HW5_urdaneta

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School

Virginia Tech *

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Course

2024

Subject

Industrial Engineering

Date

May 2, 2024

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pdf

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5

Uploaded by CommodoreMandrill4162

ISE 2024 - Homework 5 Due on April 10th, 11:59pm This assignment contains 6 questions for a total of 100 points. Note that submission is on Gradescope . 1. (16 points) If Z is N(0, 1), find • P (0 ≤ Z ≤ 0 . 87) = 0.3133 • P (−2 . 64 ≤ Z ≤ 0) P (−2 . 64 ≤ Z ≤ 0) = 0.4953 • P (−2 . 13 ≤ Z ≤ −0 . 56) = 0.2702 • P (| Z | > 1 . 39) . = 0.8354 2. (20 points) Consider the random variable X as the location of the next online delivery order requested by a customer. If X is normally distributed with a mean of 6 and a variance of 25, Find Standard Deviation = sqrt(variance) = 5 • P (6 ≤ X ≤ 12) = 0.3849 Standardizing: z₁ = (6 - 6) / 5 = 0 z₂ = (12 - 6) / 5 = 1.2 P(0 ≤ Z ≤ 1.2) • P (0 ≤ X ≤ 8) = 0.4918 Standardizing: z₁ = (0 - 6) / 5 = -1.2 z₂ = (8 - 6) / 5 = 0.4 P(-1.2 ≤ Z ≤ 0.4)
• P (−2 < X ≤ 0) . = 0.1534 Standardizing: z₁ = (-2 - 6) / 5 = -1.6 z₂ = (0 - 6) / 5 = -1.2 P(-1.6 < Z ≤ -1.2) • P (| X − 6| < 5) . = 0.6827 P(1 < X < 11) Standardizing: z₁ = (1 - 6) / 5 = -1 z₂ = (11 - 6) / 5 = 1 P(-1 < Z < 1) 3. (14 points) The strength X of a certain material is such that its distribution is found by X = e Y , where Y is N (10 , 1). Find the CDF and PDF of X , and compute P (10 , 000 < X < 20 , 000). Note: F ( x ) = P ( X ≤ x ) = P ( e Y ≤ x ) = P ( Y ≤ ln x ) so that the random variable X is said to have a lognormal distribution . F(x)=P(X≤x)=P(e Y ≤x)=P(Y≤lnx) P(10000≤X≤20000)=P(ln(10000)≤Y≤ln(20000))≈P(−0.7897≤(Y−10)≤−0.0965) ≈0.24692 4. (20 points) Assume that X ∼ N( µ,σ 2 ). Show that Var( X ) = σ 2 .
Page 1 of 2 5. (10 points) The lifetime in hours of an electronic tube is a random variable having a probability density function given by f ( x ) = xe − x ,x ≥ 0 . Compute the expected lifetime of such a tube.
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