Homework Help is Here – Start Your Trial Now!
Math
Probability
Let X₁, X₂, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values #₁, #₂, and #3 and variances ₂2, ₂2, and o32, respectively. (Round your answers to four decimal places.) USE SALT (a) If H₁ = H₂ = H3 = 60 and ₁² = 0₂² = 03² = 18, calculate P(T ≤ 198) and P(144 ≤ T ≤ 198). P(T ≤ 198) = P(144 ≤ T ≤ 198) = 0.9986 IX (b) Using the μ's and oi's given in part (a), calculate both P(54 ≤ X) and P(58 ≤ x ≤ 62). P(54 ≤X) = 0.9875 X P(58 ≤ x ≤ 62) = (c) Using the μ's and o/'s given in part (a), calculate P(-12 ≤ X₁ - 0.5X₂ - 0.5X3 ≤ 6). P(-12 ≤ X₁0.5X₂ - 0.5X3 ≤ 6) = Interpret the quantity P(-12 ≤ X₁ - 0.5X₂-0.5X3 ≤ 6). O The quantity represents the probability that the difference between X3 and the average of X₁ and X₂ is between -12 and 6. O The quantity represents the probability that the difference between X3 and the sum of X₁ and X₂ is between -12 and 6. The quantity represents the probability that X₁, X2, and X3 are all between -12 and 6. The quantity represents the probability that the difference between X₁ and the average of X₂ and X3 is between -12 and 6. The quantity represents the probability that the difference between X, and the sum of X₂ and X3 is between -12 and 6. C Let X1,... (d) If μ₁ = 50, M₂ = 60, H3 = 70, 0₁2 = 12, 0₂2 = 10, and 32 = 14, calculate P(X₁ + X₂ + X3 S 192) and also P(X₁ + X₂ 2 2X3). P(X₁ + X₂ + X3 ≤ 192) = P(X₁ + X₂ ≥ 2X3) =

Let X₁, X₂, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values #₁, #₂, and #3 and variances ₂2, ₂2, and o32, respectively. (Round your answers to four decimal places.) USE SALT (a) If H₁ = H₂ = H3 = 60 and ₁² = 0₂² = 03² = 18, calculate P(T ≤ 198) and P(144 ≤ T ≤ 198). P(T ≤ 198) = P(144 ≤ T ≤ 198) = 0.9986 IX (b) Using the μ's and oi's given in part (a), calculate both P(54 ≤ X) and P(58 ≤ x ≤ 62). P(54 ≤X) = 0.9875 X P(58 ≤ x ≤ 62) = (c) Using the μ's and o/'s given in part (a), calculate P(-12 ≤ X₁ - 0.5X₂ - 0.5X3 ≤ 6). P(-12 ≤ X₁0.5X₂ - 0.5X3 ≤ 6) = Interpret the quantity P(-12 ≤ X₁ - 0.5X₂-0.5X3 ≤ 6). O The quantity represents the probability that the difference between X3 and the average of X₁ and X₂ is between -12 and 6. O The quantity represents the probability that the difference between X3 and the sum of X₁ and X₂ is between -12 and 6. The quantity represents the probability that X₁, X2, and X3 are all between -12 and 6. The quantity represents the probability that the difference between X₁ and the average of X₂ and X3 is between -12 and 6. The quantity represents the probability that the difference between X, and the sum of X₂ and X3 is between -12 and 6. C Let X1,... (d) If μ₁ = 50, M₂ = 60, H3 = 70, 0₁2 = 12, 0₂2 = 10, and 32 = 14, calculate P(X₁ + X₂ + X3 S 192) and also P(X₁ + X₂ 2 2X3). P(X₁ + X₂ + X3 ≤ 192) = P(X₁ + X₂ ≥ 2X3) =

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
See similar textbooks
icon
Related questions
Question
Help!!!! Answer all clearly!
**Title: Statistical Analysis of Repair Times Using Normal Distribution** Let \(X_1\), \(X_2\), and \(X_3\) represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal random variables with expected values \(\mu_1\), \(\mu_2\), and \(\mu_3\) and variances \(\sigma_1^2\), \(\sigma_2^2\), and \(\sigma_3^2\), respectively. The goal is to answer the following questions with rounded answers to four decimal places. ### Part (a) If \(\mu_1 = \mu_2 = \mu_3 = 60\) and \(\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 18\), calculate the following probabilities: \[ P(T_O \le 198) \] \[ P(144 \le T_O \le 198) \] Given: \[ P(T_O \le 198) \] Solution: \[P(T_O \le 198) = 0.9986 \] ### Part (b) Using the means \(\mu_i\) and variances \(\sigma_i\) given in part (a), calculate the following probabilities: \[ P(54 \le \bar{X}) \] \[ P(58 \le \bar{X} \le 62) \] Given: \[ P(54 \le \bar{X}) \] Solution: \[ P(54 \le \bar{X}) = 0.9875 \] ### Part (c) Using the \(\mu_i\)'s and \(\sigma_i\)'s given in part (a), calculate the following probability: \[ P(-12 \le X_1 - 0.5X_2 - 0.5X_3 \le 6) \] Interpret the quantity \(P(-12 \le X_1 - 0.5X_2 - 0.5X_3 \le 6)\). Options for interpretation: 1. The quantity represents the probability that the difference between \(X_3\) and the average of \(X_1\) and \(X_2\) is between \(-12\) and 6. 2. The quantity represents
Transcribed Image Text:**Title: Statistical Analysis of Repair Times Using Normal Distribution** Let \(X_1\), \(X_2\), and \(X_3\) represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal random variables with expected values \(\mu_1\), \(\mu_2\), and \(\mu_3\) and variances \(\sigma_1^2\), \(\sigma_2^2\), and \(\sigma_3^2\), respectively. The goal is to answer the following questions with rounded answers to four decimal places. ### Part (a) If \(\mu_1 = \mu_2 = \mu_3 = 60\) and \(\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 18\), calculate the following probabilities: \[ P(T_O \le 198) \] \[ P(144 \le T_O \le 198) \] Given: \[ P(T_O \le 198) \] Solution: \[P(T_O \le 198) = 0.9986 \] ### Part (b) Using the means \(\mu_i\) and variances \(\sigma_i\) given in part (a), calculate the following probabilities: \[ P(54 \le \bar{X}) \] \[ P(58 \le \bar{X} \le 62) \] Given: \[ P(54 \le \bar{X}) \] Solution: \[ P(54 \le \bar{X}) = 0.9875 \] ### Part (c) Using the \(\mu_i\)'s and \(\sigma_i\)'s given in part (a), calculate the following probability: \[ P(-12 \le X_1 - 0.5X_2 - 0.5X_3 \le 6) \] Interpret the quantity \(P(-12 \le X_1 - 0.5X_2 - 0.5X_3 \le 6)\). Options for interpretation: 1. The quantity represents the probability that the difference between \(X_3\) and the average of \(X_1\) and \(X_2\) is between \(-12\) and 6. 2. The quantity represents
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer