the mean gas mileage for a hybrid car is 56 miles per gallon suppose that the gasoline mileage is approximately normally distributed of 3.2 miles per gallon what is the probability that a randomly selected hybrid gets more than 62 miles per gallon.
http://davidmlane.com/hyperstat/z_table.html
To find the probability that a randomly selected hybrid car gets more than 62 miles per gallon, we need to use the concept of the standard normal distribution and z-scores.
The first step is to calculate the z-score for 62 miles per gallon using the formula:
z = (x - μ) / σ
where:
x = 62 miles per gallon
μ = mean gas mileage for a hybrid car (56 miles per gallon)
σ = standard deviation (3.2 miles per gallon)
Substituting the values:
z = (62 - 56) / 3.2
z = 1.875
Now, we need to find the probability corresponding to a z-score of 1.875 using a standard normal distribution table or a calculator. This probability represents the percentage of the area under the standard normal curve that is greater than 1.875.
Using a standard normal distribution table or a calculator, we find that the probability of getting a z-score of 1.875 or greater is approximately 0.0316, or 3.16%.
Therefore, the probability that a randomly selected hybrid car gets more than 62 miles per gallon is approximately 0.0316, or 3.16%.