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Chapter 4: Problem 2

Suppose the reaction temperature \(X\left(\right.\) in \(\left.{ }^{\circ} \mathrm{C}\right)\) in a 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, meaning 40% chance of falling within any specific 4-degree interval.

Step by step solution

01

Understand the Uniform Distribution

The reaction temperature \(X\) is uniformly distributed between \(A = -5\) and \(B = 5\). This means every value between \(-5\) and \(5\) is equally likely. The length of the interval is \(B - A = 10\). Thus, the probability density function \(f(x)\) is constant with \(f(x) = \frac{1}{10}\).
02

Compute P(X < 0)

To find \(P(X < 0)\), calculate the proportion of the interval \([-5, 5]\) that is less than zero. The length from \(-5\) to \(0\) is \(5\), so:\[P(X < 0) = \text{Length of interval from -5 to 0} \times f(x) = 5 \times \frac{1}{10} = 0.5\]
03

Compute P(-2.5 < X < 2.5)

Find the proportion of the interval \([-5, 5]\) that lies between \(-2.5\) and \(2.5\). The length from \(-2.5\) to \(2.5\) is \(5\), so:\[P(-2.5 < X < 2.5) = 5 \times \frac{1}{10} = 0.5\]
04

Compute P(-2 鈮� X 鈮� 3)

Find the proportion of the interval \([-5, 5]\) that lies between \(-2\) and \(3\). The length from \(-2\) to \(3\) is \(5\), so:\[P(-2 \leq X \leq 3) = 5 \times \frac{1}{10} = 0.5\]
05

Compute P(k < X < k + 4)

For \(k\) such that \(-5 < k < k + 4 < 5\), the length of any specific interval \(\left(k, k+4\right)\) is 4, so:\[P(k < X < k + 4) = 4 \times \frac{1}{10} = 0.4\] This means that the probability of \(X\) being within any four-degree interval within \([-5, 5]\) is fixed at 0.4 due to the uniform distribution.

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Key Concepts

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

Probability Density Function
A Probability Density Function (PDF) is fundamental when dealing with continuous data, such as the reaction temperature in a chemical process. In the case of a uniform distribution, every outcome in the interval is equally likely. This means that the probability density function is constant over the range of values. Think of the PDF as a flat line in a graph showing equal probabilities for all intervals of the range.

For the uniform distribution given in the problem, the interval is ":[-5, 5]". This range has a length of 10 units. The PDF is calculated by taking the reciprocal of the length of the interval.
  • So, in this case, the PDF is given by: \( f(x) = \frac{1}{10} \).
  • This value of \( \frac{1}{10} \) implies every degree Celsius within the interval is equally probable.
Probability Calculations
Probability calculations within a uniform distribution involve finding the area under the constant PDF line over the specified interval. Since the PDF is constant, the probability is simply the length of the interval multiplied by the PDF.

Let's break down the calculations:
  • \( P(X < 0) \) needs the interval :"[-5, 0]". The length is 5. Thus, \( P(X < 0) = 5 \times \frac{1}{10} = 0.5 \).
  • Similarly, \( P(-2.5 < X < 2.5) \) covers an interval of length 5, yielding \( P(-2.5 < X < 2.5) = 0.5 \).
  • For \( P(-2 \leq X \leq 3) \), the interval length is also 5, resulting in a probability of 0.5.
  • When calculating \( P(k < X < k+4) \), the interval has a length of 4, giving \( P(k < X < k+4) = 4 \times \frac{1}{10} = 0.4 \).
Probability Distribution
A probability distribution provides the complete picture of how probabilities are spread over different values. In a uniform distribution, every value within the interval is as likely as any other.

For continuous variables, such as temperature, the uniform probability distribution shows that:
  • The probability of any exact value (e.g., precisely -2 degrees) is technically zero because we cover an infinite number of potential decimal values within any small range.
  • What matters is the probability over an interval, which is determined by the interval's size and multiplied by the PDF.
  • This uniformity ensures that as long as the length of the interval remains unchanged, the probability remains constant, regardless of where the interval is positioned within the boundary limits of ":[-5, 5]".
Interval Calculation
Calculating probabilities within specific intervals depends on the length of the interval for a uniform distribution. This calculation is straightforward due to the flat nature of the PDF.

To calculate the probability of the temperature falling within a given interval:
  • Identify the length of the specified interval. For example, for \(-5 < k < k+4\), the interval's length is 4.
  • Multiply this length by the constant PDF, \( \frac{1}{10} \), to find the probability.
  • This approach applies uniformly within any segment of ":[-5, 5]" as long as it fits within these boundaries without crossing them.
  • For example, the identified interval from \(k\) to \(k+4\) has a probability of \(0.4\) because of its length of 4 multiplied by the PDF of \( \frac{1}{10} \).
This method allows easy calculation of probabilities for various intervals across the range and is intuitive due to the equal likelihood of each potential outcome.

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Most popular questions from this chapter

A 12 -in. bar clamped at both ends is subjected to an increasing amount of stress until it snaps. Let \(Y=\) the distance from the left end at which the break occurs. Suppose \(Y\) has pdf $$ f(y)=\left\\{\begin{array}{cl} \frac{y}{24}\left(1-\frac{y}{12}\right) & 0 \leq y \leq 12 \\ 0 & \text { otherwise } \end{array}\right. $$ Compute the following: a. The cdf of \(Y\), and graph it. b. \(P(Y \leq 4), P(Y>6)\), and \(P(4 \leq Y \leq 6)\). c. \(E(Y), E\left(Y^{2}\right)\), and \(V(Y)\). d. The probability that the break point occurs more than 2 in. from the expected break point. e. The expected length of the shorter segment when the break occurs.

An ecologist wishes to mark off a circular sampling region having radius \(10 \mathrm{~m}\). However, the radius of the resulting region is actually a random variable \(R\) with pdf $$ f(r)=\left\\{\begin{array}{cl} \frac{3}{4}\left[1-(10-r)^{2}\right] & 9 \leq r \leq 11 \\ 0 & \text { otherwise } \end{array}\right. $$ What is the expected area of the resulting circular region?

Suppose that the pdf of \(X\) is $$ f(x)=\left\\{\begin{array}{cc} 5-\frac{x}{8} & 0 \leq x \leq 4 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Show that \(E(X)=\frac{4}{3}, V(X)=\frac{8}{9}\). b. The coefficient of skewness is defined as \(E\left[(X-\mu)^{3}\right] / \sigma^{3}\). Show that its value for the given pdf is \(.566\). What would the skewness be for a perfectly symmetric pdf?

If the side of a square \(X\) is random with the pdf \(f_{X}(x)=x / 8,0

Suppose only \(70 \%\) of all drivers in a state regularly wear a seat belt. A random sample of 500 drivers is selected. What is the probability that a. Between 320 and 370 (inclusive) of the drivers in the sample regularly wear a seat belt? b. Fewer than 325 of those in the sample regularly wear a seat belt? Fewer than 315 ?
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