Thank you for visiting Suppose that the miles per gallon mpg rating of passenger cars is normally distributed with a mean of 35 9 mpg and a standard deviation. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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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