The chief chemist for a major oil/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 randomly inspected 100 gallons of regular unleaded gasoline?
A).5987
B).4013
C).9938
D).0062

Question

Quizplus · Accepted Answer

We need to use the central limit theorem since we are dealing with a large sample size (100).

First, we need to calculate the standard error of the mean:

Standard error of the mean = standard deviation / square root of sample size

s/√n = 1.2/√100 = 0.12

Next, we need to find the z-score:

z = (x - μ) / (s/√n) = (4.3 - 4) / 0.12 = 2.5

Using a standard normal distribution table or calculator, we can find the probability of finding a z-score larger than 2.5:

P(z > 2.5) = 0.0062

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

The correct answer is D.

The Chief Chemist for a Major Oil/gasoline Production Company Claims