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Chapter 5: Problem 86

A student has a class that is supposed to end at 9:00 A.M. and another that is supposed to begin at 9:10 A.M. Suppose the actual ending time of the 9 A.M. class is a normally distributed rv \(X_{1}\) with mean \(9: 02\) and standard deviation \(1.5\) min and that the starting time of the next class is also a normally distributed rv \(X_{2}\) with mean \(9: 10\) and standard deviation \(1 \mathrm{~min}\). Suppose also that the time necessary to get from one classroom to the other is a normally distributed rv \(X_{3}\) with mean \(6 \mathrm{~min}\) and standard deviation \(1 \mathrm{~min}\). What is the probability that the student makes it to the second class before the lecture starts? (Assume independence of \(X_{1}, X_{2}\), and \(X_{3}\), which is reasonable if the student pays no attention to the finishing time of the first class.)

Short Answer

Expert verified
The probability that the student makes it to the second class on time is approximately 0.834 or 83.4%.

Step by step solution

01

Define Variables and Express Conditions

Let \( X = X_2 - (X_1 + X_3) \), which represents the difference between the start time of the second class and the sum of the end time of the first class and travel time. The condition to make it on time is \( X > 0 \).
02

Calculate the Mean of X

The expected value of \( X \) is given by the linearity of expectation: \[ E[X] = E[X_2] - (E[X_1] + E[X_3]) = 9:10 - (9:02 + 0:06) = 2 \text{ min} \].
03

Calculate the Variance of X

Since \( X_1, X_2, \) and \( X_3 \) are independent, the variance of \( X \) is: \[ \text{Var}(X) = \text{Var}(X_2) + \text{Var}(X_1) + \text{Var}(X_3) = 1^2 + 1.5^2 + 1^2 = 4.25 \].
04

Calculate the Standard Deviation of X

The standard deviation of \( X \) is the square root of its variance: \[ \sigma_X = \sqrt{4.25} \approx 2.06 \text{ min} \].
05

Determine Standard Normal Variable and Probability

Convert \( X \) to a standard normal variable \( Z = \frac{X - 2}{2.06} \).Find the probability that \( X > 0 \) or equivalently \( Z > -\frac{2}{2.06} \approx -0.97 \). Using a standard normal distribution table, \( P(Z > -0.97) \approx 0.8340 \).

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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 normally distributed variables, calculating probability involves determining the likelihood of an event happening based on the distribution's parameters.
In this exercise, the goal is to calculate the probability that a student arrives on time for a class given the start and end times, and traveling time between classes. To find this probability, we first define a new random variable, let's call it \(X\), that measures whether the student is on time. We are interested in the probability \(P(X > 0)\), meaning the student arrives before or when the class starts.
Using the properties of normal distributions, we calculate \(X\)'s mean and standard deviation from given values. Converting \(X\) to a standard normal distribution, allows us to use a standard normal distribution table to find \(P(Z > -0.97)\), showing us that the probability is approximately 0.8340, or 83.40% chance of being on time.
Random Variable
A random variable (RV) is a variable whose value depends on the outcomes of a random phenomenon. In this scenario, we deal with three random variables: \(X_1\), \(X_2\), and \(X_3\). They represent the time components crucial for calculating if the student will make it to class.- \( X_1 \) is the actual ending time of the first class, with \( \,mean=9:02 \) and \( \,SD=1.5 \) minutes.- \( X_2 \) is the start time of the second class, with \( \,mean=9:10 \) and \( \,SD=1 \) minute.- \( X_3 \) is the travel time between classes, with \( \,mean=6 \) minutes and \( \,SD=1 \) minute.These random variables are used collectively to calculate the probability of the student arriving on time. We assume these are normally distributed, making them part of the 'normal distribution' family, and independent so calculations related to the expected values and variances are simpler.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of a normal distribution, it helps us understand how spread out our data is around the mean.For each of the random variables \(X_1, X_2, \text{ and } X_3\), the standard deviation is given, which describes how each timing component can vary:- \(SD_{X_1} = 1.5 \text{ minutes} \)- \(SD_{X_2} = 1 \text{ minute} \)- \(SD_{X_3} = 1 \text{ minute} \)The standard deviation of \(X\), calculated by taking the square root of the summed variances of the independent random variables, is \( \sqrt{4.25} \approx 2.06 \text{ minutes} \).
This indicates how much the overall time can vary from the expected arrival time, useful for determining the probability of arriving on time when converted to a standard normal variable.
Expected Value
The expected value is a key concept in probability that gives us the average outcome we might expect from a random variable. Here, the expected value is used to find the mean time it would take for the student to arrive on time.For the new variable \( X = X_2 - (X_1 + X_3) \), the expected value, determined by subtracting the sum of expected values of \(X_1\) and \(X_3\) from \(X_2\), is calculated as:\[E[X] = 9:10 - (9:02 + 0:06) = 2 \text{ minutes}\]This \(2 \text{ minute} \) expected value translates into the student having, on average, 2 extra minutes to make it to class on time. Understanding expected value allows for better insights into how these time components combine to affect the probability of timely arrival.

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Most popular questions from this chapter

A stockroom currently has 30 components of a certain type, of which 8 were provided by supplier 1,10 by supplier 2 , and 12 by supplier 3. Six of these are to be randomly selected for a particular assembly. Let \(X=\) the number of supplier 1s components selected, \(Y=\) the number of supplier \(2 \mathrm{~s}\) components selected, and \(p(x, y)\) denote the joint pmf of \(X\) and \(Y\). a. What is \(p(3,2)\) ? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, \(p(3,2)=\) (number of outcomes with \(X=3\) and \(Y=2) /\) (total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.] b. Using the logic of part (a), obtain \(p(x, y)\). (This can be thought of as a multivariate hypergeometric distributionsampling without replacement from a finite population consisting of more than two categories.)

a. Show that \(\operatorname{Cov}(X, Y+Z)=\operatorname{Cov}(X, Y)+\operatorname{Cov}(X, Z)\). b. Let \(X_{1}\) and \(X_{2}\) be quantitative and verbal scores on one aptitude exam, and let \(Y_{1}\) and \(Y_{2}\) be corresponding scores on another exam. If \(\operatorname{Cov}\left(X_{1}, Y_{1}\right)=5, \operatorname{Cov}\left(X_{1}, Y_{2}\right)=1\), \(\operatorname{Cov}\left(X_{2}, Y_{1}\right)=2\), and \(\operatorname{Cov}\left(X_{2}, Y_{2}\right)=8\), what is the covariance between the two total scores \(X_{1}+X_{2}\) and \(Y_{1}+Y_{2} ?\)

I have three errands to take care of in the Administration Building. Let \(X_{i}=\) the time that it takes for the \(i\) th errand \((i=1,2,3)\), and let \(X_{4}=\) the total time in minutes that I spend walking to and from the building and between each errand. Suppose the \(X_{i}\) s are independent, normally distributed, with the following means and standard deviations: \(\mu_{1}=15\), \(\sigma_{1}=4, \mu_{2}=5, \sigma_{2}=1, \mu_{3}=8, \sigma_{3}=2, \mu_{4}=12, \sigma_{4}=3\). I plan to leave my office at precisely 10:00 A.M. and wish to post a note on my door that reads, "I will return by \(t\) A.M." What time \(t\) should I write down if I want the probability of my arriving after \(t\) to be \(.01\) ?

Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component 2 or component 3 functions. Let \(X_{1}, X_{2}\), and \(X_{3}\) denote the lifetimes of components 1,2 , and 3 , respectively. Suppose the \(X_{i}\) s are independent of one another and each \(X_{i}\) has an exponential distribution with parameter \(\lambda\). a. Let \(Y\) denote the system lifetime. Obtain the cumulative distribution function of \(Y\) and differentiate to obtain the pdf. [Hint: \(F(y)=P(Y \leq y)\); express the event \(\\{Y \leq y\\}\) in terms of unions and/or intersections of the three events \(\left\\{X_{1} \leq y\right\\},\left\\{X_{2} \leq y\right\\}\), and \(\left\\{X_{3} \leq y\right\\}\).] b. Compute the expected system lifetime.

Six individuals, including \(\mathrm{A}\) and \(\mathrm{B}\), take seats around a circular table in a completely random fashion. Suppose the seats are numbered \(1, \ldots, 6\). Let \(X=\) A's seat number and \(Y=\) B's seat number. If A sends a written message around the table to \(\mathrm{B}\) in the direction in which they are closest, how many individuals (including \(\mathrm{A}\) and \(\mathrm{B}\) ) would you expect to handle the message?
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