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(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?

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JeanaShupp

Answer: 0.7506

Step-by-step explanation:

Given :Mean : [tex]\mu=\text{ 7.8 grams}[/tex]

Standard deviation : [tex]\sigma =\text{ 0.2 grams}[/tex]

The formula for z -score :

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= 7.65 ,

[tex]z=\dfrac{7.65-7.8}{0.2}=-0.75[/tex]

For x= 8.2 ,

[tex]z=\dfrac{8.2-7.8}{0.2}=2[/tex]

The p-value = [tex]P(-0.75<z<2)=P(z<2)-P(z<-0.75)[/tex]

[tex]=0.9772498-0.2266274=0.750622\approx0.7506[/tex]

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

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