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Chapter 4: Problem 7

The time \(X(\mathrm{~min})\) for a lab assistant to prepare the equipment for a certain experiment is believed to have a uniform distribution with \(A=25\) and \(B=35\). a. Write the pdf of \(X\) and sketch its graph. b. What is the probability that preparation time exceeds \(33 \mathrm{~min}\) ? c. What is the probability that preparation time is within \(2 \mathrm{~min}\) of the mean time? [Hint: Identify \(\mu\) from the graph of \(f(x)\).] d. For any \(a\) such that \(25

Short Answer

Expert verified
a) The pdf is \(f(x) = \frac{1}{10}\) from \(25\) to \(35\). b) 0.2. c) 0.4. d) 0.2.

Step by step solution

01

Understanding Uniform Distribution

The uniform distribution for a random variable \(X\) which has a lower bound \(A = 25\) and an upper bound \(B = 35\) implies the probability density function (pdf) is constant over this interval. The formula for a uniform distribution pdf is \(f(x) = \frac{1}{B-A}\) for \(A \leq x \leq B\), otherwise zero.
02

Writing the Probability Density Function

Since \( A = 25 \) and \( B = 35 \), the pdf is given by: \[ f(x) = \frac{1}{35 - 25} = \frac{1}{10} \] for any \( x \) in the interval \([25, 35]\). The graph of this pdf is a constant function at height \(0.1\) from \(x = 25\) to \(x = 35\).
03

Probability of Preparation Time Exceeding 33 Minutes

To find this probability, integrate the pdf from \(x = 33\) to \(x = 35\):\[ P(X > 33) = \int_{33}^{35} \frac{1}{10} \, dx = \frac{1}{10} [x]_{33}^{35} = \frac{1}{10} (35 - 33) = 0.2 \]
04

Identifying the Mean and Calculating the Desired Probability

The mean \( \mu \) of a uniform distribution is given by \( \mu = \frac{A+B}{2} \). Substituting \(A = 25\) and \(B = 35\) gives \( \mu = \frac{25 + 35}{2} = 30 \). We want the probability that \(X\) is within \(2\) minutes of \(30\), i.e., \(28 \leq X \leq 32\):\[ P(28 \leq X \leq 32) = \int_{28}^{32} \frac{1}{10} \, dx = \frac{1}{10} [x]_{28}^{32} = \frac{1}{10} (32 - 28) = 0.4 \]
05

Probability for a Specified Range

Given any \(a\) such that \(25 < a < a+2 < 35\), the probability that \(X\) is between \(a\) and \(a+2\) is calculated as follows:\[ P(a < X < a+2) = \int_{a}^{a+2} \frac{1}{10} \, dx = \frac{1}{10} (a+2 - a) = \frac{2}{10} = 0.2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
The Probability Density Function (pdf) is a fundamental concept in probability which describes the likelihood of a continuous random variable falling within a specific range of values. In the context of a uniform distribution, the pdf is particularly straightforward. It is defined as being constant across the entire range of possible values. This means each value within the specified interval is equally likely to occur.

The formula for the pdf of a uniform distribution between two bounds, say \( A \) and \( B \), is given by:
  • \( f(x) = \frac{1}{B-A} \) for \( A \leq x \leq B \)
  • \( f(x) = 0 \) otherwise
For example, in a problem where a lab assistant's preparation time \( X \) is uniformly distributed between 25 and 35 minutes, the pdf is represented as \( f(x) = \frac{1}{10} \) within this interval. The pdf is a flat line, illustrating that every minute within the range is equally probable.
Mean of Uniform Distribution
The mean (or expected value) of a uniform distribution is a key measure of central tendency. Unlike a skewed distribution where the mean is not in the center, in a uniform distribution, the mean is exactly halfway between the lowest and highest values of the distribution interval.

The formula for finding the mean, \( \mu \), of a uniform distribution from \( A \) to \( B \) is:
  • \( \mu = \frac{A + B}{2} \)
Applying this formula, if we consider the preparation time for which \( A = 25 \) and \( B = 35 \), the mean time is calculated as \( \mu = \frac{25 + 35}{2} = 30 \) minutes. This value tells us that, on average, the time to prepare the equipment is 30 minutes, right at the center of the interval.
Integration in Probability
Integration plays a vital role in calculating probabilities for continuous random variables, such as those described by a pdf. In continuous probability, instead of summing probabilities like in discrete cases, we integrate the pdf over the desired interval.

For example, to find the probability that the lab preparation time exceeds 33 minutes, we need to evaluate the integral of the pdf from \( x = 33 \) to \( x = 35 \). This is expressed as:
  • \( P(X > 33) = \int_{33}^{35} \frac{1}{10} \, dx = 0.2 \)
The integration of the constant pdf over this interval gives us 0.2, indicating a 20% probability.

Similarly, integration will tell us the probability of the time being within 2 minutes of the mean. If the pdf is integrated from 28 to 32, the result is a probability of 0.4, reflecting a 40% chance.

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