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Suppose that the miles-per-gallon (mpg) rating of passenger cars is a normally distributed random variable with a mean and a standard deviation of 33.8 and 3.5 mpg, respectively.

a. What is the probability that a randomly selected passenger car gets more than 35 mpg?

User Azhar Khan
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Final answer:

To calculate the probability of a car getting more than 35 mpg, one would compute the Z-score and consult a standard normal distribution table or software. The question entails applying basic statistics to a normally distributed variable.

Step-by-step explanation:

The question concerns the probability that a randomly selected passenger car gets more than 35 mpg given a normally distributed random variable with a mean of 33.8 mpg and a standard deviation of 3.5 mpg. To find this probability, one would typically use the Z-score formula:

Z = (X - μ) / σ

Where X is the value of interest (35 mpg), μ is the mean, and σ is the standard deviation. After calculating the Z-score, you would look up the value in a standard normal distribution table or use statistical software to find the probability that a car gets more than 35 mpg.

However, the rest of the information provided is related to other statistical questions regarding hypothesis testing and other distributions, which are not directly related to the original question of calculating the probability for cars with more than 35 mpg. Consequently, only the calculation related to the initial question is provided here.

User Sumithran
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