Answer :
Final answer:
The probability that a randomly selected passenger car gets more than 37 mpg is 33.00%. This calculation requires converting the mpg value to a z-score and consulting a z-table or standard normal distribution calculator.
Explanation:
In order to solve this problem, you'll need to convert the mpg value to a z-score, then use a z-table or a standard normal distribution calculator to find the probability. A z-score is a standard measurement that tells how many standard deviations a particular value is from the mean.
Step 1: Calculate the z-score. The formula for a z-score is (X - μ) / σ. In this case, X = 37 mpg, μ (the mean) = 35.9 mpg, and σ (the standard deviation) = 2.5 mpg. So, the z-score is (37 - 35.9) / 2.5 = 0.44.
Step 2: Look up the z-score in the z-table to find the probability. The value for 0.44 in the z-table is 0.6700, which represents the probability that a car gets less than or equal to 37 mpg. Because you want to find the probability that a car gets more than 37 mpg, you need to subtract this number from 1. So, the probability that a car gets more than 37 mpg is 1 - 0.6700 = 0.3300, or 33.00%.
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