91Ó°ÊÓ

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?

Step by step solution

01

Define the Total Time Random Variable

Let \( Y \) be the total time spent on errands and walking, i.e., \( Y = X_1 + X_2 + X_3 + X_4 \). This defines the new random variable \( Y \) as the sum of four independent normal random variables.

02

Calculate the Mean of Total Time

Since the \( X_i \)'s are independent, the mean of \( Y \) is the sum of their means: \[ \mu_Y = \mu_1 + \mu_2 + \mu_3 + \mu_4 = 15 + 5 + 8 + 12 = 40. \]

03

Calculate the Standard Deviation of Total Time

The variance of \( Y \) is the sum of the variances of \( X_i \):\[ \sigma_Y^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2 = 4^2 + 1^2 + 2^2 + 3^2 = 16 + 1 + 4 + 9 = 30. \]The standard deviation \( \sigma_Y \) is the square root of this sum:\[ \sigma_Y = \sqrt{30}. \]

04

Determine the Time for 1% Probability

We know \( Y \sim N(40, \sqrt{30}) \). To find \( t \), we want the probability that \( Y \) is greater than \( t \) to be 0.01, which corresponds to:\[ P(Y \leq t) = 0.99. \]

05

Find the Z-Score for 99% Probability

Using the standard normal distribution table, find the \( z \)-score for 0.99 probability: \( z \approx 2.33 \).

06

Calculate the Required Time \( t \)

Use the formula \( t = \mu_Y + z \cdot \sigma_Y \):\[ t = 40 + 2.33 \times \sqrt{30} \approx 40 + 2.33 \times 5.48 \approx 52.77. \]

07

Determine the Return Time

Since time starts at 10:00 A.M., adding 52.77 minutes results in \( t \approx 10:53 \) A.M.

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.

Random Variable

In probability and statistics, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. It's a fundamental concept when dealing with uncertainties and probabilities. In the context of the exercise, each task or movement during the day is depicted as a random variable:

• The time taken for each errand is represented by individual random variables: \(X_1\), \(X_2\), \(X_3\).
• The travel time between and to/from errands is another random variable: \(X_4\).
• The total duration of errands, including walking time, is described by the random variable \(Y\), which is the sum of these four independent variables.

Since all these random variables are normally distributed, it implies that \(Y\) is also normally distributed. This means that the characteristics of a normal distribution can be applied to predict or calculate probabilities for \(Y\). Understanding the role of random variables helps in modeling real-world situations and calculating probabilities effectively.

Mean and Standard Deviation

In the normal distribution, the mean and standard deviation are critical in understanding data distribution characteristics. In our problem, they define the behavior of the random variables \(X_i\) and the overall random variable \(Y\).

• The mean refers to the average value, which is a measure of the central location of the data.
• The standard deviation indicates the level of variation or spread of data points from the mean.
• For each errand and movement time \(X_1\), \(X_2\), \(X_3\), and \(X_4\), the means were specified as 15, 5, 8, and 12 minutes respectively, while their corresponding standard deviations were 4, 1, 2, and 3.
• To find the mean of \(Y\), we summed up the means of the individual \(X_i\), resulting in \(\mu_Y = 40\).
• Similarly, the standard deviation of \(Y\) is calculated based on the sum of variances of \(X_i\). The variance, which is the square of the standard deviation, was summed up and square-rooted to find \(\sigma_Y = \sqrt{30}\).

This summarization of individual contributions to the total mean and variability is key when using the normal distribution for probability calculation.

Probability Calculation

Probability calculation in the context of the normal distribution is about determining how likely a particular outcome is, given the characteristics of the distribution. In this exercise, we use the total time \(Y\), to calculate the probability that time will exceed a certain value \(t\).

• Firstly, to find the time \(t\) with a probability of 1% that \(Y\) is greater than \(t\), we find when the cumulative probability of \(Y\) being less than or equal to \(t\) is 99%, i.e., \(P(Y \leq t) = 0.99\).
• The problem involves determining the \(z\)-score associated with this 99% probability. The \(z\)-score is a measure indicating how many standard deviations an element is from the mean.
• Using a standard normal distribution table, the \(z\)-score corresponding to 0.99 is approximately 2.33.
• This \(z\)-score can be applied in the formula: \(t = \mu_Y + z \cdot \sigma_Y\), substituting the values to find \(t\), which results in 52.77 minutes.
• Therefore, if errands start at 10:00 A.M., the probability of returning by approximately 10:53 A.M. is 0.99.

Effective probability calculation can aid in making informed decisions and preparing better for likely future events.