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If a point is randomly located in an interval \((a, b)\) and if \(Y\) denotes the location of the point, then \(Y\) is assumed to have a uniform distribution over \((a, b) .\) A plant efficiency expert randomly selects a location along a 500 -foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is a. within 25 feet of the end of the line? b. within 25 feet of the beginning of the line? c. closer to the beginning of the line than to the end of the line?

Short Answer

Expert verified
a. 0.05, b. 0.05, c. 0.5

Step by step solution

01

Understanding the Uniform Distribution

Since the point is randomly located along the assembly line, the location of the point, denoted by \(Y\), follows a uniform distribution over the interval \((0, 500)\). This means that every point along the line has an equal probability of being selected.
02

Calculate Probability for Within 25 Feet of the End (Part a)

To find the probability that the point is within 25 feet of the end of the line, we need to consider the interval \((475, 500)\) since the end of the line is at 500 feet. Probability can be calculated as \(\frac{\text{Length of Interval}}{\text{Total Length}} = \frac{25}{500} = 0.05\).
03

Calculate Probability for Within 25 Feet of the Beginning (Part b)

Similarly, for the point to be within 25 feet of the beginning of the line, the interval is \((0, 25)\). Thus, the probability is \(\frac{25}{500} = 0.05\).
04

Calculate Probability for Closer to Beginning than to End (Part c)

The location is closer to the beginning than to the end if it is less than halfway, i.e., within \(0\) to \(250\) feet. The probability is \(\frac{250}{500} = 0.5\).

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Key Concepts

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

Probability Calculation
When dealing with a uniform distribution, calculating probability involves understanding the length of the interval of interest in comparison to the total length of the distribution. In our scenario, with the 500-foot assembly line, each segment of the line is equally likely to be randomly chosen.
This leads to the simple probability formula:
  • Calculate the length of the desired interval.
  • Divide by the total length of the distribution.
For example, if we want to determine the probability that a randomly chosen point is within 25 feet of the end of the line, we calculate the probability by dividing the 25-foot segment by the total 500 feet. This gives:\[ P = \frac{25}{500} = 0.05 \] This means there is a 5% chance of selecting a point within 25 feet from the end.
In each computation, this formula holds as it reflects the core idea of a uniform distribution where all intervals of equal length are equi-probable.
Interval Analysis
Interval analysis involves a critical look into how the intervals are set within a given range, like our 500-foot line. By breaking down the line into smaller segments or intervals, we can examine specific probabilities associated with each part.
Consider the intervals analyzed in our exercise:
  • (0, 25) - within 25 feet of the start of the line
  • (475, 500) - within 25 feet of the end of the line
  • (0, 250) - closer to the beginning than the end
Each of these intervals represents a specific area of interest in random selection probability. The interval (0, 25) covers the initial part of the line, providing insights into the probability of selecting a starting location, while the interval (475, 500) reflects the ending segment.
Importantly, the interval (0, 250) is crucial, especially for part c of our original exercise. Here, each point from 0 to 250 is closer to the beginning than the end, reflecting a larger section spanning halfway along the line. This allows detailed comprehension of specific event likelihoods within a uniform distribution.
Random Selection
Random selection in the uniform distribution means any point along our interval or line has an equal chance of being chosen. In the context of a 500-foot assembly line, this implies every single foot, from 0 to 500 feet, is just as likely as any other to be selected.
This elemental concept ensures fairness and unpredictability.
  • Each spot’s probability: equal and constant
  • No point has a preference over another
  • Selection remains unbiased across the distance
A key factor is that this randomness leads to the simple probabilities calculated earlier.
By equally considering all points, we effectively mirror real-world random processes you might encounter outside mathematical exercises, letting every decision hold equal weight across the span being measured. This core randomness principle plays a significant role in many real analytical and practical applications across fields like operations research and quality control.

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Most popular questions from this chapter

Show that the normal density with parameters \(\mu\) and \(\sigma\) has inflection points at the values \(\mu-\sigma\) and \(\mu+\sigma .\) (Recall that an inflection point is a point where the curve changes direction from concave up to concave down, or vice versa, and occurs when the second derivative changes sign. Such a change in sign may occur when the second derivative equals zero.)

The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered. The delivery time. \(Y\), is uniformly distributed on the interval one to five days. The cost of a board failure and interruption includes the fixed cost \(c_{0}\) of a new board and a cost that increases proportionally to \(Y^{2} .\) If \(C\) is the cost incurred, \(C=c_{0}+c_{1} Y^{2}\). a. Find the probability that the delivery time exceeds two days. b. In terms of \(c_{0}\) and \(c_{1}\), find the expected cost associated with a single failed circuit board.

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A random variable \(Y\) is said to have a log-normal distribution if \(X=\ln (Y)\) has a normal distribution. (The symbol In denotes natural logarithm.) In this case \(Y\) must be non negative. The shape of the log-normal probability density function is similar to that of the gamma distribution, with a long tail to the right. The equation of the log-normal density function is given by $$ f(y)=\left\\{\begin{array}{ll} \left.\frac{1}{\sigma y \sqrt{2 \pi}} e^{-(\ln (y)-\mu)^{2} / 2 \sigma^{2}}\right), & y>0 \\ 0, & \text { elsewhere } \end{array}\right. $$ Because \(\ln (y)\) is a monotonic function of \(y\) $$ P(Y \leq y)=P[\ln (Y) \leq \ln (y)]=P[X \leq \ln (y)] $$ where \(X\) has a normal distribution with mean \(\mu\) and variance \(\sigma^{2} .\) Thus, probabilities for random variables with a log-normal distribution can be found by transforming them into probabilities that can be computed using the ordinary normal distribution. If \(Y\) has a log-normal distribution with \(\mu=4\) and \(\sigma^{2}=1,\) find a. \(P(Y \leq 4)\) b. \(P(Y>8)\)

Identify the distributions of the random variables with the following moment- generating functions: a. \(m(t)=(1-4 t)^{-2}\) b. \(m(t)=1 /(1-3.2 t)\) C. \(m(t)=e^{-5 t+6 t^{2}}\)

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