Question

The average retail price of gasoline (all types) for the first half of 2005 was 212.2 cents. What would the standard deviation have to be in order for a 11% probability that a gallon of gas costs less than $1.80? Round z-value calculations to two decimal places and final answer to the nearest cent.

Answer 1

Given:

Average retail price of gasoline (all types) for the first half of 2005 = 212.2 cents
Probability that a gallon of gas costs less than $1.80 = 11%
We need to find the standard deviation required for this probability
First, we need to convert the price of $1.80 to cents:
$1.80 = 180 cents

Next, we need to find the z-score corresponding to the 11th percentile (since we want the probability of a price less than $1.80):
Using a standard normal distribution table or calculator, we find that the z-score for the 11th percentile is approximately -1.22.

Now, we can use the formula for z-score to find the standard deviation:
z = (x - μ) / σ
where:
z = -1.22
x = 180 cents (price of gas)
μ = 212.2 cents (average price of gas)
σ = standard deviation

Rearranging the formula and substituting the values, we get:
σ = (x - μ) / z
σ = (180 - 212.2) / (-1.22)
σ = 26.2 / 1.22
σ ≈ 21.48

Therefore, the standard deviation required for an 11% probability that a gallon of gas costs less than $1.80 is approximately 21.48 cents.