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4. [-/25 Points] DETAILS MY NOTES DEVORESTAT9 5.AE.032. ASK YOUR TEACHER Example 5.32 The total revenue from the sale of the three grades of gasoline on a particular day was $Y = 3.0X_1 + 3.2X_2 + 3.4X_3$, and it was determined that $\mu_Y = 5620$ and (assuming independence) $\sigma_Y = 429.46$. If the $X_i$'s are ---Select--- distributed, the probability that revenue exceeds 4100 is $P(Y > \text{_}) = P(Z > \frac{4100 - 5620}{429.46})$ $= P(Z > -3.54) = 1 - \Phi(\text{_}) = 0.9998.$

          4. [-/25 Points]
DETAILS
MY NOTES
DEVORESTAT9 5.AE.032.
ASK YOUR TEACHER
Example 5.32 The total revenue from the sale of the three grades of gasoline on a particular day was $Y = 3.0X_1 + 3.2X_2 + 3.4X_3$, and it was determined that $\mu_Y = 5620$
and (assuming independence) $\sigma_Y = 429.46$. If the $X_i$'s are ---Select--- distributed, the probability that revenue exceeds 4100 is
$P(Y > \text{_____}) = P(Z > \frac{4100 - 5620}{429.46})$
$= P(Z > -3.54) = 1 - \Phi(\text{_____}) = 0.9998.$

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4. [-/25 Points]
DETAILS
MY NOTES
DEVORESTAT9 5.AE.032.
ASK YOUR TEACHER
Example 5.32 The total revenue from the sale of the three grades of gasoline on a particular day was Y = 3.0X1 + 3.2X2 + 3.4X3, and it was determined that = 5620
and (assuming independence) = 429.46. If the Xi's are —Select— distributed, the probability that revenue exceeds 4100 is
P(Y > ) = P(Z > (4100 - 5620)/(429.46))
= P(Z > -3.54) = 1 - Φ() = 0.9998.

Added by Mark A.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Supreeta N

11/03/2023

Step 1

Given that Y = 3.0X₁ + 3.2X₂ + 3.4X₃, and μ(Y) = 5620 and σ(Y) = 429.46, we can determine the mean and standard deviation of each Xᵢ by dividing the coefficients in Y by their respective standard deviations. This will give us the mean and standard deviation of Show more…

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Points] DEVORESTAT95.AE.032. Example 5.32: The total revenue from the sale of the three grades of gasoline on a particular day was Y = 3.0X₁ + 3.2X₂ + 3.4X₃, and it was determined that μ(Y) = 5620 and (assuming independence) σ(Y) = 429.46. If the Xᵢ's are distributed, the probability that revenue exceeds 4100 is P(Y > 4100) = P(Z > (4100 - 5620) / (429.46)) = P(Z > -3.54) = 1 - Φ(-3.54) = 0.9998

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