1) Suppose that the miles-per-gallon (mpg) rating of passenger cars is a normall

Question

1) 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 39.6 and 3.8 mpg, respectively. Use Table 1.

What is the probability that a randomly selected passenger car gets more than 41 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

What is the probability that the average mpg of three randomly selected passenger cars is more than 41 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

If three passenger cars are randomly selected, what is the probability that all of the passenger cars get more than 41 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

  Probability

1) 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 39.6 and 3.8 mpg, respectively. Use Table 1.

Explanation / Answer

First we get the probabilty of getting mpg greater than 41 from the normal distribution

here Z = (41- 39.6) / 3.8 = 1.4/ 3.8 =0.3684 = 0.37

P (Z> 0.3684) can be found from normal distribution tables as 0.3563

Hence, the required answers are

1) 0.3563

2)

3) 0.3563^3 = 0.0452

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