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
