/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Suppose the reaction temperature... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose the reaction temperature \(X\) (in \({ }^{\circ} \mathrm{C}\) ) in a certain chemical process has a uniform distribution with \(A=-5\) and \(B=5 .\) a. Compute \(P(X<0)\). b. Compute \(P(-2.5

Short Answer

Expert verified
a. 0.5; b. 0.5; c. 0.5; d. 0.4.

Step by step solution

01

Understanding the Uniform Distribution

The uniform distribution means that every interval of the same length has the same probability. It is described by the parameters \( A \) and \( B \), where \( A = -5 \) and \( B = 5 \). The probability density function (PDF) for a uniform distribution is \( f(x) = \frac{1}{B-A} \) for \( A \leq x \leq B \). In this case, \( f(x) = \frac{1}{5 - (-5)} = \frac{1}{10} \).
02

Calculating P(X

The probability \( P(X < 0) \) is the area under the PDF from \( A = -5 \) to \( 0 \). Calculate it as: \[ P(X < 0) = \frac{0 - (-5)}{10} = \frac{5}{10} = 0.5 \]
03

Calculating P(-2.5

The probability \( P(-2.5 < X < 2.5) \) is the area under the PDF from \( -2.5 \) to \( 2.5 \). Calculate it as: \[ P(-2.5 < X < 2.5) = \frac{2.5 - (-2.5)}{10} = \frac{5}{10} = 0.5 \]
04

Calculating P(-2≤X≤3)

The probability \( P(-2 \leq X \leq 3) \) is the area under the PDF from \( -2 \) to \( 3 \). Calculate it as: \[ P(-2 \leq X \leq 3) = \frac{3 - (-2)}{10} = \frac{5}{10} = 0.5 \]
05

Calculating P(k

Given the condition \( -5 < k < k+4 < 5 \), calculate the probability \( P(k < X < k+4) \). This simplifies to: \[ P(k < X < k+4) = \frac{k+4 - k}{10} = \frac{4}{10} = 0.4 \] This is true for any \( k \) in the specified range.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Density Function
The Probability Density Function (PDF) for a uniform distribution helps to determine the likelihood of a random variable falling within a particular range. For a uniform distribution, every outcome in the interval is equally likely to occur. This means the graph of the PDF is a flat, horizontal line.

In a uniform distribution defined between two points, such as our example with bounds \(A = -5\) and \(B = 5\), the PDF is given by the formula \( f(x) = \frac{1}{B-A} \) for \( A \leq x \leq B \). For our parameters, this simplifies to a constant value of \(\frac{1}{10}\) across the interval.

This PDF can be used to compute probabilities by calculating the area under the curve. Since the curve is flat, the probability between any two points is simply the length of the interval divided by \(10\), which is the total length from \(A\) to \(B\).
Continuous Probability Distribution
Continuous Probability Distributions apply to continuous random variables, which can take any value within a given range. Unlike discrete distributions, which concern specific values, continuous distributions deal with intervals. In our problem, we consider the interval from \(-5\) to \(5\).

A uniform distribution is one of the simplest types of continuous probability distributions. It is characterized by a constant PDF, indicating that every value within this range is equally probable. This means the probability of the variable taking any single, specific value is literally zero; instead, we consider probabilities across intervals.
  • The length of the interval directly determines the probability under a uniform distribution; larger intervals have higher probabilities.
  • The total area under the PDF curve between \(A\) and \(B\) sums up to \(1\), representing the total certainty that the variable is somewhere within this range.
Understanding continuous distributions is foundational to solving probability tasks involving such variables.
Uniform Probability Calculations
Calculating probabilities in uniform distributions is straightforward since each outcome in the specified interval is equally likely. The key is to properly understand the range of interest and apply the basic formula for uniform probabilities.

In the given exercise, we compute probabilities by finding the proportion of the interval over the entire length from \(A\) to \(B\). For instance, to find \(P(X < 0)\), calculate the length from \(-5\) to \(0\) and divide by total interval length \(10\):
\[ P(X < 0) = \frac{0 - (-5)}{10} = 0.5 \]

Similarly, for intervals \(P(-2.5 < X < 2.5)\) and \(P(-2 \leq X \leq 3)\), the procedure remains the same:
  • Find the length of the desired interval.
  • Divide by the total interval length of \(10\).
For a flexible approach, like finding \(P(k < X < k+4)\), ensure the specified range \(k\) satisfies given conditions: \(-5 < k < k+4 < 5\). Then, apply the same method for probability, leading to a constant result for any \(k\) within this range, which is \(0.4\).

This methodology highlights the simplicity and elegance of uniform probability calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Extensive experience with fans of a certain type used in diesel engines has suggested that the exponential distribution provides a good model for time until failure. Suppose the mean time until failure is 25,000 hours. What is the probability that a. A randomly selected fan will last at least 20,000 hours? At most 30,000 hours? Between 20,000 and 30,000 hours? b. The lifetime of a fan exceeds the mean value by more than 2 standard deviations? More than 3 standard deviations?

The special case of the gamma distribution in which \(\alpha\) is a positive integer \(n\) is called an Erlang distribution. If we replace \(\beta\) by \(1 / \lambda\) in Expression (4.8), the Erlang pdf is $$ f(x, \lambda, n)=\left\\{\begin{array}{cc} \frac{\lambda(\lambda x)^{n-1} e^{-\lambda x}}{(n-1) !} & x \geq 0 \\ 0 & x<0 \end{array}\right. $$ It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter \(\lambda\), then the total time \(X\) that elapses before all of the next \(n\) events occur has pdf \(f(x ; \lambda, n)\). a. What is the expected value of \(X\) ? If the time (in minutes) between arrivals of successive customers is exponentially distributed with \(\lambda=.5\), how much time can be expected to elapse before the tenth customer arrives? b. If customer interarrival time is exponentially distributed with \(\lambda=.5\), what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next \(30 \mathrm{~min}\) ? c. The event \(\\{X \leq t\\}\) occurs iff at least \(n\) events occur in the next \(t\) units of time. Use the fact that the number of events occurring in an interval of length \(t\) has a Poisson distribution with parameter \(\lambda t\) to write an expression (involving Poisson probabilities) for the Erlang cdf \(F(t, \lambda, n)=\) \(P(X \leq t)\)

Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5 (information in the article "Mathematical Model of Chloride Concentration in Human Blood," J. of Med. Engr. and Tech., 2006: 25-30, including a normal probability plot as described in Section 4.6, supports this assumption). a. What is the probability that chloride concentration equals 105 ? Is less than 105 ? Is at most 105 ? b. What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? Does this probability depend on the values of \(\mu\) and \(\sigma\) ? c. How would you characterize the most extreme . \(1 \%\)

Let \(X\) have a standard gamma distribution with \(\alpha=7\). Evaluate the following: a. \(P(X \leq 5)\) b. \(P(X<5)\) c. \(P(X>8)\) d. \(P(3 \leq X \leq 8)\) e. \(P(36)\)

Suppose that \(10 \%\) of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let \(X\) denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that \(X\) is a. At most 30 ? b. Less than 30 ? c. Between 15 and 25 (inclusive)?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.