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

Standard normal:The standard normal distribution, also called the z-distribution, is a special…

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) =

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter13: Probability And Calculus

Section13.2: Expected Value And Variance Of Continuous Random Variables

Problem 10E