Suppose the amount of protein is at least 8.6 grams. What is the probability that it is more than 8.7 grams

This question is incomplete, the complete question is;

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 weighs 45 grams. The mean amount of protein in each bar is 8 grams. Suppose the amount of protein in the bars have a normal distribution with a standard deviation of 0.32 grams and a random Kashi bar is selected.  

Suppose the amount of protein is at least 8.6 grams. What is the probability that it is more than 8.7 grams

the required probability is 0.472

Step-by-step explanation:

Given the data in the question;

Let x represent the random variable that shows the protein in the bar.

{ normal distribution }

mean μ = 8

standard deviation σ =  0.32

Now, Suppose the amount of protein is at least 8.6 grams. What is the probability that it is more than 8.7 grams

first we get the z-score for x = 8.6

z = x - μ / σ

z = ( 8.6 - 8 ) / 0.32

z = 0.6 / 0.32

z = 1.875

so

P( x ≥ 8.6 ) = P( z ≥ 1.875 ) = 1 - 0.9696 = 0.0304

Also for, x = 8.7

z = x - μ / σ

z = ( 8.7 - 8 ) / 0.32

z = 0.7 / 0.32

z = 2.1875

so

P( x > 8.7 ∩ x ≥ 8.6 ) = P( x > 8.7 ) = P( z > 2.1875 ) = 1 - 0.98565 = 0.01435

Now, the required probability will be;

P( x > 8.7 | x ≥ 8.6 ) = [P( x > 8.7 ∩ x ≥ 8.6 )] / [ P( x ≥ 8.6 ) ]

= 0.01435 / 0.0304

= 0.472

Therefore, the required probability is 0.472

the answer is c)8

step-by-step explanaction:

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it