Suppose that the mpg rating of Passengers Cars is a normally distributed random variable with a mean and a standard deviation of 33,8 mpg and 3,5 mpg. Question : an automobiles manufacturer wants to build a new Passenger car with an mpg rating that improves upon 99% of existing Cars. What is the minimum mpg that would achieve this goal ?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.99) and its Z score. Insert data into above equation and solve for score.
To find the minimum mpg rating that would improve upon 99% of existing cars, we need to calculate the z-score associated with the desired percentile (99%).
We can find the z-score using the formula:
z = (x - μ) / σ
where z is the z-score, x is the value we want to find, μ is the mean, and σ is the standard deviation.
In this case, the mean (μ) is 33.8 mpg, and the standard deviation (σ) is 3.5 mpg.
To find the z-score associated with the 99th percentile, we need to find the z-value for which the cumulative area to the left is 0.99.
Using a standard normal distribution table or a z-score calculator, we find that the z-value for a cumulative area of 0.99 is approximately 2.33.
Now, we can rearrange the formula to solve for the desired value (x):
x = (z * σ) + μ
Plugging in the values, we get:
x = (2.33 * 3.5) + 33.8
Calculating the expression:
x ≈ 8.155 + 33.8
x ≈ 41.955
Therefore, the minimum mpg rating that would achieve the goal of improving upon 99% of existing cars is approximately 41.955 mpg.