Mileage ratings for cars and trucks generally come with a qu

\"Mileage ratings for cars and trucks generally come with a qualifier stating actual mileage will depend on driving conditions and habits. Ford is stating the Ecoboost F-150 pickup truck will get 19 miles per gallon with combined town and country driving. Assume the mean stated by Ford is the actual average, and the distribution has a standard deviation of 3 mpg. a. Given the above mean and standard deviation, what
is the probability that 100 drivers will get more than
19.2 miles per gallon average? b. Suppose 1,000 drivers were randomly selected.
What is the probability the average obtained by
these drivers would exceed 19.2 mpg?\"

Please explain how to got the answerts. Thanks.

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    19.2      
u = mean =    19      
n = sample size =    100      
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0.666666667      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.666666667   ) =    0.252492538 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    19.2      
u = mean =    19      
n = sample size =    1000      
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.108185107      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.108185107   ) =    0.017507491 [ANSWER]