Mathematics 215: Introduction to Statistics

Study Guide

Unit 2: Self-Test Answer Key

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

1. The following survey was recently sent to all 500 employees of a large hospital.

   “All hospital employees should get a flu shot each year. Circle ONE of the following:

   Strongly agree  Agree  Disagree  No opinion”

   The responses to this survey are summarized in the following frequency distribution.

   Response	Strongly agree (SA)	Agree (A)	Disagree (D)	No opinion (NO)
   Frequency	310	120	40	30

   1. If an employee is randomly selected from this hospital, find the probability that

      1. the employee will strongly agree with the flu shot statement.

         Solution:

         ${\displaystyle P\left(SA\right)=\left(\frac{310}{500}\right)=0.62}$

      2. the employee will either strongly agree or agree with the flu shot statement.

         Solution:

         Since both events are mutually exclusive:

         ${\displaystyle \begin{array}{ll}P\left(SA\text{ or }A\right)\hfill & =P\left(SA\right)+P\left(A\right)\hfill \\ \hfill & ={\displaystyle \frac{\left(310+120\right)}{500}}\hfill \\ \hfill & ={\displaystyle \frac{430}{500}}\hfill \\ \hfill & =0.86\hfill \end{array}}$

      3. the employee will neither strongly agree nor agree with the flu shot statement. Use your answer to part ii. above to help answer this.

         Solution:

         $\begin{array}{ll}P\left(\text{neither}\left(SA\text{ or }A\right)\right)\hfill & =1-P\left(SA\text{ or }A\right)\hfill \\ \hfill & =1-0.86\hfill \\ \hfill & =0.14\hfill \end{array}$

   2. Are the two events strongly agree and agree mutually exclusive? Explain.

      Answer: Yes, the events are mutually exclusive, as for the same randomly selected employee the events SA and A cannot both happen at the same time. It has to be one or the other.

2. Circle True (T) or False (F) for each of the following statements:
   1. True is correct answerTF

      The probability that the sample space will occur from the generation of a given experiment is equal to 1.
   2. TFalse is correct answerF

      An event that includes one and only one of the (final) outcomes for the generation of a given experiment is called a compound event.
   3. TFalse is correct answerF

      For any given event based on the generation of an experiment, the probability of the event will typically exceed 1.
   4. True is correct answerTF

      Assigning probabilities to a compound event by dividing the number of simple outcomes associated with the event by the total possible number of simple outcomes in the sample space is consistent with the classical concept of probability.
   5. TFalse is correct answerF

      If two events $A$ and $B$ are mutually exclusive, then we can state that $P\left(A⁄B\right)=P\left(A\right)$.
   6. TFalse is correct answerF

      If $P\left(A⁄B\right)=P\left(B\right)$, we can conclude that $A$ and $B$ are independent events.
   7. True is correct answerTF

      If $A$ and $B$ are complementary events, then $P\left(A\right)+P\left(B\right)=1$.
   8. TFalse is correct answerF

      Depending on the experiment, it is possible to have the numeric probability of an event exceed 1.
   9. True is correct answerTF

      If the joint probability of events $A$ and $B$ equals zero, then $P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$.
   10. True is correct answerTF

       If two events $A$ and $B$ are independent, then we can compute $P\left(A\text{and}B\right)$ as follows: $P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$.
   11. TFalse is correct answerF

       If two events $A$ and $B$ are NOT independent, then we can compute $P\left(A\text{and}B\right)$ as follows: $P\left(A\text{and}B\right)=P\left(A\right)\times P\left(A⁄B\right)$.

3. A car dealership located on the outskirts of a large subdivision surveyed 200 of its regular customers in order to compare the level of customer satisfaction of its urban customers with that of its rural customers. The survey responses are summarized in the two-way classification table below.

   	Very satisfied (VS)	Satisfied (S)	Not satisfied (NS)	Total
   Rural (R)	10	35	65	110
   Urban (U)	60	25	5	90
   Total	70	60	70	200

   1. If one customer is selected at random from the 200 customers that were surveyed, find the probability that this customer: (show your answers to 2 decimals)

      1. is an urban customer.

         Solution: ${\displaystyle P\left(U\right)=\frac{90}{200}=0.45}$

      2. is a rural customer or is very satisfied with the dealership’s service.

         Solution:

         ${\displaystyle \begin{array}{ll}P\left(R\text{ or }VS\right)\hfill & =P\left(R\right)+P\left(VS\right)-P\left(R\text{ and }VS\right)\hfill \\ \hfill & ={\displaystyle \frac{\left(110+70-10\right)}{200}}\hfill \\ \hfill & ={\displaystyle \frac{170}{200}}\hfill \\ \hfill & =0.85\hfill \end{array}}$

      3. is an urban customer and is not satisfied with the dealership’s service.

         Solution: ${\displaystyle P\left(U\text{ and }NS\right)=\frac{5}{200}=0.025}$

      4. is very satisfied with the dealership’s service, given that the customer is from an urban area.

         Solution: $P\left(VS|U\right)$

      5. is from a rural area, given that the customer is not satisfied with the dealership’s service.

         Solution: ${\displaystyle P\left(R|NS\right)=\frac{65}{70}=0.9286}$

   2. Are the events Urban (U) and Very Satisfied (VS) mutually exclusive? Explain.

      Answer: No, because a randomly selected customer can be both urban as well as very satisfied at the same time.

   3. Are the events Urban (U) and Very Satisfied (VS) independent? Perform the appropriate math proof.

      Solution:

      Compare ${\displaystyle P\left(U|VS\right)=\frac{60}{70}=0.8571}$ with ${\displaystyle P\left(U\right)=\frac{90}{200}=0.45}$

      Since these two events are not equal, they are NOT independent.

      OR

      Compare ${\displaystyle P\left(VS|U\right)=\frac{60}{90}=0.6667}$ with ${\displaystyle P\left(VS\right)=\frac{70}{200}=0.35}$

      Since these two events are not equal, they are NOT independent.

4. When a new student registers in a Statistics 101 course at a local university, they can decide to take the course online or in a classroom. In the past, 65% of students have taken the course in the classroom and 35% have taken it online. Given that the student takes the course in the classroom, they have an 80% chance of passing the course. Given that the student takes the course online for convenience, they have a 75% chance of passing the course.

   1. A new student has just registered in Statistics 101. Draw a tree diagram describing the different possible outcomes that face the student, in terms of whether the student takes the course in class or online, as well as whether the student ends up passing the course or not. At the end of the tree, display all possible joint probabilities. Keep your work to 4 decimals.

      Solution:

      Let $C=\text{In Class}$ and $\overline{C}=\text{Online}$ and $P=\text{Passes}\text{Course}$ and $\overline{P}=\text{Fails}\text{Course}$

      

   2. Compute the probability that the student will pass the course.

      Solution:

      $P(\text{Passes Course}) = $

      \begin{align*}P(C \: \text{and} \: P) = 0.65 \times 0.80 = 0.52 \end{align*}

      OR 

      \begin{align*} P(\overline C \: \text{and} \: P) & = 0.35 \times 0.75 = 0.2625\\ & = 0.52 + 0.2625 = 0.7825 \end{align*}

   3. Are the events “takes the course in class” and “student will pass” independent events? Explain by making the appropriate math computations.

      Solution:

      Take $P\left(\text{Passes}\text{Course}\right)=0.7825$ from part b. above.

      Take $P\left(\text{Passes}\text{Given}\text{Takes}\text{Course}\text{in}\text{Class}\right)=0.80$ from the question.

      Since $P\left(\text{Passes}\text{Course}\right)$ is NOT EQUAL to $P\left(\text{Passes}\text{Given}\text{Takes}\text{Course}\text{in}\text{Class}\right)$, the events “$C$ ” and “$P$” are NOT independent.

5. Consider the following experiment. You draw one card at random from a full deck of playing cards and observe whether it is a “hearts” card or not. You then replace the card in the deck, shuffle the deck, and then draw a second card. You observe whether this card is a hearts card or not. Note that in a deck of 52 cards there are 13 hearts cards.

   1. Draw a tree diagram to describe all possible outcomes. At the end of the tree, display all possible joint probabilities in one full play of this experiment. Keep your work to 4 decimals.

      Solution:

      Let $H=\text{Hearts}\text{Selected}$ and $\overline{H}=\text{Hearts}\text{Not}\text{Selected}$

      Note that ${\displaystyle P\left(H\right)=\frac{13}{52}=0.25}$ and $P(\overline{H})=0.75$.

      

   2. Find the probability that two hearts cards will be drawn in one full play of this experiment.

      Solution:

      $=P\left(H\text{ and }H\right)=0.25\times 0.25=0.0625$

   3. Find the probability that exactly one hearts card will be drawn in one full play of this experiment.

      Solution:

      $\begin{array}{l}\left[P\left(H\text{ and }\overline{H}\right)=0.25\times 0.75=0.1875\right]\hfill \\ +\left[P\left(\overline{H}\text{ and }H\right)=0.75\times 0.25=0.1875\right]\hfill \\ =0.1875+0.1875\hfill \\ =0.3750\hfill \end{array}$

   4. Find the probability that no hearts card will be drawn in one full play of this experiment

      Solution:

      $=P\left(\overline{H}\text{ and }\overline{H}\right)=0.75\times 0.75=0.5625$

6. For two events $A$ and $B$:  

   $\begin{array}{l}P\left(A\right)=0.30\\ P(B)=0.40\\ P(B|A)=0.50\end{array}$

   1. Find $P\left(A\text{and}B\right)$.

      Solution:

      $=P\left(A\right)\times P\left(B|A\right)=\left(0.30\right)\times \left(0.50\right)=0.15$

   2. Find $P\left(A|B\right)$.

      Solution:

      ${\displaystyle =\frac{\left[P\left(A\text{ and }B\right)\right]}{P\left(B\right)}=\frac{\left(0.15\right)}{0.40}=0.3750}$

   3. Are the events $A$ and $B$ independent? Explain.

      Solution:

      Compare the two probabilities:

      $P\left(A\right)=0.30$, while $P\left(A|B\right)=0.3750$.

      Since $P\left(A\right)$ is not equal to $P\left(A|B\right)$, the two events are NOT independent.

7. Consider an experiment in which you draw 4 cards, without replacement, from a standard deck of 52 cards. How many distinct 4-card hands could you draw? Show your calculations.

   Solution:

   Since order does not matter, we are dealing with a combinations problem. Find ${}_{52}{C}_4$.

   ${\displaystyle \begin{array}{ll}{}_{52}\text{C}{}_4\hfill & ={\displaystyle \frac{\left(52!\right)}{\left(4!\right)\left(48!\right)}}\hfill \\ \hfill & =\left(52\times 51\times 50\times 49\right)\left(4\times 3\times 2\right)\hfill \\ \hfill & =270,725\text{ distinct hands}\hfill \end{array}}$