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

Step-by-step explanation:

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

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

The formula for z -score :

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

For x= 7.65 ,

$z=\dfrac{7.65-7.8}{0.2}=-0.75$

For x= 8.2 ,

$z=\dfrac{8.2-7.8}{0.2}=2$

The p-value = $P(-0.75$

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

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