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Chapter 6: Problem 41

A professor has 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 she spends 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: \(\mu_{1}=15, \quad \sigma_{1}=4\), \(\mu_{2}=5, \quad \sigma_{2}=1, \quad \mu_{3}=8, \quad \sigma_{3}=2, \quad \mu_{4}=12\), \(\sigma_{4}=3 .\) She plans to leave her office at precisely \(10=00\) a.m. and wishes to post a note on her door that reads, "I will return by \(t\) a.m." What time \(t\) should she write down if she wants the probability of her arriving after \(t\) to be \(.01\) ?

Short Answer

Expert verified
She should write 10:53 a.m. on the note.

Step by step solution

01

Identify Total Time

Calculate the total time spent by combining the time for all errands and the walking time. Express this as \( T = X_1 + X_2 + X_3 + X_4 \) where each \( X_i \) is an independent normal variable.
02

Calculate Mean and Standard Deviation of Total Time

Since the \( X_i \)鈥檚 are independent, the mean of the total time \( T \) is the sum of the means: \( \mu_T = \mu_1 + \mu_2 + \mu_3 + \mu_4 = 15 + 5 + 8 + 12 = 40 \). Likewise, the variance of \( T \) is the sum of the variances: \( \sigma_T^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2 = 4^2 + 1^2 + 2^2 + 3^2 = 16 + 1 + 4 + 9 = 30 \). Hence, the standard deviation is \( \sigma_T = \sqrt{30} \approx 5.48 \).
03

Determine the Time t

To find \( t \) such that the probability of spending more than \( t \) minutes is 0.01, we look for \( t \) where \( P(T > t) = 0.01 \). Using the standard normal distribution, find the z-score for the 99th percentile, \( z_{0.99} \). This typically corresponds to \( z = 2.33 \).
04

Transform to Time Domain

Use the z-score formula \( z = \frac{t - \mu_T}{\sigma_T} \). Here, \( z = 2.33 \), \( \mu_T = 40 \), and \( \sigma_T = 5.48 \). Solve for \( t \): \[ t = z \cdot \sigma_T + \mu_T = 2.33 \cdot 5.48 + 40 \] Calculating this gives \( t \approx 52.78 \).
05

Convert Minutes to AM Time

Since the professor leaves at 10:00 a.m., add the calculated \( t \) to 10:00 a.m. \( t = 52.78 \) minutes after 10 a.m. is 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.

Independent Random Variables
Understanding independent random variables is crucial when dealing with multiple processes or events that occur separately. Random variables are considered independent if the outcome of one does not affect the outcomes of others. In the given exercise, the four random variables \( X_1, X_2, X_3, \) and \( X_4 \) represent the time taken for various errands and walking.

Since they are independent, this implies that the time spent by the professor on one task does not influence the time spent on another task.

This property is beneficial in calculations because it allows us to simply add their individual effects, such as means and variances, without considering any interaction between them. The entire duration of the professor's trip is just the sum of all these individual times.
Mean and Variance Calculation
Calculating the mean and variance of the total time \( T \) taken by the professor involves understanding both central tendency (mean) and variability (variance).

The mean for each task is given, and by the property of independent variables, the mean of their sum is the sum of their means:
  • \( \mu_T = \mu_1 + \mu_2 + \mu_3 + \mu_4 \)
  • Calculation yields \( \mu_T = 15 + 5 + 8 + 12 = 40 \) minutes.
For variance, since the tasks are independent, the variance of the sum is the sum of their variances:
  • \( \sigma_T^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2 \)
  • Calculation yields \( \sigma_T^2 = 16 + 1 + 4 + 9 = 30 \).
  • The standard deviation \( \sigma_T \) is the square root of the variance: \( \sigma_T = \sqrt{30} \approx 5.48 \).
The mean and standard deviation give us a complete picture of expected central time and its spread due to randomness.
Probability and Percentiles
Probability and percentiles help in determining how likely an event is given its normal distribution. The professor wants to know by what time she should return so that she arrives on time in 99% of cases. Since normal distributions are symmetrical and bell-shaped, we can use z-scores to find percentile information.

For this exercise, use a z-score which corresponds to the 99th percentile, typically \( z = 2.33 \), which tells us how many standard deviations away from the mean our target time \( t \) should be:
  • Using the formula \( z = \frac{t - \mu_T}{\sigma_T} \), we can solve for \( t \).
  • Plug in: \( t = z \cdot \sigma_T + \mu_T = 2.33 \cdot 5.48 + 40 \).
  • Calculate \( t \approx 52.78 \) minutes.
Thus, the professor should post that she will return by approximately 10:53 a.m., ensuring she will be on time 99% of the time.

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Most popular questions from this chapter

a. Use the general formula for the variance of a linear combination to write an expression for \(V(a X+Y)\). Then let \(a=\sigma_{Y} / \sigma_{X}\), and show that \(\rho \geq-1\). [Hint: Variance is always \(\geq 0\), and \(\left.\operatorname{Cov}(X, Y)=\sigma_{X} \cdot \sigma_{Y} \cdot \rho \cdot\right]\) b. By considering \(V(a X-Y)\), conclude that \(\rho \leq 1 .\) c. Use the fact that \(V(W)=0\) only if \(W\) is a constant to show that \(\rho=1\) only if \(Y=a X+b\).

Show that \(F_{p, v_{1}, v_{2}}=1 / F_{1-p, v_{2}, v_{1}}\).

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 \(X_{5}\) 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 ry \(Y\) 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 \(P(-1 \leq Y \leq 1)\). \(\left[\right.\) Hint: \(Y=a_{1} X_{1}+\cdots+a_{5} X_{5}\), with \(a_{1}=\frac{1}{2}, \ldots,\), \(\left.a_{5}=-\frac{1}{3} .\right]\)

Show that if \(X\) has a gamma distribution and \(c\) \((>0)\) is a constant, then \(c X\) has a gamma distribution. In particular, if \(X\) is chi-squared distributed, then \(c X\) has a gamma distribution.

The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value \(10 \mathrm{~min}\) and standard deviation \(2 \mathrm{~min}\). If five individuals fill out a form on 1 day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?
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