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
