OPTIMA Statistics

The random variable has six values, \(X = 1, 2, 3, 4, 5, \text{ and } 6\text{.}\) So, for each of these values, \(P(X=x) = \frac{1}{6}\text{.}\) Therefore, this is a discrete uniform distribution with \(k=6\text{.}\) To help visualize this, we construct the following probability histogram.

To show that the mean and standard deviation really are as stated above, we will do the computation two ways.

1. First, we use the probability distribution method of Section 3.1:

   \(x\)	\(P(X=x)\)	\(x\times P(X=x)\)	\((x-\mu)^2\times P(X=x)\)\)
   \(1\)	\(\frac{1}{6} \approx 0.1667\)	\(1(0.1667) = 0.1667\)	\((1-3.5)^2(0.1667) = 1.0417\)
   \(2\)	\(\frac{1}{6} \approx 0.1667\)	\(2(0.1667) = 0.3333\)	\((2-3.5)^2(0.1667) = 0.3750\)
   \(3\)	\(\frac{1}{6} \approx 0.1667\)	\(3(0.1667) = 0.5000\)	\((3-3.5)^2(0.1667) = 0.0417\)
   \(4\)	\(\frac{1}{6} \approx 0.1667\)	\(4(0.1667) = 0.6667\)	\((4-3.5)^2(0.1667) = 0.0417\)
   \(5\)	\(\frac{1}{6} \approx 0.1667\)	\(5(0.1667) = 0.8333\)	\((5-3.5)^2(0.1667) = 0.3750\)
   \(6\)	\(\frac{1}{6} \approx 0.1667\)	\(6(0.1667) = 1.0000\)	\((6-3.5)^2(0.1667) = 1.0417\)
   		\(\mu = 3.5\)	\(\sigma^2= 2.9168\)

   So from the table to the left, we can read off the mean (expected value) and variance, which gives:

   \begin{align*} \mu \amp = 3.5\\ \sigma \amp = \sqrt{2.9168} = 1.71 \end{align*}

2. Next, we use the methods from Subsection 1.3.4.

   We first compute the mean:

   \begin{equation*} \mu = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \end{equation*}

   Then we can construct the table below to find the standard deviation.

   \(x\)	\((x-\mu)\)	\((x-\mu)^2\)
   \(1\)	\(-2.5\)	\(6.25\)
   \(2\)	\(-1.5\)	\(2.25\)
   \(3\)	\(-0.5\)	\(0.25\)
   \(4\)	\(0.5\)	\(0.25\)
   \(5\)	\(1.5\)	\(2.25\)
   \(6\)	\(2.5\)	\(6.25\)
   Total:		\(17.5\)

   Based on that table,

   \begin{align*} \sigma^2 \amp = \frac{17.5}{6} \approx 2.9167\\ \Rightarrow \sigma \amp = \sqrt{2.9167} \approx 1.71\text{.} \end{align*}

Note that the answers found agreed.

There are only two outcomes, so this is a Bernoulli trial. We could choose to call flipping a heads a success and a tails a failure. In that case, \(p = \frac{2}{3}\) and \(q = 1 - \frac{2}{3} = \frac{1}{3}\text{.}\)

While the underlying experiment is the same, the result variable is different in these two cases.

1. If we note the color drawn, then the results are either \(R\text{,}\) \(G\text{,}\) \(B\text{,}\) or \(W\text{.}\) Since there are more then two possible outcomes, this is not a Bernoulli trial.

2. In this case the outcomes are categorized as either a success (\(R\) or \(W\)) or failure (\(G\) or \(B\)). Therefore this is a Bernoulli trial with \(p = \frac{8}{15}\) and \(q = 1 - \frac{8}{15} = \frac{7}{15}\text{.}\)

To verify that this is a binomial process, we go through the four characteristics mentioned above.

1. There are 20 questions and answering each question is a Bernoulli trial since we are either right (success) or wrong (failure).

2. Since we are guessing, the probability of getting a question right is \(p=\frac{1}{4}\) and the probability of getting question wrong is \(q = \frac{3}{4}\text{.}\)

3. How we answer one question does not affect how we answer the next question (remember, we are guessing) so the trials are independent.

4. Finally, the variable \(X\) represents the number of questions we get right—that is the number of successes in our 20 trials.

This is a binomial process and \(X\) is a binomial random variable.

We again verify the four characteristics of a binomial process.

1. There are 100 individuals who are asked if they own are truck. They will either answer “yes” or “no”, so this is a binomial process with \(n = 100\) trials.

2. We do not know what the probability of a success is, but since the population is large, it is safe to assume that \(p\) will not change as we ask each person (see below).

3. How one person answers is not affected by how another person answers the question. Therefore the trials are independent.

4. Again, the random variable \(Y\) is the number of successes (people who had a truck).

Therefore this is a binomial process and \(Y\) is a binomial random variable.

Let's review the four questions again.

1. We are repeating the act of drawing a marble a fixed number of times (\(n = 5\)).

2. There are only two options, red or blue, and since we are counting the number of red marbles drawn, we will count drawing a red marble as a success. On the first draw, the probability of a success is \(\frac{10}{20}\) since there are 10 red marbles.

3. However, the trials are not independent and the probabilities change. To see this, note that on the second draw the probability of a success is either \(\frac{9}{19}\text{,}\) if we drew a red on the first draw, or \(\frac{10}{19}\text{,}\) if we drew a blue marble on the first draw. So the probabilities change and the new probabilities depend on what happened on the previous draw.

4. Our random variable is the number of successes (red marbles).

Because the probabilities change between trials and depend on what happend on previous trials, this process does not meet criteria 2 or 3. It is therefore not a binomial process.

The batch must contain at least \(N\) widgets so that 50 is less than 5% of \(N\text{.}\) We find \(N\) by solving the inequality

If we do not have more than 1000 widgets to select from, then the trials of selecting individual widgets without replacement will not be independent and this will not be a binomial process.

Notice that each flip is a Bernoulli trial with \(p = \frac{2}{3}\) and \(q = \frac{1}{3}\) when we consider getting a heads to be a success. We repeat this trial four times, and each time \(p\) and \(q\) remain unchanged. Because each flip is independent and \(X\) is the number of successes (heads), \(X\) is a binomial random variable.

• Finding \(P(X=1)\).

  Since we want \(X=1\) head, we compute the probability to be

  \begin{equation*} \frac{2}{3}\times\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3} = \left(\frac{2}{3}\right)^1\left(\frac{1}{3}\right)^3\text{.} \end{equation*}

  The problem with this is that it only counts the probability of HTTT. We could also get THTT, TTHT, and TTTH. In fact, a little counting shows that there are \(C(4,1)\) ways to pick the one toss out of 4 that will be heads. So the total probability is:

  \begin{equation*} C(4,1)\left(\frac{2}{3}\right)^1\left(\frac{1}{3}\right)^3 = \frac{8}{81} \approx 0.0988\text{.} \end{equation*}

• Finding \(P(X=2)\).

  This analysis is very similar except that we get a heads twice and a tails twice and we have to choose 2 of 4 four flips that will be our tails. This results in the probability:

  \begin{equation*} C(4,2)\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^2 = \frac{24}{81} \approx 0.2963\text{.} \end{equation*}

First note that \(p = 0.40\text{,}\) \(q = 0.60\text{,}\) and \(n = 10\) in this binomial process. Let the random variable \(Y\) equal the number of plants that are infected in the 10 that we select. Then, the probabilities (rounded to four decimal places) are:

1. \begin{equation*} P(Y=0) = C(10,0)(0.40)^0(0.60)^{10} = 1(1)(0.0060) = 0.0060 \end{equation*}
2. \begin{equation*} P(Y=5) = C(10,5)(0.40)^5(0.60)^5 = 252(0.0102)(0.0778) \approx 0.2007 \end{equation*}
3. \begin{equation*} P(Y=10) = C(10,10)(0.40)^{10}(0.60)^0 = 1(0.0001)(1) = 0.0001 \end{equation*}

Note that this is a binomial process with \(n=20\) trials, assuming that the number of widgets produced is large. If we let a success mean getting a defective widget, then \(p = 0.04\) and \(q = 0.96\text{.}\) Let the random variable \(X\) be the number of defective widgets in our sample of 20 widgets.

We wish to find the probably that \(X\) is less than or equal to 2. This means that \(X\) can be 0, 1, or 2. We use the binomial probability formula to compute \(P(X=0)\text{,}\) \(P(X=1)\text{,}\) and \(P(X=2)\text{.}\) These are mutually exclusive events (we can&t have exactly one defective widget and exactly two defective widgets at the same time), so we then use the addition rule to get our final probability. The computation is shown below.

First note that this is a binomial process with \(n=10\) fish. The true proportion of contaminated fish is \(p = 0.20\text{,}\) so \(q = 0.80\text{.}\) We wish to know the probability that the biologist finds at least 30% of his 10 fish contaminated. That means, we want the probability that \(X\) is at least 3.

We could compute this directly by finding \(P(X=3)\text{,}\) \(P(X=4)\text{,}\) \(P(X=5)\text{,}\) \(P(X=6)\text{,}\) \(P(X=7)\text{,}\) \(P(X=8)\text{,}\) \(P(X=9)\text{,}\) and \(P(X=10)\text{.}\) However, this is a lot of work! The complement of “at least 3” is “less than 3”. That would mean \(X\) could be 0, 1, or 2. This includes only three probabilities to compute, which is a lot less work. So we will use the complement rule as follows.

The number of shots made, we will call it \(X\text{,}\) is a binomial random variable with \(n = 10\text{,}\) and \(p = 0.70\text{.}\) To find the expected value of \(X\)—the expected number of shots the player will make, we use the probability distribution method shown below.

Summing the final column, we get \(E(X) = 7\text{.}\) So the mean of this binomial random variable is 7 and we expect the player to make 7 out of 10 shots.

1. To determine the number of shots we expect, we compute the mean of the random variable.

   \begin{equation*} \mu = n\times p = (350)(0.70) = 245\text{.} \end{equation*}

2. Recall that in Subsection 1.4.3 we used z-scores to decide if a particular value of a variable was unusual. If that value had a z-score bigger than 3 or less than -3, then it is more than 3 standard deviations away from the mean and should be considered unusual. The formula for a z-score, using the formula above for standard deviation, is:

   \begin{equation*} z = \frac{x - \mu}{\sigma} = \frac{300 - 245}{\sqrt{350(0.7)(0.3)}} \approx \frac{55}{8.5732} \approx 6.42\text{.} \end{equation*}

   So yes, making 300 or more shots would definitely be unusual. If this happens, we might suspect that he has raised his free-throw percentage above the 70% we were given.