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Consider a large ferry that can accommodate cars and buses. The toll for cars is \(\$ 3\), and the toll for buses is \(\$ 10\). Let \(x\) and \(y\) denote the number of cars and buses, respectively, carried on a single trip. Cars and buses are accommodated on different levels of the ferry, so the number of buses accommodated on any trip is independent of the number of cars on the trip. Suppose that \(x\) and \(y\) have the following probability distributions: $$ \begin{array}{lrrrrrr} x & 0 & 1 & 2 & 3 & 4 & 5 \\ p(x) & .05 & .10 & .25 & .30 & .20 & .10 \\ y & 0 & 1 & 2 & & & \\ p(y) & .50 & .30 & .20 & & & \end{array} $$ a. Compute the mean and standard deviation of \(x\). b. Compute the mean and standard deviation of \(y\). c. Compute the mean and variance of the total amount of money collected in tolls from cars. d. Compute the mean and variance of the total amount of money collected in tolls from buses. e. Compute the mean and variance of \(z=\) total number of vehicles (cars and buses) on the ferry. f. Compute the mean and variance of \(w=\) total amount of money collected in tolls.

Short Answer

Expert verified
The mean and variance of \(x\) are 2.70 and 1.19 respectively. The mean and variance of \(y\) are 0.70 and 0.51 respectively. The mean and variance of the total amount of money collected from cars are \$8.10 and \$10.71. The mean and variance of the total amount of money collected in tolls from buses are \$7.00 and \$51.00. The mean and variance of total number of vehicles are 3.40 and 1.70 respectively. The mean and variance of total tolls collected are \$15.10 and \$61.71.

Step by step solution

01

- Calculate the mean of x

Calculate the mean of x using the formula: \(E[X] = \sum xp(x)\). Substituting the known values into the formula gives, \(E[X] = 0*0.05 + 1*0.10 + 2*0.25 + 3*0.30 + 4*0.20 + 5*0.10 = 2.70\)
02

- Calculate the variance of x

Calculate the variance of x using the formula: Var(x) = E[(X-E[X])^2] = E[X^2] - (E[X])^2. First, calculate E[X^2] = \(\sum x^2p(x) = 0^2*0.05 + 1^2*0.10 + 2^2*0.25 + 3^2*0.30 + 4^2*0.20 + 5^2*0.10 = 7.80\). Then substitute E[X^2] and E[X] values into the formula gives Var(X)=7.80 - (2.70)^2 = 1.19
03

- Calculate the mean of y

Calculate the mean of y using the formula similar to step 1: \(E[Y] = \sum y*p(y) = 0*0.50 + 1*0.30 + 2*0.20 = 0.70\)
04

- Calculate the variance of y

Calculate the variance of y using the method similar to step 2. First calculate E[Y^2] = \(\sum y^2*p(y) = 0^2*0.50 + 1^2*0.30 + 2^2*0.20 = 0.90\). Then substitute E[Y^2] and E[Y] values into the formula gives Var(Y)=0.90 - (0.70)^2 = 0.51
05

- Compute the mean and variance of tolls collected from cars

Let Tc represents tolls collected from cars. Then we have Tc = 3x. E[Tc] = \$3E[X]. Substituting E[X]=2.70, E[Tc]= \$3*2.70= \$8.10. Var(Tc) = (3^2)Var(X)= 9*1.19 = \$10.71
06

- Compute the mean and variance of tolls collected from buses

Let Tb represents tolls collected from buses. Then we have Tb = \$10y. E[Tb] = \$10E[Y]. Substituting E[Y] = 0.70, we get E[Tb] = \$10*0.70= \$7.00. Var(Tb) = (10^2)Var(Y) = 100*0.51 = \$51.00
07

- Compute the mean and variance of total number of vehicles

Let Z = x + y which represents the total number of vehicles. So, E[Z] = E[X] + E[Y] = 2.70 + 0.70 = 3.40, and Var(Z) = Var(X) + Var(Y) = 1.19 + 0.51 = 1.70
08

- Compute the mean and variance of total amount of tolls collected

Let W = Tc + Tb which represents the total tolls collected. So, E[W] = E[Tc] + E[Tb] = \$8.10 + \$7.00 = \$15.10, and Var(W) = Var(Tc) + Var(Tb) = \$10.71 + \$51.00 = \$61.71

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean and Standard Deviation
Understanding the mean and standard deviation of a probability distribution is essential in statistics as they provide insights into the data's central tendency and variability, respectively. The mean, often referred to as the expected value, is the average outcome one expects if an experiment is repeated multiple times.

As demonstrated in the solution, computing the mean involves multiplying each possible outcome by its probability and summing these products. For example, the mean number of cars, represented as E[X], was calculated by adding the products of each number of cars and their respective probability. Similarly, the mean number of buses, or E[Y], was found in the same way.

The standard deviation, on the other hand, measures the spread of the data around the mean and is the square root of variance. The variance of a distribution is a numerical measure of how the data values are dispersed. As seen in the provided solution, variance calculation (Var(X)) involves determining E[X2] and then subtracting the square of the mean. For a discrete probability distribution, this is accomplished by squaring each possible outcome, multiplying by the probability, summing those values, and finally subtracting the squared mean.
Variance Calculation
The variance of a probability distribution quantifies the variability or spread of the data. A higher variance indicates that the data points are more spread out from the mean, while a lower variance suggests they are closer to the mean.

To calculate the variance, as detailed in the solution steps for both the number of cars (x) and the number of buses (y), one must first calculate the expected value of the square of the random variable (E[X2] or E[Y2]). The next step involves subtracting the square of the mean (E[X] or E[Y]) from this value.

In practical exercises, such as the toll collection problem presented, variance plays a crucial role in understanding the uncertainty related to the sum collected. For the tolls from cars (Tc) and buses (Tb), variance was computed considering the toll amounts and respective variances. It is interesting to note the direct relationship between the variance of the number of vehicles and the variance of the collected tolls. Since tolls are proportional to the number of vehicles, their variances are also scaled by the square of the proportionality constant.
Expected Value
The expected value is another term for the mean of a probability distribution and is a fundamental concept in probability and statistics, representing the long-term average if an experiment is repeated multiple times.

The expected value is calculated similarly to the mean. As for the tolls collected from cars (E[Tc]) and buses (E[Tb]), it is a matter of applying the linearity of the expectation, which is clearly shown in the problem's solutions. The total expected value of tolls collected (E[W]) can be found by summing the individual expected values, reflecting the property that the expected value of the sum of two independent random variables is the sum of their expected values.

In real-world terms, if a ferry operates many times, the average toll collected from cars and buses would approximate the calculated expected values, which are integral in making financial projections and business decisions related to the ferry's operation.

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