SOLUTION: 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 t
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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?
(b) What is the probability that the demand for gas in a day will be between 1500 and 2000 gallons?
(c) What is the probability that the demand for gas will exceed 1500 gallons on any day?
(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%?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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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