Question 672426
The daily demand for gas at Good’s Gas station is normally distributed with a mean of 1812 gallons and a standard deviation of 254 gallons. 
(a) What is the probability that the demand for gas will exceed 2000 gallons on any day?
z(2000) = (2000-1812)/254 = 0.7402
P(x > 2000) = P(z > 0.7402) = normalcdf(0.7402,100) = 0.2296
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(b) What is the probability that the demand for gas in a day will be between 1500 and 2000 gallons?
z(1500) = (1500-1812)/254 = -1.2283
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P(1500<= x <=2000) = P(-1.2283<= z <=0.2296) = 0.4811
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(c) What is the probability that the demand for gas will exceed 1500 gallons on any day?
Ans: P(-1.2283,100) = 0.8903
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(d) how much gasoline should the station have on hand at the beginning of the day so that the probability of running out of gas that day is only 1%?
Find the z-value with a right tail of 1%.
invNorm(0.99) = 2.3263
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Find the corresponding gas value using x = z*s+u
x = 2.3263*254+1812 = 2402.89 gallons
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Cheers,
Stan H.
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