An automobile manufacturer introduces a new model that has an advertised mean in

Question

An automobile manufacturer introduces a new model that has an advertised mean in-city mileage of 27 miles per gallon and the standard deviation of 3 miles per gallon. Let the random variable X be in-city mileage and assume that it has a normal distribution. (a) What is the probability that a car of this model has less than 22 miles per gallon for in-city driving? (b) Find a value x such that 25% of cars of this model have in-city mileage greater than x miles. (c) Within what limits (centered about the mean) would you expect that X to fall 88% of time? (d) We select 25 cars at random. What is the probability that the average in-city mileage of these 25 cars will be less than 28.5 miles per gallon? (e) What is the probability that the average in-city mileage of these 25 cars will be between 26.4 to 28.5 miles per gallon?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    22      
u = mean =    27      
          
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) / s =    -1.666666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.666666667   ) =    0.047790352 [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.25 =   0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    27      
z = the critical z score =    0.67448975      
s = standard deviation =    3      
          
Then          
          
x = critical value =    29.02346925   [ANSWER]  

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C)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.88      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.06      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.554773595      
By symmetry,          
z2 =    1.554773595      
          
As          
          
u = mean =    27      
s = standard deviation =    3      
          
Then          
          
x1 = u + z1*s =    22.33567922   [ANSWER]  
x2 = u + z2*s =    31.66432078   [ANSWER]

Hence, between 22.336 and 31.664.

**********************  

d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    28.5      
u = mean =    27      
n = sample size =    25      
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.5      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2.5   ) =    0.993790335 [ANSWER]

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e)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    26.4      
x2 = upper bound =    28.5      
u = mean =    27      
n = sample size =    25      
s = standard deviation =    3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.993790335      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.835135081   [ANSWER]  

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