Question

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?

Answer 1

Answer:

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 = P(x > 4.3) = 0.00621

Step-by-step explanation:

This is a normal distribution problem

The mean of the sample = The population mean

μₓ = μ = 4 ounces

But the standard deviation of the sample is related to the standard deviation of the population through the relation

σₓ = σ/√n

where n = Sample size = 100

σₓ = 1.2/√100

σₓ = 0.12

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 = P(x > 4.3)

To do this, we first normalize/standardize the 4.3 ounces

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (4.3 - 4)/0.12 = 2.5

To determine 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 = P(x > 4.3) = P(z > 2.5)

We'll use data from the normal probability table for these probabilities

P(x > 4.3) = P(z > 2.5) = 1 - P(z ≤ 2.5) = 1 - 0.99379 = 0.00621