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The mean weight of luggage checked by a randomly selected tourist-class passenger flying between two cities on a certain airline is\({\bf{40}}\)lb, and the standard deviation is\({\bf{10}}\)lb. The mean and standard deviation for a business class passenger is\({\bf{30}}\)lb and\({\bf{6}}\)lb, respectively.

a. If there are\({\bf{12}}\)business-class passengers and\({\bf{50}}\)tourist-class passengers on a particular flight, what is the expected value of total luggage weight and the standard deviation of total luggage weight?

b. If individual luggage weights are independent, normally distributed RVs, what is the probability that total luggage weight is at most\({\bf{2500}}\)lb?

Short Answer

Expert verified

\(\begin{array}{l}{\rm{a}}{\rm{. }}E\left( {{T_0}} \right) = 2360;{\sigma _{{T_0}}} = 73.7021;{\rm{ }}\\{\rm{b}}{\rm{. }}P\left( {{T_0} \le 2500} \right) = 0.9713\end{array}\)

Step by step solution

01

Definition of Variance

In probability and statistics, variance is the expected value of the squared variation of a random variable from its mean value. Informally, variance is a measure of how far a group of numbers (random) deviates from its mean value. The square of standard deviation, which is another important tool, equals the value of variance.

02

Calculation for the determination of expected value in part a.


(a):

There are two classes-the business class and the tourist class. Let random variables\({X_i},i = 1,2, \ldots ,12\)denote weights of luggage for business class and random variables\({Y_j},j = 1,2, \ldots ,50\)denote weights of luggage for tourist class. Then the total weight can be represented as the sum of the random variables\({X_i},{Y_j}\)

\({T_0} = {X_1} + {X_2} + \ldots {X_{12}} + {Y_1} + {Y_2} + \ldots + {Y_{50}}\)

The mean and standard deviation of random variables\({X_i},{Y_j}\)are given in the exercise

\(\begin{array}{*{20}{l}}{{\mu _{{X_i}}} = 30,}&{i = 1,2, \ldots ,12,}\\{{\sigma _{{X_i}}} = 6,}&{i = 1,2, \ldots ,12,}\\{{\mu _{{Y_i}}} = 40,}&{j = 1,2, \ldots ,50}\\{{\sigma _{{Y_i}}} = 6,}&{j = 1,2, \ldots ,50}\end{array}\)

Therefore, the expected value is

\(\begin{aligned}E\left( {{T_0}} \right) &= E\left( {{X_1} + {X_2} + \ldots {X_{12}} + {Y_1} + {Y_2} + \ldots + {Y_{50}}} \right)\\ &= E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + \ldots E\left( {{X_{12}}} \right) + E\left( {{Y_1}} \right) + E\left( {{Y_2}} \right) + \ldots E\left( {{Y_{50}}} \right)\\ &= 12 \cdot 30 + 50 \cdot 40\\ &= 2360\end{aligned}\)

03

Calculation for the determination of expected value in part a.

The variance is

\(\begin{aligned}\sigma _{{T_0}}^2 &= V\left( {{T_0}} \right) = V\left( {{X_1} + {X_2} + \ldots {X_{12}} + {Y_1} + {Y_2} + \ldots + {Y_{50}}} \right)\\\ &= V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + \ldots V\left( {{X_{12}}} \right) + V\left( {{Y_1}} \right) + V\left( {{Y_2}} \right) + \ldots V\left( {{Y_{50}}} \right)\\ &= 12 \cdot {6^2} + 50 \cdot {10^2}\\ &= 5432\end{aligned}\)

(1): because random variables are independent.

Finally, the standard deviation is

\({\sigma _{{T_0}}} = \sqrt {5432} = 73.7021\)

04

Calculation for the determination of probability in part b.

(b):

By assuming the normality of random variables and the independence, Central Limit Theorem can be used. Standardize random variables\({T_0}\)in order to obtain the probabilities.

\(\begin{array}{c}P\left( {{T_0} \le 2500} \right) = P\left( {\frac{{{T_0} - E\left( {{T_0}} \right)}}{{{\sigma _{{T_0}}}}} \le \frac{{2500 - 2360}}{{73.7021}}} \right)\\ = P(Z \le 1.9)\mathop = \limits^{(2)} 0.9713\end{array}\)

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

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