Chapter PIV, Problem 3RE

(a)

To determine

To explain: a random variable and then use it to show the total amount that the company would have to pay to send the two meetings to these delegations.

Explanation of Solution

Given:

The number of officials sent to China for a meeting by the company = 3

The number of officials being sent France for a meeting by the organisation = 5

The total price of a ticket to China is 1000 dollars.

And the standard deviation = 150 dollars

Mean price for tickets to France = $500 = $500

And the standard deviation = $100

A random variable X is the amount spent on sending officials to China, and a random variable Y is the amount spent on sending officials to France by the organisation for the conference.

Now, to send these delegations to the two meetings, the total amount the organisation would have to pay is:

  $Z=3X+5Y$

(b)

To determine

To find: the mean and standard deviation of this total price.

Answer to Problem 3RE

  $\begin{array}{c}E\left(Z\right)=\$5500\\ V\left(Z\right)=\$452500\\ SD\left(Z\right)=\$672.6812\end{array}$

Explanation of Solution

Formula used:

  $\begin{array}{c}E\left(Z\right)=E\left(X+Y\right)\\ =E\left(X\right)+E\left(Y\right)\\ V\left(Z\right)=V\left(X+Y\right)\\ =V\left(X\right)+V\left(Y\right)\\ SD\left(Z\right)=\sqrt{V\left(Z\right)}\end{array}$

Calculation:

The mean is

  $\begin{array}{c}E\left(Z\right)=E\left(3X+5Y\right)\\ =E\left(3X\right)+E\left(5Y\right)\\ =3E\left(X\right)+5E\left(Y\right)\\ =\left(3\times 1000\right)+\left(5\times 500\right)\\ =3000+2500\\ =\$5500\end{array}$

The variance is

  $\begin{array}{c}V\left(Z\right)=V\left(3X+5Y\right)\\ =V\left(3X\right)+V\left(5Y\right)\\ =9V\left(X\right)+25V\left(Y\right)\\ =9{\left(SD\left(X\right)\right)}^2+25{\left(SD\left(Y\right)\right)}^2\\ =\left(9\times {150}^2\right)+\left(25\times {100}^2\right)\\ =\left(9\times 22500\right)+\left(25\times 10000\right)\\ =202500+250000\\ =\$452,500\end{array}$

The standard deviation is

  $\begin{array}{c}SD\left(Z\right)=\sqrt{V\left(Z\right)}\\ =\sqrt{452,500}\\ =\$672.6812\end{array}$

(c)

To determine

To Find: the mean and standard deviation between the price difference between the ticket to China and the ticket for France.

Answer to Problem 3RE

  $\begin{array}{c}E\left(W\right)=\$500\\ V\left(W\right)=32500\\ SD\left(W\right)=180.2776\end{array}$

Explanation of Solution

Formula used:

  $\begin{array}{c}E\left(W\right)=E\left(X-Y\right)\\ =E\left(X\right)-E\left(Y\right)\\ V\left(W\right)=V\left(X-Y\right)\\ =V\left(X\right)+V\left(Y\right)\\ SD\left(W\right)=\sqrt{V\left(W\right)}\end{array}$

Calculation:

Suppose, the price gap for a ticket to China and for a ticket to France is W.

W = X − Y

The mean is:

  $\begin{array}{c}E\left(W\right)=E\left(X-Y\right)\\ =E\left(X\right)-E\left(Y\right)\\ =1000-500\\ =\$500\end{array}$

The variance is:

  $\begin{array}{c}V\left(W\right)=V\left(X-Y\right)\\ =V\left(X\right)+V\left(Y\right)\\ ={\left(SD\left(X\right)\right)}^2+{\left(SD\left(Y\right)\right)}^2\\ =\left({150}^2\right)+\left({100}^2\right)\\ =22500+10000\\ =\$32,500\end{array}$

The standard deviation is:

  $\begin{array}{c}SD\left(W\right)=\sqrt{V\left(W\right)}\\ =\sqrt{32,500}\\ =\$180.2776\end{array}$

(d)

To determine

To find: that is there required for any assumptions in estimating the means and about the standard deviation.

Explanation of Solution

In determining the standard deviation, the assumptions to be taken are:

1. For a country, all the tickets must be priced at the same price.
2. The ticket price should be independent for various countries.

In calculating the means, no such assumptions are needed.