91Ó°ÊÓ

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.

(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

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

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

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

(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.