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Chapter 5: Q60E (page 241)

Refer back to Example, Two cars with six-cylinder engines and three with four-cylinder engines are to be driven over a \(300\)-mile course. Let \({X_1}, . . . {X_5}\)denote the resulting fuel efficiencies (mpg). Consider the linear combination

\(Y = \left( {{X_1} + {X_2}} \right)/2 - \left( {{X_3} + {X_4} + {X_5}} \right)/3\)

which is a measure of the difference between four-cylinder and six-cylinder vehicles. Compute \(P\left( {0 \le Y} \right)\)and\(P(Y > - 2)\).

Short Answer

Expert verified

\(\begin{array}{l}P(Y \ge 0) = 0.2877\\P( - 1 \le Y \le 1) = 0.3686\end{array}\)

Step by step solution

01

Definition of Standard Deviation

The standard deviation is a statistical measure of how much a set of values varies or disperses. A low standard deviation implies that the values are close to the set's mean (also known as the anticipated value), whereas a large standard deviation shows that the values are spread out over a wider range.

02

Calculation for finding the value of probability.

The given random variable can be written as

\(Y = \frac{1}{2}{X_1} + \frac{1}{2}{X_2} - \frac{1}{3}{X_3} - \frac{1}{3}{X_4} - \frac{1}{3}{X_5}\)

It is normally distributed with mean value of

\(\begin{aligned}E(Y) &= E\left( {\frac{1}{2}{X_1} + \frac{1}{2}{X_2} - \frac{1}{3}{X_3} - \frac{1}{3}{X_4} - \frac{1}{3}{X_5}} \right)\\ &= \frac{1}{2}E\left( {{X_1}} \right) + \frac{1}{2}E\left( {{X_2}} \right) - \frac{1}{3}E\left( {{X_3}} \right) - \frac{1}{3}E\left( {{X_4}} \right) - \frac{1}{3}E\left( {{X_5}} \right)\\ &= - 1\end{aligned}\)

(1): the expected values are given in the mentioned example.

The variance of random variable Y is

\(\begin{aligned}V(Y) &= V\left( {\frac{1}{2}{X_1} + \frac{1}{2}{X_2} - \frac{1}{3}{X_3} - \frac{1}{3}{X_4} - \frac{1}{3}{X_5}} \right)\\ &= \frac{1}{4}V\left( {{X_1}} \right) + \frac{1}{4}V\left( {{X_2}} \right) + \frac{1}{9}V\left( {{X_3}} \right) + \frac{1}{9}V\left( {{X_4}} \right) + \frac{1}{9}V\left( {{X_5}} \right)\\ &= 3.167\end{aligned}\)

(2): the variances are given in the mentioned example.

03

Calculation for finding the value of probability.

The standard deviation of random variable Y is

\({\sigma _Y} = \sqrt {V(X)} = \sqrt {3.167} = 1.7795\)

The probability of event \(\{ Y \ge 0\} \)is

\(\begin{aligned}P(Y \ge 0) &= P\left( {\frac{{Y - E(Y)}}{{{\sigma _Y}}} \ge \frac{{0 - ( - 1)}}{{1.7795}}} \right)\\ &= P(Z \ge 0.56)\\ &= 1 - P(Z < 0.56)\\ &= 0..2877\end{aligned}\)

(3): from the normal probability table in the appendix. The probability can also be computed with software.

04

Calculation for finding the value of probability.

Similarly, the probability of the event \(\{ - 1 \le Y \le 1\} \)is

\(\begin{aligned}P( - 1 \le Y \le 1) &= P\left( {\frac{{ - 1 - ( - 1)}}{{1.7795}} \le Z \le \frac{{1 - ( - 1)}}{{1.7795}}} \right)\\ &= P(0 \le Z \le 1.12)\\ &= 0.3686,\end{aligned}\)

(3): from the normal probability table in the appendix. The probability can also be computed with a software.

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Most popular questions from this chapter

A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (\({\rm{1000}}\)s of dollars) is as follows:

\(\begin{array}{*{20}{c}}{{\rm{ Office }}}&{\rm{1}}&{\rm{1}}&{\rm{2}}&{\rm{2}}&{\rm{3}}&{\rm{3}}\\{{\rm{ Employee }}}&{\rm{1}}&{\rm{2}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}\\{{\rm{ Salary }}}&{{\rm{29}}{\rm{.7}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{30}}{\rm{.2}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{25}}{\rm{.8}}}&{{\rm{29}}{\rm{.7}}}\end{array}\)

a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary\({\rm{\bar X}}\).

b. Suppose one of the three offices is randomly selected. Let\({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\)denote the salaries of the two employees. Determine the sampling distribution of\({\rm{\bar X}}\).

c. How does \({\rm{E(\bar X)}}\) from parts (a) and (b) compare to the population mean salary\({\rm{\mu }}\)?

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \({\rm{\bar X}}\) when the population distribution is lognormal with \({\rm{E(ln(X)) = 3}}\) and\({\rm{V(ln(X)) = 1}}\). Consider the four sample sizes\({\rm{n = 10,20,30}}\), and\({\rm{50}}\), and in each case use \({\rm{1000}}\) replications. For which of these sample sizes does the \({\rm{\bar X}}\) sampling distribution appear to be approximately normal?

Annie and Alvie have agreed to meet between\({\rm{5:00 P}}{\rm{.M}}\). and\({\rm{6:00 P}}{\rm{.M}}\). for dinner at a local health-food restaurant. Let\({\rm{X = }}\)Annie's arrival time and\({\rm{Y = }}\)Alvie's arrival time. Suppose\({\rm{X}}\)and\({\rm{Y}}\)are independent with each uniformly distributed on the interval\({\rm{(5,6)}}\).

a. What is the joint pdf of\({\rm{X}}\)and\({\rm{Y}}\)?

b. What is the probability that they both arrive between\({\rm{5:15}}\)and\({\rm{5:45}}\)?

c. If the first one to arrive will wait only \({\rm{10\;min}}\) before leaving to eat elsewhere, what is the probability that they have dinner at the health-food restaurant? (Hint: The event of interest is\({\rm{A = \{(x,y):|x - y|拢1/6\}}}\).)

Question: The number of customers waiting for gift-wrap service at a department store is an rv X with possible values \({\rm{0,1,2,3,4}}\)and corresponding probabilities \({\rm{.1,}}{\rm{.2,}}{\rm{.3,}}{\rm{.25,}}{\rm{.15}}{\rm{.}}\)A randomly selected customer will have \({\rm{1,2}}\),or \({\rm{3}}\) packages for wrapping with probabilities \({\rm{.6,}}{\rm{.3,}}\)and \({\rm{.1,}}\)respectively. Let \({\rm{Y = }}\)the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer).

a. Determine \({\rm{P(X = 3,Y = 3)}}\), i.e., \({\rm{P(3,3)}}\).

b. Determine \({\rm{p(4,11)}}\).

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 \({\rm{2}}\) or component \({\rm{3}}\)functions. Let \({{\rm{X}}_{{\rm{1,}}}}{{\rm{X}}_{\rm{2}}}\), and \({{\rm{X}}_{\rm{3}}}\) denote the lifetimes of components \({\rm{1}}\), \({\rm{2}}\), and \({\rm{3}}\), respectively. Suppose the \({{\rm{X}}_{\rm{i}}}\) 鈥檚 are independent of one another and each \({{\rm{X}}_{\rm{i}}}\) has an exponential distribution with parameter \({\rm{\lambda }}\).

a. Let \({\rm{Y}}\) denote the system lifetime. Obtain the cumulative distribution function of \({\rm{Y}}\)and differentiate to obtain the pdf. (Hint: \({{\rm{F}}_{\left( {\rm{Y}} \right)}}{\rm{P}}\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\); express the event \(\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\)in terms of unions and/or intersections of the three events \(\left\{ {{{\rm{X}}_{\rm{i}}} \le {\rm{y}}} \right\}\), \(\left\{ {{{\rm{X}}_{\rm{2}}} \le {\rm{y}}} \right\}\), and \(\left\{ {{{\rm{X}}_3} \le {\rm{y}}} \right\}\).)

b. Compute the expected system lifetime
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