OPTIMA Statistics

In each case, we ask ourselves if the value we measure is a result of a random or deterministic process.

1. Since your grandmother can be assumed to have the same height each time you measure her (within a set time-frame of course), this is a deterministic process. Thus, her height is not a random variable.

2. When you randomly choose a grandmother to measure, you will get different grandmothers resulting in different heights. The underlying process is random and therefore, this is a random variable.

3. Flipping a coin is by nature a random event. The number of heads could be anything from 0-10 in this process. Therefore, this variable is a random variable.

4. Since the coin has a heads on both sides, we know for certain that one head will appear. The process is deterministic, and therefore the variable is not random.

The number of heads in ten flips of a coin must take on one of the values

This is therefore a discrete random variable.

The height of a grandmother can take on any value within a “reasonable” range (we don't expect 10ft tall grandmothers, nor do we expect 1ft tall grandmothers). It is therefore a continuous random variable.

Remembering that “\(X=x\)” stands for a set of outcomes, we find that:

1. “\(X=1\)” is the set \(\lbrace HTT, THT, TTH \rbrace\) — in other words, the set of outcomes with \(X=1\) head in the 3 flips.

2. “\(X=3\)” is the set \(\lbrace HHH \rbrace\) — this is the only outcome in which \(X\) (the number of heads) is 3.

3. \(P(X=2)\) is the probability of the event \(\lbrace HHT, HTH, THH \rbrace\text{.}\) Since the coin is fair and each flip is independent of the previous flip, each of these outcomes has probability \(\frac{1}{2}\times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\text{.}\) Therefore,

   \begin{equation*} P(X=2) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}\text{.} \end{equation*}

The first step in constructing our probability distribution is to list the possible values of \(X\text{.}\) Since we only draw two marbles, the possible number of white marbles drawn is 0, 1, or 2.

Our first thought might be to construct a table for this probability distribution similar to what we did in the last example. The problem we run into is that there are an infinite number of possible values (or at least, very large considering we don't know how big the town is). So instead, we will construct a formula.

\(Y=y\) means that it took \(y\) attempts to find the first dog owner. So the first \(y-1\) people that we asked were not dog owners. The probability that a given person is not a dog owner is \(\frac{2}{3}\text{.}\) The last person we ask does have a dog, and that probability is \(\frac{1}{3}\text{.}\) This yields the formula:

Based on the distribution we found, which is repeated below, we construct the histogram shown.

\(x\)	\(P(X=x)\)
\(0\)	\(\frac{10}{21} \approx 0.4762\)
\(1\)	\(\frac{10}{21} \approx 0.4762\)
\(2\)	\(\frac{1}{21} \approx 0.0476\)

The area principle plays an important role in this histogram. Recall that the area principle states that the area of a bar in a histogram must accurately reflect the proportion of the class represented by that bar in the frequency distribution. In a probability histogram, this relationship is even stronger. The area of each bar must be equal to the probability of the value of \(X\) that it represents. Since each bar has a height equal to the probability and a width equal to one, this is indeed the case.

Limiting ourselves to just the first ten values of the variable, we construct the histogram shown to the right. A few things to note about this histogram:

• It still obeys the area principle. That is, each bar has area equal to the probability of that random variable value.

• If we were to add together all the bars (an infinite number), the sum would be 1 because the probability of the sample space is 1.

• In this instance, the bars become so short that they can not be seen after about \(Y=10\text{.}\)

To construct a probability distribution, we first list the possible values of \(X\text{.}\)

• If we draw a red ball, we win nothing \(X = 0.\)

• If we draw a white ball, we win $5 so \(X = 5\text{.}\)

• Finally, if we are lucky enough to draw the gold ball, we win $50, so \(X=50\text{.}\)

This yields the following probability distribution.

In order to determine if the charity will make money, we need to factor in both the payouts as well as the probability. We will do this by computing the average value of one of the 100 balls in the urn. We know that there is one ball worth $50 and there are four worth $5. All of the others are worth $0. So, the average is:

So if we were to play this game over and over again, our average winnings would be $0.70. But we paid a dollar to play. So our average payout would be \(0.70-1 = -0.30\text{.}\) So we end up loosing thirty cents every time we play. So the charity does in fact make money on the game.

Your first step is to construct a probability distribution by listing all of the values of \(Y\) and determining their probabilities.

• If we draw two red balls, then \(Y = 0 + 0 - 2 = -2\text{.}\)

• If we draw one red and one white, then \(Y = 0 + 5 - 2 = 3\text{.}\)

• If we draw two white, then \(Y = 5 + 5 - 2 = 8\text{.}\)

• If we draw a red and a gold, then \(Y = 0 + 50 - 2 = 48\text{.}\)

• And finally, a white and a gold gives \(Y = 5 + 50 - 2 = 53\text{.}\)

We can then use combinations to compute the probability of each of these values. For example, the probability of winning $3 by drawing a red and a white ball is

because we draw 1 of the 95 red balls and 1 of the 4 white balls in an experiment which involves drawing 2 out of 100 possible balls. The entire probability distirbution is shown below.

Now to compute the expected value, we must multiply each value of \(Y\) by its probability and then add those together. One easy way to keep track of this is to add a third column to our probability distribution table, into which we place these products. The sum of that third column will be our expected value.

Since the expected value of \(Y\) is negative, the average outcome will result in the player losing money on the game. That menas that, in the long run, the charity will still make money.

1. Our first step is to create a probability distribution for \(Y\text{,}\) which is done below. note that the probabilities are found by observing the number of outcomes that gives each sum. For example, \(X=4\) is the set \(\lbrace (1,3), (2,2), (3,1) \rbrace\text{,}\) so \(P(X=4) = \frac{3}{36}\text{.}\)

   \(x\)	\(P(X=x)\)
   \(2\)	\(\frac{1}{36}\)
   \(3\)	\(\frac{2}{36}\)
   \(4\)	\(\frac{3}{36}\)
   \(5\)	\(\frac{4}{36}\)
   \(6\)	\(\frac{5}{36}\)
   \(7\)	\(\frac{6}{36}\)
   \(8\)	\(\frac{5}{36}\)
   \(9\)	\(\frac{4}{36}\)
   \(10\)	\(\frac{3}{36}\)
   \(11\)	\(\frac{2}{36}\)
   \(12\)	\(\frac{1}{36}\)

   

   We note that the histogram is symmetric about a single mode. In other words, it is mound shaped.

2. Next we need to compute the mean (expected value) and standard deviation. We can then use the empirical rule on this mound shaped distribution to determine the range of 95% of the values of \(X\text{.}\)

   \(x\)	\(P(X=x)\)	\(x\times P(X=x)\)	\((x-\mu)^2\times P(X=x)\)
   \(2\)	\(\frac{1}{36} \approx 0.0278\)	\(2(0.0278) = 0.0556\)	\((2-7)^2(0.0278) = 0.6950\)
   \(3\)	\(\frac{2}{36} \approx 0.0556\)	\(3(0.0556) = 0.1668\)	\((3-7)^2(0.0556) = 0.8896\)
   \(4\)	\(\frac{3}{36} \approx 0.0833\)	\(4(0.0833) = 0.3332\)	\((4-7)^2(0.0833) = 0.7497\)
   \(5\)	\(\frac{4}{36} \approx 0.1111\)	\(5(0.1111) = 0.5555\)	\((5-7)^2(0.1111) = 0.4444\)
   \(6\)	\(\frac{5}{36} \approx 0.1389\)	\(6(0.1389) = 0.8334\)	\((6-7)^2(0.1389) = 0.1389\)
   \(7\)	\(\frac{6}{36} \approx 0.1667\)	\(7(0.1667) = 1.1669\)	\((7-7)^2(0.1667) = 0.0000\)
   \(8\)	\(\frac{5}{36} \approx 0.1389\)	\(8(0.1389) = 1.1112\)	\((8-7)^2(0.1389) = 0.1389\)
   \(9\)	\(\frac{4}{36} \approx 0.1111\)	\(9(0.1111) = 0.9999\)	\((9-7)^2(0.1111) = 0.4444\)
   \(10\)	\(\frac{3}{36} \approx 0.0833\)	\(10(0.0833) = 0.8330\)	\((10-7)^2(0.0833) = 0.7497\)
   \(11\)	\(\frac{2}{36} \approx 0.0556\)	\(11(0.0556) = 0.6116\)	\((11-7)^2(0.0556) = 0.8896\)
   \(12\)	\(\frac{1}{36} \approx 0.0278\)	\(12(0.0278) = 0.3336\)	\((12-7)^2(0.0278) = 0.6950\)
   		\(\mu = E(X) = 7\)	\(\sigma^2 = 5.8352\)

   So the mean is \(\mu = 7\) and the standard deviation is \(\sigma = \sqrt{5.8352} \approx 2.4156\text{.}\) According to the empirical rule, 95% of the data should like within two standard deviations of the mean in a mound shaped distribution. That is:

   \begin{align*} 7 - 2(2.4156) \amp\lt \text{ 95% of data } \lt 7 + 2(2.4156)\\ 2.1688 \amp\lt \text{ 95% of data } \lt 11.83 \end{align*}

   Since the variable is discrete, we round to get the range of 3 to 11. We expect to get between 3 and 11 about 95% of the time.