/*! 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 72 I have three errands to take car... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 72

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 \(X_{l}\) 's are independent, and 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 \(.01\) ?

Short Answer

Expert verified
I should write 鈥淚 will return by 10:53 A.M.鈥� on my door.

Step by step solution

01

Define the Problem Variables

We have four random variables, \(X_1\), \(X_2\), \(X_3\), and \(X_4\), each representing time spent in minutes for different tasks. The total time \(T\) is the sum of these random variables, i.e., \(T = X_1 + X_2 + X_3 + X_4\). Given the mean (\(\mu\)) and standard deviation (\(\sigma\)) of each \(X_i\), we need to determine \(t\) such that \(P(T > t) = 0.01\).
02

Calculate Mean and Standard Deviation of Total Time

Since the \(X_i\)'s are independent normal variables, \(T\) is also normally distributed with mean \(\mu_T = \mu_1 + \mu_2 + \mu_3 + \mu_4\) and variance \(\sigma_T^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2\). Calculate \(\mu_T = 15 + 5 + 8 + 12 = 40\) minutes, and \(\sigma_T^2 = 4^2 + 1^2 + 2^2 + 3^2 = 16 + 1 + 4 + 9 = 30\). Thus, \(\sigma_T = \sqrt{30}\).
03

Use the Normal Distribution to Find Time

For a normal distribution, we know that \(P(T > t) = 0.01\) implies \(P(T \leq t) = 0.99\). Use the standard normal distribution table or calculator to find the z-score corresponding to 0.99, which is approximately 2.33. The formula for \(t\) becomes \( t = \mu_T + z \cdot \sigma_T \). Therefore, \( t = 40 + 2.33 \times \sqrt{30} \approx 52.76 \) minutes.
04

Calculate Return Time

Since you leave your office at 10:00 A.M., and want to ensure you return by time \(t\) calculated in minutes, add 52.76 minutes to 10:00 A.M. This results in approximately 10:53 A.M.

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Key Concepts

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

Random Variables
In the context of probability and statistics, a random variable is a numerical variable whose value depends on the outcomes of a random phenomenon. Random variables are used to quantify the outcomes of an experiment or process. They are essential in statistical modeling as they allow us to exploit the rich mathematical structure of probability distributions to analyze random processes.

In the given exercise, the random variables are defined as follows:
  • \( X_1 \), \( X_2 \), and \( X_3 \): These represent the time taken for each of the three errands, respectively.
  • \( X_4 \): This represents the time spent walking to, from, and between errands.
Each random variable \( X_i \) is characterized by a specific mean \( \mu_i \) and standard deviation \( \sigma_i \), highlighting the uncertainty in time for each task. Importantly, these random variables are assumed to be independent, meaning the time taken for one errand does not influence the time taken for another.The independence assumption simplifies calculations involving these variables because independent normal random variables, when summed, result in another normal random variable. It's this beauty of normal distribution that allows analysts to model complex systems with relative ease.
Expected Value
The expected value, often denoted by \( E[X] \) or \( \mu \), is a fundamental concept in probability that represents the average or mean value of a random variable over numerous trials of an experiment. It's essentially what you would "expect" to happen on average in a large number of experiments.

In our exercise, the expected value is the average total time \( T \) spent on all tasks, including walking. Since \( T = X_1 + X_2 + X_3 + X_4 \), and all variables are normal, the expected value of the sum \( T \) is simply the sum of the expected values of the individual random variables:
  • \( \mu_T = \mu_1 + \mu_2 + \mu_3 + \mu_4 \)
  • For this problem, \( \mu_T = 15 + 5 + 8 + 12 = 40 \) minutes.

This expected value signals that, on average, you would spend about 40 minutes completing all errands. However, it's crucial to note that this is an average, and actual outcomes may vary due to the inherent variability represented by the standard deviation.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. In the context of a normal distribution, it reflects how spread out the values of a random variable are from the mean. The larger the standard deviation, the more spread out the data points are around the mean.

In the exercise, each errand time \( X_i \) has its standard deviation \( \sigma_i \), indicating variability in completion time for each task. For the cumulative time \( T \), the standard deviation \( \sigma_T \) is calculated from the variances of the individual random variables:
  • Variance of \( T \), \( \sigma_T^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2 \)
  • So, \( \sigma_T^2 = 16 + 1 + 4 + 9 = 30 \)
  • \( \sigma_T = \sqrt{30} \approx 5.48 \)
This computation gives a standard deviation for the total errand time, indicating that 68% of your total errand times will likely fall within \( 5.48 \) minutes of the mean time of 40 minutes. Understanding this variation helps in planning by allowing for consideration of possible delays or expedited times.

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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 l's components selected, \(Y=\) the number of supplier 2 '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 distribution-sampling without replacement from a finite population consisting of more than two categories.)

Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\), and \(X_{6}\) denote the numbers of blue, brown, green, orange, red, and yellow \(\mathrm{M} \& \mathrm{M}\) candies, respectively, in a sample of size \(n\). Then these \(X_{l}\) 's have a multinomial distribution. According to the M\&M Web site, the color proportions are \(p_{1}=.24, p_{2}=.13, p_{3}=.16\), \(p_{4}=.20, p_{5}=.13\), and \(p_{6}=.14\). a. If \(n=12\), what is the probability that there are exactly two M\&Ms of each color? b. For \(n=20\), what is the probability that there are at most five orange candies? [Hint: Think of an orange candy as a success and any other color as a failure.] c. In a sample of \(20 \mathrm{M} \& \mathrm{Ms}\), what is the probability that the number of candies that are blue, green, or orange is at least 10 ?

Suppose that when the \(\mathrm{pH}\) of a certain chemical compound is \(5.00\), the \(\mathrm{pH}\) measured by a randomly selected beginning chemistry student is a random variable with mean \(5.00\) and standard deviation 2. A large batch of the compound is subdivided and a sample given to each student in a morning lab and each student in an afternoon lab. Let \(\bar{X}=\) the average \(\mathrm{pH}\) as determined by the morning students and \(\bar{Y}=\) the average \(\mathrm{pH}\) as determined by the afternoon students. a. If \(\mathrm{pH}\) is a normal variable and there are 25 students in each lab, compute \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\). [Hinr: \(\bar{X}-\bar{Y}\) is a linear combination of normal variables, so is normally distributed. Compute \(\mu_{\bar{X}-\bar{Y}}\) and \(\sigma_{\bar{X}-\bar{Y}}\).] b. If there are 36 students in each lab, but pH determinations are not assumed normal, calculate (approximately) \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\).

You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in 1000 s of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most 1000 hours (i.e., \(X \leq 1\) and \(Y \leq 1\) )? c. What is the probability that the total lifetime of the two bulbs is at most 2 ? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0, y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and 2 ?

Five automobiles of the same type are to be driven on a 300 mile trip. The first two will use an economy brand of gasoline, and the other three will use a name brand. Let \(X_{1}, X_{2}\), \(X_{3}, X_{4}\), and \(X_{3}\) be the observed fuel efficiencies (mpg) for the five cars. Suppose these variables are independent and normally distributed with \(\mu_{1}=\mu_{2}=20, \mu_{3}=\mu_{4}=\mu_{5}=21\), and \(\sigma^{2}=4\) for the economy brand and \(3.5\) for the name brand. Define an rv \(Y\) by $$ Y=\frac{X_{1}+X_{2}}{2}-\frac{X_{3}+X_{4}+X_{5}}{3} $$ so that \(Y\) is a measure of the difference in efficiency between economy gas and name-brand gas. Compute \(P(0 \leq Y)\) and \(P(-1 \leq Y \leq 1)\). [Hint: \(Y=a_{1} X_{1}+\ldots+a_{5} X_{5}\), with \(\left.a_{1}=\frac{1}{2}, \ldots, a_{5}=-\frac{1}{3} .\right]\)
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