a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 1.7 seconds after the person arrives is P(x = 1.7) =
a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 1.7 seconds after the person arrives is P(x = 1.7) =
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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Question
![**Wave Crashing Time Analysis**
Today, the waves are crashing onto the beach every 4.5 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follow a Uniform distribution from 0 to 4.5 seconds. Round to 4 decimal places where possible.
a. **Mean of the Distribution**
The mean of this distribution is \(\dfrac{0 + 4.5}{2} = 2.25\) seconds.
b. **Standard Deviation**
The standard deviation is calculated using the formula for a Uniform distribution:
\[
\sigma = \sqrt{\dfrac{(b - a)^2}{12}}
\]
Where \(a\) is the minimum value (0 seconds) and \(b\) is the maximum value (4.5 seconds).
Therefore,
\[
\sigma = \sqrt{\dfrac{(4.5 - 0)^2}{12}} = \sqrt{\dfrac{20.25}{12}} = \sqrt{1.6875} \approx 1.2990
\]
Answer: **1.2990 seconds**
c. **Probability of Wave Crashing Exactly at 1.7 Seconds**
The probability that a wave will crash onto the beach exactly 1.7 seconds after the person arrives is \(P(x = 1.7)\).
For a continuous uniform distribution, the probability at a specific point is effectively 0.
Answer: **0**
d. **Probability of Wave Crashing Between 1.5 and 3.9 Seconds**
Calculate the probability that the wave will crash onto the beach between 1.5 and 3.9 seconds after the person arrives \(P(1.5 < x < 3.9)\):
Using the properties of a Uniform distribution:
\[
P(a < x < b) = \dfrac{b - a}{B - A}
\]
Here, \(a = 1.5\), \(b = 3.9\), \(A = 0\), and \(B = 4.5\).
\[
P(1.5 < x < 3.9) = \dfrac{3.9 - 1.5}{4.5 - 0} = \dfrac{2.4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5480d4d1-ea31-4a53-998f-6e5504770643%2Fcc4c54fe-eb4f-47ac-ba2a-d62c34a9d5c6%2Fcaws6f_processed.png&w=3840&q=75)
Transcribed Image Text:**Wave Crashing Time Analysis**
Today, the waves are crashing onto the beach every 4.5 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follow a Uniform distribution from 0 to 4.5 seconds. Round to 4 decimal places where possible.
a. **Mean of the Distribution**
The mean of this distribution is \(\dfrac{0 + 4.5}{2} = 2.25\) seconds.
b. **Standard Deviation**
The standard deviation is calculated using the formula for a Uniform distribution:
\[
\sigma = \sqrt{\dfrac{(b - a)^2}{12}}
\]
Where \(a\) is the minimum value (0 seconds) and \(b\) is the maximum value (4.5 seconds).
Therefore,
\[
\sigma = \sqrt{\dfrac{(4.5 - 0)^2}{12}} = \sqrt{\dfrac{20.25}{12}} = \sqrt{1.6875} \approx 1.2990
\]
Answer: **1.2990 seconds**
c. **Probability of Wave Crashing Exactly at 1.7 Seconds**
The probability that a wave will crash onto the beach exactly 1.7 seconds after the person arrives is \(P(x = 1.7)\).
For a continuous uniform distribution, the probability at a specific point is effectively 0.
Answer: **0**
d. **Probability of Wave Crashing Between 1.5 and 3.9 Seconds**
Calculate the probability that the wave will crash onto the beach between 1.5 and 3.9 seconds after the person arrives \(P(1.5 < x < 3.9)\):
Using the properties of a Uniform distribution:
\[
P(a < x < b) = \dfrac{b - a}{B - A}
\]
Here, \(a = 1.5\), \(b = 3.9\), \(A = 0\), and \(B = 4.5\).
\[
P(1.5 < x < 3.9) = \dfrac{3.9 - 1.5}{4.5 - 0} = \dfrac{2.4
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