Let X represent the amount of gasoline gallons purchased by

Let X represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the mean value and standard deviation of X are 11.5 and 4.0, respectively.

a. In a sample of 50 randomly selected customers,

what is the approximate probability that

the total amount of gasoline purchased is at

most 600 gallons.

b. What is the approximate value of the 95th

percentile for the total amount purchased by

50 randomly selected customers.

Solution

a)

Asking at most a sum of 600 for 50 people is like asking the probability of the mean of this sample to be at most 600/50 = 12.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    12      
u = mean =    11.5      
n = sample size =    50      
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0.883883476      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   0.883883476   ) =    0.811620441 [answer]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    11.5      
z = the critical z score =    1.644853627      
s = standard deviation =    4      
n = sample size =    50      
Then          
          
x = critical value =    12.43046972      

Thus, the total amount of 50 customers is this multiplied by 50,

95th percentile of sum of 50 customers = 621.5234861 [ANSWER]