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The expectation of a random variable's squared deviation from its population mean or sample mean is known as variance in probability theory and statistics. Variance is a measure of dispersion, or how far a group of numbers deviates from its mean value.

(a):

Define random variable\({X_1},{X_2}, \ldots ,{X_5}\)as morning times and\({X_6},{X_7}, \ldots ,{X_{10}}\)as waiting time in the evening.

By the rule of expected value, the following holds

\(\begin{aligned}E\left( {{X_1} + {X_2} + \ldots + {X_{10}}} \right) &= E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + \ldots + E\left( {{X_{10}}} \right)\\ &= 5E\left( {{X_1}} \right) + 5E\left( {{X_6}} \right)\\ &= 5 \cdot 4 + 5 \cdot 5\\ & = 45\end{aligned}\)

(1): random variables\({X_i},i = 1,2, \ldots ,5\)have the same distribution. Random variables\({X_i},i = 6,7, \ldots ,10\)have the same distribution as well,

(2): expected values of uniformly distributed random variables on an interval are simply the midpoint. For example, the midpoint of interval \((0,10)\)is\(5\). It is possible to calculate the expected value by the definition as well.

(b):

The random variables are independent, therefore the following holds

\(\begin{array}{l}V\left( {{X_1} + {X_2} + \ldots + {X_{10}}} \right)\mathop = \limits^{(3)} V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + \ldots + V\left( {{X_{10}}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 5V\left( {{X_1}} \right) + 5V\left( {{X_6}} \right)\end{array}\)

(3): here we use the independence. The variance of uniformly distributed random variable X on interval (0, a) is

\(\begin{aligned}V(X) &= E\left( {{X^2}} \right) - {(E(X))^2} &= \int_0^a {{x^2}} \frac{1}{a}dx - {\left( {\frac{a}{2}} \right)^2}\\ &= \left. {\frac{1}{a}\frac{{{x^3}}}{3}} \right|_0^a - \frac{{{a^2}}}{4} &= \frac{{{a^3}}}{{3a}} - \frac{{{a^2}}}{4}\\ &= \frac{{4{a^2} - 3{a^2}}}{{12}} &= \frac{{{a^2}}}{{12}}\end{aligned}\)

Substitute\(a = 8\)and\(a = 10\)to obtain the variances\(V\left( {{X_1}} \right)\)and\(V\left( {{X_6}} \right)\). Continuing from above, the following holds

\(\begin{array}{l}V\left( {{X_1} + {X_2} + \ldots + {X_{10}}} \right)\mathop = \limits^{(4)} V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + \ldots + V\left( {{X_{10}}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 5V\left( {{X_1}} \right) + 5V\left( {{X_6}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 5 \cdot \frac{{64}}{{12}} + 5 \cdot \frac{{100}}{{12}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 68.33\end{array}\)

(4): independent random variables.

(c):

Look at random variable\({X_1} - {X_6}\)and compute the expectation and variance. The following holds

\(E\left( {{X_1} - {X_6}} \right) = E\left( {{X_1}} \right) - E\left( {{X_6}} \right) = 4 - 5 = - 1\)

The variance is

\(V\left( {{X_1} - {X_6}} \right) = V\left( {{X_1}} \right) + {( - 1)^2}V\left( {{X_6}} \right) = \frac{{64}}{{12}} + \frac{{100}}{{12}} = 13.67\)

Total morning waiting time can be represented with a random variable

\({T_m} = {X_1} + {X_2} + {X_3} + {X_4} + {X_5},\)

and total evening waiting time can be represented with a random variable

\({T_e} = {X_6} + {X_7} + {X_8} + {X_9} + {X_{10}}.\)

The expected value and the variance of the difference between total waiting times are

\(\begin{array}{c}E\left( {{T_m} - {T_e}} \right) = E\left( {{X_1} + {X_2} + {X_3} + {X_4} + {X_5} - \left( {{X_6} + {X_7} + {X_8} + {X_9} + {X_{10}}} \right)} \right)\\ = 5 \cdot E\left( {{X_1}} \right) - 5 \cdot E\left( {{X_6}} \right)\\ = 5 \cdot 4 - 5 \cdot 5\\ = - 5\end{array}\)

And

\(\begin{array}{c}V\left( {{T_m} - {T_e}} \right) = V\left( {{X_1} + {X_2} + {X_3} + {X_4} + {X_5} - \left( {{X_6} + {X_7} + {X_8} + {X_9} + {X_{10}}} \right)} \right)\\ = V\left( {{X_1} + {X_2} + {X_3} + {X_4} + {X_5}} \right) + {( - 1)^2}V\left( {{X_6} + {X_7} + {X_8} + {X_9} + {X_{10}}} \right)\\\mathop = \limits^{(1)} 68.33\end{array}\)