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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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Answer

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c. Use software to obtain the exact probabilities in (a) and (b) and compare to their approximations.
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