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

Question

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 36.6 and 3.7 mpg, respectively. Use Table 1.

a. What is the probability that a randomly selected passenger car gets more than 37 mpg?

What is the probability that the average mpg of three randomly selected passenger cars is more than 37 mpg? (Round intermediate calculations to 4 decimal places, “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 37 mpg? (Round intermediate calculations to 4 decimal places, “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 37 mpg? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    37      
u = mean =    36.6      
          
s = standard deviation =    3.7      
          
Thus,          
          
z = (x - u) / s =    0.11      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.11   ) =    0.4562 [ANSWER]
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b)

For 3 cars, the new standard deviation for the mean is sigma/sqrt(n) = 3.7/sqrt(3) = 2.136195996.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    37      
u = mean =    36.6      
          
s = standard deviation =    2.136195996      
          
Thus,          
          
z = (x - u) / s =    0.19      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.19   ) =    0.4247 [ANSWER]

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c)

From part 1, the probability for a single car is 0.4562.

Hence, for 3 cars,

P(3 cars greater than 37) = 0.4562^3 = 0.094943632 [ANSWER]

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