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Variance is the expected squared variation of a random variable from its population mean or sample mean in probability theory and statistics. Variance is a measure of dispersion, or how far a set of numbers deviates from its average value.

(a):

The following is the marginal pmf of X

and the marginal pmf of Y

Compute expected values as follows

\(\begin{array}{l}E(X) = 0 \cdot 0.05 + 1 \cdot 0.1 + 2 \cdot 0.25 + 3 \cdot 0.3 + 4 \cdot 0.2 + 5 \cdot 0.1 = 2.8\\E(Y) = 0 \cdot 0.5 + 1 \cdot 0.3 + 2 \cdot 0.2 = 0.7\end{array}\)

The variances of the given random variables are

\(\begin{aligned}V(X) &= E\left( {{X^2}} \right) - {(E(X))^2}\\ &= {0^2} \cdot 0.05 + {1^2} \cdot 0.1 + {2^2} \cdot {0.25^2} + 3 \cdot 0.3 + {4^2} \cdot 0.2 + {5^2} \cdot 0.1 - {2.8^2}\\ &= 1.66\\V(Y) &= E\left( {{Y^2}} \right) - {(E(Y))^2}\\ &= {0^2} \cdot 0.5 + {1^2} \cdot 0.3 + {2^2} \cdot 0.2 - {0.7^2}\\& = 0.61\end{aligned}\)

The total number of vehicles is a random variable

\(Z = X + Y\)

therefore, the mean value is

\(\begin{aligned}E(Z) &= E(X + Y)\\ &= E(X) + E(Y)\\ &= 2.8 + 0.7\\ &= 3.5\end{aligned}\)

the variance is

\(\begin{aligned}V(Z) &= V(X + Y)\mathop = \limits^{(1)} V(X) + V(Y)\\ &= 1.66 + 0.61\\ &= 2.27\end{aligned}\)

where (1) stands because of the independence. Finally, the standard deviation of the total number of vehicles is

\({\sigma _Z} = \sqrt {V(Z)} = \sqrt {2.27} = 1.51.\)

(b):

We look at random variable

\(U = 3X + 10Y\)

therefore, the mean value is

\(\begin{aligned}E(U) &= E(3X + 10Y)\\ &= 3E(X) + 10E(Y)\\ &= 3 \cdot 2.8 + 10 \cdot 0.7\\ &= 15.4\end{aligned}\)

the variance is

\(\begin{aligned}V(U) &= V(3X + 10Y)\mathop = \limits^{(1)} {3^2}V(X) + {10^2}V(Y)\\ &= 9 \cdot 1.66 + 100 \cdot 0.61\\ &= 75.94\end{aligned}\)

where (1) stands because of the independence. Finally, the standard deviation of the random variable U is\({\sigma _U} = \sqrt {V(U)} = \sqrt {75.94} = 8.71.\)