### 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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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

### Probability Calculation of Play Completion Time

Consider 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 minutes

We aim to find the probability that the total performance (play plus intermission) is completed within 140 minutes.

### Aggregated Distribution of Performance Time

Define:
- $X$: Random variable representing the play time
- $Y$: Random variable representing the intermission time

These 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 Calculation

We need to calculate:
\[ P(Z < 140) \]

### Standardizing the Normal Variable

Convert $Z$ 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 Table

Determine the probability corresponding to

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