The chief chemist for a major oil and gasoline production company claims that the regular unleaded gasoline produced by the company contains on average 4 ounces of a certain ingredient.The chemist further states that the distribution of this ingredient per gallon of regular unleaded gasoline is normal and has a standard deviation of 1.2 ounces.What is the probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline?
A).5987
B).4013
C).9938
D).0062

Question

Quizplus · Accepted Answer

Here, we need to use the Central Limit Theorem because the sample size is large (n=100) and the population standard deviation (sigma = 1.2) is known.

Let X̄ be the sample mean of the ingredient per gallon of regular unleaded gasoline. Then, X̄ is normally distributed with mean = 4 ounces and standard deviation = 1.2/sqrt(100) = 0.12 ounces.

We need to find P(X̄ > 4.3).

z-score = (4.3 - 4) / 0.12 = 2.5

Using a standard normal distribution table or calculator, we find P(Z > 2.5) = 0.0062

Therefore, the probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline is 0.0062 or 0.62%.

Hence, the best choice is (D).

The Chief Chemist for a Major Oil and Gasoline Production