Mathematics 215: Introduction to Statistics

Study Guide

Unit 3: Self-Test Answer Key

Show all your work and keep your calculations to four decimal places, unless otherwise stated.

1. Jacky’s Sportspub operates a happy hour between 4 PM and 6 PM every day. Currently, the pub is collecting data regarding the daily happy hour sales of its popular spicy beef taco appetizer plate. Over the past 40 happy hours, the pub has recorded the following daily sales of its spicy beef taco plates.

   	# of spicy beef taco plates per happy hour ($\mathit{x}$)	# of happy hours (frequency)
   	$10$	6
   	$15$	10
   	$20$	8
   	$25$	12
   	$30$	4
   Total	$100$	40

   1. Let $x$ be the number of spicy beef taco plates ordered by customers in any one happy hour period. Construct a probability distribution for $x$, based on an analysis of the past 40 happy hours as described in the table above. Keep your calculations to two decimals.

      Solution:

      $\mathit{x}\mathit{=}\mathit{\text{\#}}$ taco plates	$\mathit{P}\left(\mathit{x}\right) \mathit{=}$ $\mathit{\text{freq}}\mathbf{⁄}\mathit{n} \mathit{=}$ $\mathit{\text{freq}}\mathbf{⁄}\mathit{40}\mathit{=}$
      $10$	${\displaystyle \frac{6}{40}=0.15}$
      $15$	${\displaystyle \frac{10}{40}=0.25}$
      $20$	${\displaystyle \frac{8}{40}=0.2}$
      $25$	${\displaystyle \frac{12}{40}=0.3}$
      $30$	${\displaystyle \frac{4}{40}=0.1}$
      $100$	$\sum P\left(x\right)=1.0$

   2. Based on the probability distribution constructed in part a. above, compute the mean number of spicy beef taco plates ordered per happy hour.

      Solution:

      Based on the table below, the mean $=19.75=\sum xP\left(x\right)$.

      $\mathit{x}\mathbf{=}\mathit{\text{\#}}$ of taco plates	$\mathit{P}\left(\mathit{x}\right)\mathbf{=}$ $\mathit{\text{freq}}⁄\mathit{n}\mathbf{=}$ $\mathit{\text{freq}}⁄\mathit{40}$	$\mathit{x}\mathit{*}\mathit{P}\left(\mathit{x}\right)$	${\mathit{x}}^{\mathrm{<mn>2</mn>}}$	${\mathit{x}}^{\mathrm{<mn>2</mn>}}\mathit{P}\left(\mathit{x}\right)$
      $10$	$0.15$	$1.5$	$100$	$15.00$
      $15$	$0.25$	$3.75$	$225$	$56.25$
      $20$	$0.2$	$4$	$400$	$80.00$
      $25$	$0.3$	$7.5$	$625$	$187.5$
      $30$	$0.1$	$3$	$900$	$90.00$
      $100$	$1$	$\sum xP\left(x\right)=19.75$		$\sum x^{2}P\left(x\right)=428.75$

   3. Based on the probability distribution constructed in part a. above, compute the standard deviation of $x$.

      Solution:

      Based on the table above in part b., the standard deviation is:

      $\begin{array}{ll}\sqrt{{\displaystyle \sum x^2\left(Px\right)-{\text{Mean}}^2}}\hfill & =\sqrt{428.75-{19.75}^2}\hfill \\ \hfill & =\sqrt{38.6875}\hfill \\ \hfill & =6.2199\hfill \end{array}$

   4. The owners of the pub are planning to follow the strategy of using the mean computed in part b. above to help them decide how many spicy beef taco plates to serve per happy hour in the next month. Under what conditions—a large standard deviation or a small standard deviation—would they be more likely to experience severe problems relating to: either preparing insufficient taco plates to satisfy daily customer demand, or preparing far too many taco plates (resulting in significant food waste)? Explain.

      Answer: Under a large standard deviation, the daily number of beef taco plates ordered per happy hour will either be far above or far below the mean of 19.75 plates per day per happy hour. This will result in either extreme shortages or extreme surpluses.

2. Scientific research has established that 64% of individuals suffering migraine headaches receive instant pain-relief from taking an over-the-counter medication called Painfree. In a random sample of 8 individuals experiencing migraine headaches, use the appropriate formulae to compute the probability that

   1. exactly three individuals receive instant pain-relief from Painfree.

      Solution:

      This is a binomial distribution problem with $n=8\text{ trials}$ (8 people);
      “success” = “instant pain-relief”; $p=0.64$;
      $x=$ number of people in the sample of 8 who get instant pain-relief.
      Since $p=0.64$ is not found in the binomial tables, it’s necessary to use the binomial formula.

      ${\displaystyle \begin{array}{ll}P\left(x=3\right)\hfill & ={}_{8}C{}_3{\left(0.64\right)}^3{\left(0.36\right)}^5\hfill \\ \hfill & ={\displaystyle \frac{8!}{\left(3!\times 5!\right)}}\times 0.2621\times 0.0060\hfill \\ \hfill & ={\displaystyle \frac{\left(8\times 7\times 6\right)}{\left(3\times 2\right)}}\times 0.2621\times 0.0060\hfill \\ \hfill & =\left(56\right)\times 0.2621\times 0.0060\hfill \\ \hfill & =0.0881\hfill \end{array}}$

   2. more than six individuals receive instant pain-relief from Painfree.

      Solution:

      ${\displaystyle \begin{array}{ll}P\left(x>6\right)\hfill & =P\left(x=7\right)+P\left(x=8\right)\hfill \\ \hfill & ={}_{8}C{}_7{\left(0.64\right)}^7{\left(0.36\right)}^1+{}_{8}C{}_8{\left(0.64\right)}^8{\left(0.36\right)}^0\hfill \\ \hfill & =\left[{\displaystyle \frac{8!}{\left(7!1!\right)}}\left({0.64}^7\right)\left({0.36}^1\right)\right]+\left[\left({\displaystyle \frac{8!}{8!}}\right){\left(0.64\right)}^8{\left(0.36\right)}^0\right]\hfill \\ \hfill & =\left(8\times 0.0440\times 0.36\right)+\left(1\times 0.0281\times 1\right)\hfill \\ \hfill & =0.1267+0.0281\hfill \\ \hfill & =0.1548\hfill \end{array}}$

   3. at most two individuals receive instant pain-relief from Painfree.

      Solution:

      ${\displaystyle \begin{array}{ll}P\left(x\le 2\right)\hfill & =P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)\hfill \\ \hfill & =\left[{\displaystyle \frac{8!}{\left(0!8!\right)}}\left({0.64}^0\right)\left({0.36}^8\right)\right]+\left[\left({\displaystyle \frac{8!}{1!7!}}\right){\left(0.64\right)}^1{\left(0.36\right)}^7\right]\hfill \\ \hfill & +\left[\left({\displaystyle \frac{8!}{2!6!}}\right){\left(0.64\right)}^2{\left(0.36\right)}^6\right]\hfill \\ \hfill & =0.0003+0.0040+0.0250\hfill \\ \hfill & =0.0293\hfill \end{array}}$

3. Circle True (T) or False (F) for each of the following:
   1. TFalse is correct answerF

      The following table, which lists all the different values of $x$, along with their respective probabilities, is a valid probability distribution.
      

      $\mathit{x}$	$\mathit{P}$($\mathit{x}$)
      $1$	$0.40$
      $2$	$0.40$
      $3$	$0.10$

   2. TFalse is correct answerF

      A binomial experiment consists of $n$ independent and identical trials where, for each trial, there are three possible outcomes.
   3. True is correct answerTF

      A random variable that can assume any value contained in one or more intervals is called a continuous random variable.
   4. True is correct answerTF

      For a continuous random variable, the probability that $x$ exceeds “$a$” is equal to the probability that $x$ is at least “$a$”.
   5. TFalse is correct answerF

      The standard normal variable is computed as follows: $z=\left(x-\text{standard}\text{deviation}\right)⁄\text{mean}$
   6. True is correct answerTF

      The total area under the normal curve equals 1.
4. Protection Plus Insurance Corporation charges its policy-holders (homeowners) a $500 annual premium to be protected from the following three types of home fire damage:
   • If the homeowner experiences 100% fire damage, the company will pay out (to the homeowner) the full value of the home.
   • If the homeowner experiences 50% fire damage, the company will pay out (to the homeowner) one-half of the value of the home.
   • If the homeowner experiences 25% fire damage, the company will pay out (to the homeowner) one-quarter of the full value of the home.

   Past records indicate that the chances of the three types of fire damage for any insured home are 0.0002 for full damage, 0.0004 for 50% damage, and 0.001 for 25% damage.

   Let the random variable $x$ be the annual profit, in dollars, that Protection Plus earns from charging an annual premium of $500 for policy-holders who own a home valued at $400,000. The annual profit equals the annual premium minus the dollar amount that the insurance company pays out to cover home fire damages.

   1. Construct the probability distribution for $x$.

      Solution:

      Event	$\mathit{x}\mathbf{=}\mathbf{\text{profit}}$	$\mathit{P}\left(\mathit{x}\right)$
      100% fire damage	$\begin{array}{l}500-400,000\hfill \\ =-399,500\hfill \end{array}$	$0.0002$
      50% fire damage	$\begin{array}{l}500-200,000\hfill \\ =-199,500\hfill \end{array}$	$0.0004$
      25% fire damage	$\begin{array}{l}500-100,000\hfill \\ =-99,500\hfill \end{array}$	$0.001$
      No fire damage	$500$	$\begin{array}{l}1-\left[0.0002+0.0004+0.001\right]\hfill \\ =1-0.0016=0.9984\hfill \end{array}$

   2. Compute the expected value of $x$. Interpret your answer.

      Solution:

      $\mathit{x}\mathbf{=}\mathbf{\text{profit}}$	$\mathit{P}\left(\mathit{x}\right)$	$\mathit{x}\mathit{*}\mathit{P}\left(\mathit{x}\right)$
      $-399,500$	$0.0002$	$-79.9000$
      $-199,500$	$0.0004$	$-79.8000$
      $-99,500$	$0.001$	$-99.5000$
      $500$	$0.9984$	$499.2000$
      	$\sum =1$	$\sum \left(x*P\left(x\right)\right)=\text{\$}240$

      As described in the table above, $E\left(x\right)=\sum \left(x*P\left(x\right)\right)=\text{\$}240$. This means that if the company sells this type of fire insurance policy to many, many clients, on average the company can expect to earn a profit of $240 per individual client.

5. A statistics quiz consists of 20 multiple choice questions, where each question has five answer options. If a student is going to randomly guess the correct answer for each of the 20 questions, use the appropriate table from Appendix B in the eText to compute the probability that

   1. the student correctly answers exactly 5 questions.

      Solution:

      This is a binomial distribution problem with $n=20$ trials (20 questions);
      “success” = “correct answer”; ${\displaystyle p=\left(\frac{1}{5}\right)=0.20}$;
      $x$ = number of questions in the sample of 20 that are answered correctly.
      Since the combination $p=0.20$ and $n=20$ is found in the binomial tables, you can use the tables to obtain the following:
      $P\left(x=5\right)=0.1746$

   2. the student correctly answers at least 5 questions.

      Solution:

      $\begin{array}{ll}P\left(x\ge 5\right)\hfill & =P\left(x=5\right)+P\left(x=6\right)+P\left(x=7\right)\hfill \\ \hfill & +\text{. }\text{. }\text{.}+P\left(x=20\right)\hfill \\ \hfill & =1-P\left(x\le 4\right)\hfill \\ \hfill & =1-\left[P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)\right.\hfill \\ \hfill & +\left.P\left(x=3\right)+P\left(x=4\right)\right]\hfill \\ \hfill & =1-\left[0.0115+0.0576+0.1369\right.\hfill \\ \hfill & \left.+0.2054+0.2182\right]\hfill \\ \hfill & =1-\left[0.6296\right]\hfill \\ \hfill & =0.3704\hfill \end{array}$

   3. the student passes the quiz.

      Solution:

      $\begin{array}{ll}P\left(\text{Passes}\right)\hfill & =P\left(x\ge 10\right)\hfill \\ \hfill & =P\left(x=10\right)+P\left(x=11\right)+P\left(x=12\right)\hfill \\ \hfill & +\text{. }\text{. }\text{.}+P\left(x=20\right)\hfill \\ \hfill & =0.0020+0.0005+0.0001+0.0000+\text{. }\text{. }\text{.}\hfill \\ \hfill & =0.0026\hfill \end{array}$

6. The length of French fries manufactured by McCann Company is normally distributed with a mean length of 4 inches, with a standard deviation of 0.5 inches.
   1. If a McCann French fry is selected at random off the production line, find the probability that the fry

      1. will be at least 4.5 inches long.

         Solution:

         ${\displaystyle \begin{array}{ll}P\left(x\ge 4.5\right)\hfill & =P\left(z\ge {\displaystyle \frac{\left(4.5-4\right)}{0.5}}\right)\hfill \\ \hfill & =P\left(z\ge 1\right)\hfill \\ \hfill & =1-0.8413\hfill \\ \hfill & =0.1587\hfill \end{array}}$

         

      2. will be between 3.5 and 4.5 inches long.

         Solution:

         ${\displaystyle \begin{array}{ll}P\left(3.5\le x\le 4.5\right)\hfill & ={\displaystyle P\left(\frac{\left(3.5-4\right)}{0.50}\le z\le \frac{\left(4.5-4\right)}{0.50}\right)}\hfill \\ \hfill & =P\left(-1\le z\le 1\right)\hfill \\ \hfill & =0.8413-0.1587\hfill \\ \hfill & =0.6826\hfill \end{array}}$

         

      3. will be at most 4.5 inches long.

         Solution:

         ${\displaystyle \begin{array}{ll}P\left(x\le 4.5\right)\hfill & =P\left(z\le {\displaystyle \frac{\left(4.5-4\right)}{0.5}}\right)\hfill \\ \hfill & =P\left(z\le 1\right)\hfill \\ \hfill & =0.8413\hfill \end{array}}$

         

   2. McCann Company is in the process of trying to sell its fries to a very large customer, who prefers to buy longer fries, and wants to make sure that there is at least a 95% chance that a McCann fry will be longer than 3 inches. Do the fries manufactured by McCann meet this customer’s requirement?

      Solution:

      Step 1: Find the $z$ value where the area under the normal curve to the left of $z$ equals 0.05.

      

      Based on the standard normal table, $z=-1.645$.

      Step 2: Find the value of $x$, where

      $\begin{array}{ll}x\hfill & =\mu +z\cdot \sigma \hfill \\ \hfill & =4-1.645\left(0.50\right)\hfill \\ \hfill & =4-0.8225\hfill \\ \hfill & =3.1775\hfill \end{array}$

      Conclusion: Since at least 95% of McCann fries exceed 3.1775 inches, then we know at least 95% of the fries exceed 3 inches, which is the customer’s requirement. Therefore, the fries manufactured by McCann do meet this large customer’s requirement.

7. Past studies indicate that 80% of all households own smartphones. A random sample of 300 households has just been selected. Compute the probability that among these 300 households selected

   1. exactly 250 households own a smartphone.

      Solution:

      Since both $np$ and $nq$ exceed 5, this is a normal approximation of the binomial distribution problem.

      $\mu =np=300\times 0.80=240$

      $\sigma =\sqrt{npq}=\sqrt{300\times 0.80\times 0.20}=6.9282$

      Let $x=$ number of households that own a smartphone.

      $\begin{array}{ll}P\left(249.5\le x\le 250.5\right)\hfill & =P\left(1.3712\le z\le 1.5155\right)\hfill \\ \hfill & =P\left(1.37\le z\le 1.52\right)\hfill \\ \hfill & =0.9357-0.9147\hfill \\ \hfill & =0.0210\hfill \end{array}$
      (see graph below)

      

   2. at most 230 households own a smartphone.

      Solution:

      $\begin{array}{ll}P\left(x\le 230.5\right)\hfill & =P\left(z\le -1.371\right)\hfill \\ \hfill & =0.0853\hfill \end{array}$

      

   3. more than 250 households own a smartphone.

      Solution:

      $\begin{array}{ll}P\left(X>250.5\right)\hfill & =P\left(z>1.52\right)\hfill \\ \hfill & =1-0.9357\hfill \\ \hfill & =0.0643\hfill \end{array}$