The inside diameter of a randomly selected piston ring is a random variable with mean value 8 cm and standard deviation 0.03 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.)(a) Calculate P(7.99 ≤ X ≤ 8.01) when n = 16. P(7.99 ≤ X ≤ 8.01) =(b) How likely is it that the sample mean diameter exceeds 8.01 when n = 25? P(X ≥ 8.01) =

a) P(7.99 ≤ X ≤ 8.01) = 0.8164

b) P(X ≥ 8.01) = 0.0475.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , the sample means with size n can be approximated to a normal distribution with mean

In this problem, we have that:

(a) Calculate P(7.99 ≤ X ≤ 8.01) when n = 16.

n = 16, so

This probability is the pvalue of Z when X = 8.01 subtracted by the pvalue of Z when X = 7.99. So

X = 8.01

Applying the Central Limit Theorem

has a pvalue of 0.9082

X = 7.99

has a pvalue of 0.0918

0.9082 - 0.0918 = 0.8164

P(7.99 ≤ X ≤ 8.01) = 0.8164

(b) How likely is it that the sample mean diameter exceeds 8.01 when n = 25? P(X ≥ 8.01) =

n = 25, so

This is 1 subtracted by the pvalue of Z when X = 8.01. So

has a pvalue of 0.9525

1 - 0.9525 = 0.0475

P(X ≥ 8.01) = 0.0475.

ddd

step-by-step explanation: