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
Answer:
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
