A gas station sells two grades of gasoline: regular and super. These are priced at $2.00 and $3.00 per gallon, respectively. Let
X₁ and X₂ denote the amounts of these grades purchased (gallons) on a particular day. Suppose X₁ and X₂ are independent Normal random variables with E(X₁) = 1000, sd(X₁) = 90, E(X₂) = 800, and sd(X₂) = 45. The revenue from sales is Y = 2X₁ + 3X₂. (a) Find the mean and standard deviation of Y, i.e., E(Y) and sd(Y). (b) Find the probability that the revenue Y does not exceed 4500. Hint: Y is Normally distributed. (c) Find the probability that the gas station sells regular gasoline more, i.e., pr (X₁ > X₂). Attach the R codes or write out the formula(s) you used for full credit.
2 answers:
Answer:
(a) E(Y) = 4400
sd (Y) =225
(b) P(Y ≤ 4500) = 0.67003
(c) P (X₁ > X₂) = 0.31744
Step-by-step explanation:
(a) Here we have
Y = 2·X₁ + 3·X₂
Therefore E(Y) = 2·E(X₁) + 3·E(X₂) = 2000 + 2400 = 4400
sd(Y) is given by
Variance Y = (sd (Y))² = 2²·(sd (X₁))² + 3²·(sd (X₂))²
= 4·8100 + 9·2025 = 50625
sd (Y) = √50625 = 225
(b) The probability that the revenue does not exceed 4500 is given by
P(Y ≤ 4500) = P(z ≤0.444)
z =
z = = 0.444
Therefore from the normal distribution table, we have
P = 0.67003
(c) The probability that the P(X₁ > X₂)
Since the gas station sells 2 portions of X₁ to 3 portions of X₂
Therefore, the probability that the gas station sells more of X₁ is given by
₅C₀ × 2/5⁰×3/5⁵ = 0.07776
₅C₁ × 2/5¹×3/5⁴= 0.2592
₅C₂ × 2/5²×3/5³ = 0.3456
P (X₁ > X₂) = 1 - 0.68256 = 0.31744
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