Today, the waves are crashing onto the beach every 4.9 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.9 seconds. Round to 4 decimal places where possible.. a. The mean of this distribution is 2.45 O b. The standard deviation is 1.4145 O c. The probability that wave will crash onto the beach exactly 3.7 seconds after the person arrives is P(x = 3.7) = 0 d. The probability that the wave will crash onto the beach between 1.6 and 3 seconds after the person: arrives is P(1.6 < x < 3) = 0.2857 O e. The probability that it will take longer than 1.78 seconds for the wave to crash onto the beach after the person arrives is P(x > 1.78) = 0.6367 ✓ OF f. Suppose that the person has already been standing at the shoreline for 0.1 seconds without a wave crashing in. Find the probability that it will take between 1.6 and 2.4 seconds for the wave to crash onto the shoreline. 0.7576 xg. 67% of the time a person will wait at least how long before the wave crashes in? seconds. h. Find the minimum for the upper quartile. seconds.
Today, the waves are crashing onto the beach every 4.9 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.9 seconds. Round to 4 decimal places where possible.. a. The mean of this distribution is 2.45 O b. The standard deviation is 1.4145 O c. The probability that wave will crash onto the beach exactly 3.7 seconds after the person arrives is P(x = 3.7) = 0 d. The probability that the wave will crash onto the beach between 1.6 and 3 seconds after the person: arrives is P(1.6 < x < 3) = 0.2857 O e. The probability that it will take longer than 1.78 seconds for the wave to crash onto the beach after the person arrives is P(x > 1.78) = 0.6367 ✓ OF f. Suppose that the person has already been standing at the shoreline for 0.1 seconds without a wave crashing in. Find the probability that it will take between 1.6 and 2.4 seconds for the wave to crash onto the shoreline. 0.7576 xg. 67% of the time a person will wait at least how long before the wave crashes in? seconds. h. Find the minimum for the upper quartile. seconds.
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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Help with f,g and h
![**Understanding Uniform Distribution in Wave Crashing Intervals**
Today, we will explore a real-world example of uniform distribution by observing the timing of waves crashing onto the beach. Waves crash onto the beach every 4.9 seconds. The time from when a person arrives at the shoreline until a crashing wave is observed follows a uniform distribution from 0 to 4.9 seconds. Below, we'll address various calculations based on this scenario, rounding to four decimal places where necessary.
**a. The Mean of This Distribution**
The mean of this uniform distribution is:
\[ \mu = 2.45 \]
---
**b. The Standard Deviation**
The standard deviation of this distribution is:
\[ \sigma = 1.4145 \]
---
**c. Exact Probability of Wave Crashing**
The probability that a wave will crash onto the beach exactly 3.7 seconds after the person arrives is:
\[ P(x = 3.7) = 0 \]
Explanation: In a continuous uniform distribution, the probability of observing any single exact point is zero.
---
**d. Probability of Wave Crashing Between 1.6 and 3 Seconds**
To find the probability that a wave crashes between 1.6 and 3 seconds after a person arrives, we calculate:
\[ P(1.6 < x < 3) = 0.2857 \]
---
**e. Probability of Wave Crashing After 1.78 Seconds**
The probability that it will take longer than 1.78 seconds for the wave to crash onto the beach after the person arrives is:
\[ P(x > 1.78) = 0.6367 \]
---
**f. Conditional Probability with Initial Time Standing**
Suppose that the person has already been standing at the shoreline for 0.1 seconds without a wave crashing in. We need to find the probability that it will take between 1.6 and 2.4 seconds for the wave to crash onto the shoreline:
\[ P(1.6 < x < 2.4) = 0.7576 \]
---
**g. Minimum Time for the Upper Quartile**
67% of the time a person will wait at least how long before the wave crashes in?
\[ \text{Minimum for the upper quartile} = \text{seconds} \]
**Summary of Calculations:**
1. Mean: \(2.45\)
2. Standard Deviation:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc3b72c7-eb99-406f-83fe-82fc41b2a8b5%2F007c38df-034b-46f9-888b-37fd21acf57e%2F3jp3o7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Uniform Distribution in Wave Crashing Intervals**
Today, we will explore a real-world example of uniform distribution by observing the timing of waves crashing onto the beach. Waves crash onto the beach every 4.9 seconds. The time from when a person arrives at the shoreline until a crashing wave is observed follows a uniform distribution from 0 to 4.9 seconds. Below, we'll address various calculations based on this scenario, rounding to four decimal places where necessary.
**a. The Mean of This Distribution**
The mean of this uniform distribution is:
\[ \mu = 2.45 \]
---
**b. The Standard Deviation**
The standard deviation of this distribution is:
\[ \sigma = 1.4145 \]
---
**c. Exact Probability of Wave Crashing**
The probability that a wave will crash onto the beach exactly 3.7 seconds after the person arrives is:
\[ P(x = 3.7) = 0 \]
Explanation: In a continuous uniform distribution, the probability of observing any single exact point is zero.
---
**d. Probability of Wave Crashing Between 1.6 and 3 Seconds**
To find the probability that a wave crashes between 1.6 and 3 seconds after a person arrives, we calculate:
\[ P(1.6 < x < 3) = 0.2857 \]
---
**e. Probability of Wave Crashing After 1.78 Seconds**
The probability that it will take longer than 1.78 seconds for the wave to crash onto the beach after the person arrives is:
\[ P(x > 1.78) = 0.6367 \]
---
**f. Conditional Probability with Initial Time Standing**
Suppose that the person has already been standing at the shoreline for 0.1 seconds without a wave crashing in. We need to find the probability that it will take between 1.6 and 2.4 seconds for the wave to crash onto the shoreline:
\[ P(1.6 < x < 2.4) = 0.7576 \]
---
**g. Minimum Time for the Upper Quartile**
67% of the time a person will wait at least how long before the wave crashes in?
\[ \text{Minimum for the upper quartile} = \text{seconds} \]
**Summary of Calculations:**
1. Mean: \(2.45\)
2. Standard Deviation:
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