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Section 7.5 Exercises

A January 2008 Gallup poll on consumer confidence asked the question "How would you rate economic conditions in this country today" and 22% responded "Excellent" or "Good", 45% responded "Only Fair", and 33% responded "Poor". If you pick a representative sample of 25 people and ask them the same question, if X measures the number of responses that are "Excellent" or "Good", determine

  • P(X is at most 5).
  • P(X is at least 5).
  • the expected number of Excellent or Good responses.

Solution

This is a Binomial distribution with n=25 and p = 0.22.

  • P(X is at most 5) = F(5) = f(0)+f(1)+f(2)+f(3)+f(4)+f(5) = 0.51843
  • P(X is at least 5) = 1 - F(4) = 0.67183
  • \(\mu = np = 25 \cdot .22 = 5.5\)

You keep on rolling a pair of dice and let X be the number of rolls needed until you get a sum of 7 or 11 for the second time. Determine:

  1. P(7 or 11 on one roll)
  2. The expected number of rolls until you get the second 7 or 11 sum.
  3. P(X = 12)
  4. \(P(X \ge 4)\)

Solution

For this problem, use the Negative Binomial Distribution when looking for the number of trials till the 2nd success. p is determined in the first answer.

  1. P(7 or 11 on one roll) = 8/36 = 2/9, using equally likely outcomes.
  2. \(\mu = \frac{r}{p} = \frac{2}{2/9} = 9\)
  3. \(P( X = 12) = \binom{11}{1} (7/9)^10 \cdot (2/9)^2 = 0.044\)
  4. \begin{align*} P(X \ge 4) & = 1- P(x \le 3) \\ & = 1 - [ f(2) + f(3) ]\\ & = 1 - \left[ \binom{1}{1} (2/9)^2 + \binom{2}{1} (7/9)^1 \cdot (2/9)^2 \right ]\\ & = 1 - 0.1262 = 0.8738. \end{align*}

You love to eat at Chick-Fil-A with your kids and want to collect all of the five new book titles that come randomly included with each kids meal. If the promotion with these books starts today, determine:

  1. The probability that you get a book you don't have when purchasing the first children's meal.
  2. The probability that you it takes more than four purchases in order to get a second title.
  3. The expected total number of children's meals you would expect to purchase in order to get all five titles.

Solution

  1. One. The first meal will certainly have a book that you have not received yet.
  2. This is a geometric distribution with p=4/5. P(X > 4) = 1 - F(4) = \((1 - 4/5)^4 = \frac{1}{625}\) which is very small. Note, in this case you would have needed to receive the same title randomly for all of the first four kids meal purchases. If this were to ever happen, please let the people at the counter know and it will be their pleasure to swap out for a new title.
  3. Use the geometric distribution five times with changing values for p. For the first book p = 1 means you are certain to get a new title. For the second book title the probability of success is p=4/5; for the third book title the probability of success is p=3/5; for the fourth the probability is p=2/5; and for the last the probability is p = 1/5. Using the mean as 1/p in each case and accumulating these gives the total expected number of meals to purchase as
    \begin{align*} & 1 + \frac{1}{\frac{4}{5}} + \frac{1}{\frac{3}{5}} + \frac{1}{\frac{2}{5}} + \frac{1}{\frac{1}{5}} \\ & = 1 + \frac{5}{4} + \frac{5}{3} + \frac{5}{2} + \frac{5}{1} \\ & = \frac{12 + 15 + 25 + 30 + 60}{12} \\ & = \frac{142}{12} = 11.833 \end{align*}
    and so you would need 12 kids meals. If this were to happen, please be certain to donate the "extra" books to an organization that works with kids or directly to some kids that you might know.

Suppose you roll a standard pair of 6-sided dice 20 times and let X measure the number of outcomes which result in a sum of 7 or 11. Determine:

  1. the expected number of rolls which have a sum of 7 or 11
  2. P(X=5)
  3. P( X > 5)
  4. P( X < 5)

Suppose you roll a standard pair of 6-sided dice X times until you get a sum of 7 or 11 a third time. Determine:

  1. the expected number of rolls needed on average.
  2. P(X=5)
  3. P( X > 5)
  4. P( X < 5)

Given p = 0.3 determine the following:

  1. For Binomial with n = 50, \(P(\mu - 2\sigma \le X \le \mu + 2\sigma)\)
  2. For Negative Binomial with r = 2, \(P(\mu - 2\sigma \le X \le \mu + 2\sigma)\)

Solution

For Binomial with p = 0.3 and n = 50, \(\mu = n \cdot p = 15\) and \(\sigma^2 = n \cdot p \cdot (1-p) = 10.5\text{.}\) So, \(\sigma = \sqrt{10.5}\text{.}\) Therefore,

\begin{align*} P(\mu - 2\sigma \le X \le \mu + 2\sigma) & = P(15 - 2 \sqrt{10.5} \le X \le 15 + 2 \sqrt{10.5})\\ & P( X \in \{9, 10, 11, ... , 19, 20, 21 \} ) \end{align*}

Then

\begin{equation*} F(21) - F(8) \approx 0.97491 - 0.01825 = 0.95666 \end{equation*}

For Negative Binomial with p = 0.3 and r = 2, \(\mu = \frac{2}{0.3} = \frac{20}{3}\) and \(\sigma^2 = 2 \frac{0.7}{0.3^2} = \frac{140}{9}\) and so \(\sigma \approx 3.9\text{.}\) Therefore,

\begin{align*} P(\mu - 2\sigma \le X \le \mu + 2\sigma) & = P(6.7 - 7.8 \le X \le 6.7 + 7.8)\\ & P( X \in \{2, 3, ... , 14, 15, 16 \} ) \end{align*}

Then,

\begin{equation*} F(16) \approx 0.973888 \end{equation*}

Take a die and roll it 200 times, keeping track on paper of the outcome for each roll. You can use a standard 6-sided die or find a more exotic one with more sides.