Under usual conditions a distillation unit in a refinery can

Under usual conditions, a distillation unit in a refinery can process a mean of 135,000 barrels per day of crude petroleum, with a standard deviation of 6,000 barrels per day. You may assume a normal distribution. a. Find the probability that more than 135,000 barrels will be produced on a given day. b. Find the probability that more than 130,000 barrels will be produced on a give day. c. Find the probability that more than 150,000 barrels will be produced on a given day. d. Find the probability that less than 125,000 barrels will be produced on a given day. e. Find the probability that less than 100,000 barrels will be produced on a given day.

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    135000      
u = mean =    135000      
          
s = standard deviation =    6000      
          
Thus,          
          
z = (x - u) / s =    0      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0   ) =    0.5 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    130000      
u = mean =    135000      
          
s = standard deviation =    6000      
          
Thus,          
          
z = (x - u) / s =    -0.833333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.833333333   ) =    0.797671619 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    150000      
u = mean =    135000      
          
s = standard deviation =    6000      
          
Thus,          
          
z = (x - u) / s =    2.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.5   ) =    0.006209665 [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    125000      
u = mean =    135000      
          
s = standard deviation =    6000      
          
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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e)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    100000      
u = mean =    135000      
          
s = standard deviation =    6000      
          
Thus,          
          
z = (x - u) / s =    -5.833333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -5.833333333   ) =    2.71654*10^-9 [ANSWER]