Question

The amount of cream sauce on the fettuccine at Al Fred-O's follows a Normal distribution, with a mean of 3.78 ounces and a standard deviation of 0.14 ounce. A random sample of 12 plates of fettuccine is selected every day and the sauce is measured. What is the probability that the mean weight will exceed 3.81 ounces

Answer 1

Correct answer is 0.2290

Explanation:

Answer 2

22.66% probability that the mean weight will exceed 3.81 ounces

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
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 3.78, \sigma = 0.14, n = 12, s = (0.14)/(√(12)) = 0.04`

What is the probability that the mean weight will exceed 3.81 ounces

This probability is 1 subtracted by the pvalue of Z when X = 3.81. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (3.81 - 3.78)/(0.04)`

`Z = 0.75`

`Z = 0.75` has a pvalue of 0.7734

1 - 0.7734 = 0.2266

22.66% probability that the mean weight will exceed 3.81 ounces