### Probability Calculation of Play Completion TimeConsider a play that is being performed regularly at a local theatre. The duration of the play and its intermission can both be described using normal distributions.#### Given Data:1. **Play Time:** - Mean (\(\mu_{\text{play}}\)): 130 minutes - Standard Deviation (\(\sigma_{\text{play}}\)): 3 minutes 2. **Intermission Time:** - Mean (\(\mu_{\text{intermission}}\)): 15 minutes - Standard Deviation (\(\sigma_{\text{intermission}}\)): 5 minutesWe aim to find the probability that the total performance (play plus intermission) is completed within 140 minutes.### Aggregated Distribution of Performance TimeDefine:- \(X\): Random variable representing the play time- \(Y\): Random variable representing the intermission timeThese variables are independently normally distributed:- \(X \sim N(130, 3^2)\)- \(Y \sim N(15, 5^2)\)The total performance time \(Z\) is the sum of \(X\) and \(Y\):\[ Z = X + Y \]The mean (\(\mu_Z\)) and variance (\( \sigma_Z^2 \)) of \(Z\) are given by:\[ \mu_Z = \mu_{\text{play}} + \mu_{\text{intermission}} = 130 + 15 = 145 \]\[ \sigma_Z^2 = \sigma_{\text{play}}^2 + \sigma_{\text{intermission}}^2 = 3^2 + 5^2 = 9 + 25 = 34 \]\[ \sigma_Z = \sqrt{34} \]Thus, \(Z\) follows a normal distribution:\[ Z \sim N(145, \sqrt{34}) \]### Probability CalculationWe need to calculate:\[ P(Z < 140) \]### Standardizing the Normal VariableConvert \(Z\) to the standard normal variable \(Z'\):\[ Z' = \frac{Z - \mu_Z}{\sigma_Z} = \frac{140 - 145}{\sqrt{34}} = \frac{140 - 145}{5.831} = -0.857 \]### Using the Standard Normal Distribution TableDetermine the probability corresponding to

We have given that, Play time be modelled are normally distributed with mean (Mu) = 130 minutes…

3 A play is enjoying a long run at a theatre. It is found that the play time may be modelled as a normal variable with mean 130 minutes and standard deviation 3 minutes, and that the length of the intermission in the middle of the performance may be modelled by a normal variable with mean 15 minutes and standard deviaton 5 minutes. Find the probability that the performance is completed in less than 140 minutes.

3 A play is enjoying a long run at a theatre. It is found that the play time may be modelled as a normal variable with mean 130 minutes and standard deviation 3 minutes, and that the length of the intermission in the middle of the performance may be modelled by a normal variable with mean 15 minutes and standard deviaton 5 minutes. Find the probability that the performance is completed in less than 140 minutes.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 515 and a standard deviation of 175. The grading process of this year's exam has just begun. The average score of the 40 exams graded so far is 518. What is the probability that a sample of 40 exams will have a mean score of 518 or more if the exam scores follow the same distribution as in the past? Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places. Sub Continue
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