d. The probability that the wave will crash onto the beach between 0.2 and 3.3 seconds after the person arrives is P(0.2 < x < 3.3)= 0.0738 e. The probability that it will take longer than 2.84 seconds for the wave to crash onto the beach after the person arrives is P(x 2.84) = f. Find the minimum for the upper quartile. seconds.

d. The probability that the wave will crash onto the beach between 0.2 and 3.3 seconds after the person arrives is P(0.2 < x < 3.3)= 0.0738 e. The probability that it will take longer than 2.84 seconds for the wave to crash onto the beach after the person arrives is P(x 2.84) = f. 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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Question

Parts d,e,f

### Understanding Uniform Distribution in Real-Life Applications

**Scenario:**
The waves at the beach are crashing every 4.2 seconds. The time interval between a person's arrival at the shoreline and the next crashing wave follows a uniform distribution from 0 to 4.2 seconds. In this scenario, we analyze various statistical measures related to the time intervals and round answers to four decimal places where applicable.

**Key Points:**

**a. Mean of the Distribution:**

- The mean of this distribution is \(2.1\).

**b. Standard Deviation:**

- The standard deviation is approximately \(1.2124\).

**c. Probability of a Specific Time Interval:**

- The probability that a wave will crash exactly \(1.8\) seconds after the person arrives is \(0\).

**d. Probability of a Wave Crashing Within an Interval:**

- The probability that a wave crashes between \(0.2\) and \(3.3\) seconds after a person arrives is \(0.0738\).

**e. Probability of a Wave Taking Longer Than a Certain Time:**

- The probability that it will take longer than \(2.84\) seconds for the wave to crash after the person arrives is \( P(x \geq 2.84) \).

**f. Upper Quartile:**

- To find the minimum for the upper quartile, we solve this as a uniform distribution property.

**Graphical Representation:**

In this scenario, a uniform distribution is represented visually as a rectangle over the interval from \(0\) to \(4.2\) seconds. The height of the rectangle is determined by the fact that the total area under the distribution curve must equal \(1\).

- **Mean and Standard Deviation:** The mean of a uniform distribution on \([a, b]\) is given by \( \mu = \frac{a+b}{2} \).
- **Probability Calculation:** The probability of any single point in a continuous distribution, such as the exact \(1.8\) seconds, is zero. The probability over an interval \([c, d]\) within \([a, b]\) is calculated as proportional to the length of the interval, i.e., \( P(c \leq x \leq d) = \frac{d-c}{b-a} \).

We urge you to understand these principles thoroughly and apply them to various real-life situations, demonstrating

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