`=0.9772498-0.2266274=0.750622\approx0.7506`

Hence, the probability that the amount of protein is between 7.65 and 8.2 grams=0.7506.