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 = 5620 and (assuming independence) = 429.46. distributed, the probability that revenue exceeds 4300 is If the X's are ---Select--- P(x>\ 43005620 429.46 = P(Z > -3.07) = 1 - ( ])=P(z > = 0.9989.
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 = 5620 and (assuming independence) = 429.46. distributed, the probability that revenue exceeds 4300 is If the X's are ---Select--- P(x>\ 43005620 429.46 = P(Z > -3.07) = 1 - ( ])=P(z > = 0.9989.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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![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 y = 5620 and (assuming independence)
If the X's are ---Select---
distributed, the probability that revenue exceeds 4300 is
p(Y> [
]) = P(Z >
43005620
429.46
= P(Z > -3.07) = 1
- $(1
= 0.9989.
= 429.46.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05306099-d17b-4786-a552-6aca0323abbc%2F626767aa-f311-4519-a7a2-10cbc7deeb60%2Fjv2804_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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 y = 5620 and (assuming independence)
If the X's are ---Select---
distributed, the probability that revenue exceeds 4300 is
p(Y> [
]) = P(Z >
43005620
429.46
= P(Z > -3.07) = 1
- $(1
= 0.9989.
= 429.46.
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