AP Statistics 2.7 Independent Events and Unions of Events- Exam Style Questions - FRQs - New Syllabus

Question

Ms. Fey is a manager at a restaurant. To improve the dining experience for her customers, she uses a digital music service to create a playlist of songs that will be played in the restaurant. The playlist contains 1,000 songs and consists of four different types of music in the following quantities: 200 country songs, 400 pop songs, 100 rock songs, and 300 jazz songs. The digital music service will select songs at random from the playlist to be played in the restaurant. Any song can be replayed at any time.

i. Suppose one song is selected at random to be played. What is the probability that the song is a rock song? Show your work.
ii. Suppose two songs are selected at random to be played. What is the probability that both songs are rock songs? Show your work.

B. In every one-hour period, 20 songs will be played at random and any song can be replayed at any time. Ms. Fey is interested in how many rock songs will be played in a typical one-hour period.

i. Define the random variable of interest to Ms. Fey, and state how the random variable is distributed.
ii. What is the expected value for the random variable in part B (i)? Show your work.

C. Recall that in every one-hour period, 20 songs will be played at random and any song can be replayed at any time.

i. Determine the probability that 4 or more rock songs in a particular one-hour period will be played. Show your work.
ii. Suppose 4 rock songs are played during a particular one-hour period. Does this provide strong evidence that the song selection process was not truly random? Justify your answer without performing an inference procedure.

Most-appropriate topic codes (AP Statistics):

• Topic $2.4$ — Introduction to Probability (Part $ \mathrm{A} $)
• Topic $2.7$ — Independent Events and Unions of Events (Part $ \mathrm{A} $)
• Topic $2.10$ — The Binomial Distribution (Parts $ \mathrm{B} $, $ \mathrm{C} $)

▶️ Answer/Explanation

A. i.
Let $R$ represent selecting a rock song.
$P(R) = \dfrac{\text{Number of rock songs}}{\text{Total number of songs}} = \dfrac{100}{1,000}$
$\boxed{P(R) = 0.10}$

A. ii.
Because any song can be repeated, the selection of the second song is independent of the first.
$P(\text{Both Rock}) = P(R) \times P(R) = 0.10 \times 0.10$
$\boxed{P(\text{Both Rock}) = 0.01}$

B. i.
• Let $X$ be the random variable representing the number of rock songs played in a one-hour period.
• The random variable $X$ follows a binomial distribution, expressed as $X \sim \text{Binomial}(n = 20, p = 0.10)$.

B. ii.
The expected value of a binomial distribution is given by the formula $E(X) = n \cdot p$.
$E(X) = 20 \times 0.10$
$\boxed{E(X) = 2\text{ songs}}$

C. i.
We want to calculate $P(X \ge 4) = 1 – P(X \le 3)$.
Using the binomial cumulative distribution formula, $P(X \le 3) = \sum_{k=0}^{3} \binom{20}{k} (0.10)^k (0.90)^{20-k} \approx 0.8670$.
$P(X \ge 4) = 1 – 0.8670$
$\boxed{P(X \ge 4) = 0.1330}$

C. ii.
• No, this does not provide strong evidence that the song selection process was not truly random.
• The calculated probability of playing 4 or more rock songs is $0.1330$, which is greater than conventional significance thresholds like $\alpha = 0.05$, meaning an outcome of 4 rock songs is a relatively common chance occurrence.

Question

A medical researcher surveyed a large group of men and women about whether they take medicine as prescribed.

The responses were categorized as never, sometimes, or always. The relative frequency of each category is shown in the table.

(a) One person from those surveyed will be selected at random.

i. What is the probability that the person selected will be someone whose response is never and who is a woman?

ii. What is the probability that the person selected will be someone whose response is never or who is a woman?

iii. What is the probability that the person selected will be someone whose response is never given that the person is a woman?

(b) For the people surveyed, are the events of being a person whose response is never and being a woman independent? Justify your answer.

(c) Assume that, in a large population, the probability that a person will always take medicine as prescribed is $0.54$. If $5$ people are selected at random from the population, what is the probability that at least $4$ of the people selected will always take medicine as prescribed? Support your answer.

Most-appropriate topic codes (AP Statistics):

• Topic $2.6$ — Conditional Probability (Part $ \mathrm{a} $)
• Topic $2.7$ — Independent Events and Mutually Exclusive Events (Part $ \mathrm{b} $)
• Topic $2.10$ — The Binomial Distribution (Part $ \mathrm{c} $)

▶️ Answer/Explanation

(a)(i)
From the given two-way table, we can find the joint probability directly by looking at the intersection of the “Women” row and “Never” column.
$P(\text{never and woman}) = 0.0636$

(a)(ii)
To find this probability, we use the general addition rule by adding the marginal probability of “never” to the marginal probability of “woman”, and subtracting their intersection.
$P(\text{never or woman}) = P(\text{never}) + P(\text{woman}) – P(\text{never and woman})$
$P(\text{never or woman}) = 0.1200 + 0.5300 – 0.0636 = 0.5864$

(a)(iii)
For conditional probability, we divide the joint probability of both events occurring by the marginal probability of the given condition (“woman”).
$P(\text{never} \mid \text{woman}) = \dfrac{P(\text{never and woman})}{P(\text{woman})}$
$P(\text{never} \mid \text{woman}) = \dfrac{0.0636}{0.5300} = 0.12$

(b)
We can check for independence by seeing if the conditional probability of an event equals its marginal probability.
Since $P(\text{never} \mid \text{woman}) = 0.12$ and the overall probability $P(\text{never}) = 0.12$, the probabilities are identical.
Yes, this indicates that the event of responding “never” is perfectly independent of the event of being a “woman”.

(c)
This scenario can be modeled using a binomial distribution with $n=5$ trials and a success probability of $p=0.54$.
We need to find the probability of getting at least $4$ successes, which means finding the sum of $P(X=4)$ and $P(X=5)$.
$P(X \ge 4) = \binom{5}{4}(0.54)^4(0.46)^1 + \binom{5}{5}(0.54)^5(0.46)^0$
$P(X \ge 4) \approx 0.19557 + 0.04592 \approx 0.24149$

Question

Approximately $3.5$ percent of all children born in a certain region are from multiple births (that is, twins, triplets, etc.). Of the children born in the region who are from multiple births, $22$ percent are left-handed. Of the children born in the region who are from single births, $11$ percent are left-handed.

(a) What is the probability that a randomly selected child born in the region is left-handed?

(b) What is the probability that a randomly selected child born in the region is a child from a multiple birth, given that the child selected is left-handed?

(c) A random sample of $20$ children born in the region will be selected. What is the probability that the sample will have at least $3$ children who are left-handed?

Most-appropriate topic codes (AP Statistics):

• Topic $2.6$ — Conditional Probability (Parts $ \mathrm{a} $, $ \mathrm{b} $)
• Topic $2.7$ — Independent Events and Mutually Exclusive Events (Parts $ \mathrm{a} $, $ \mathrm{b} $)
• Topic $2.10$ — The Binomial Distribution (Part $ \mathrm{c} $)

▶️ Answer/Explanation

(a)
Let $L$ denote the event that a child is left-handed, $M$ denote the event of a multiple birth, and $S$ denote the event of a single birth.
We are given: $P(M) = 0.035$, $P(S) = 0.965$, $P(L \mid M) = 0.22$, and $P(L \mid S) = 0.11$.
Apply the Law of Total Probability:
$P(L) = P(M) \cdot P(L \mid M) + P(S) \cdot P(L \mid S)$
$P(L) = (0.035)(0.22) + (0.965)(0.11)$
$P(L) = 0.0077 + 0.10615$
$\boxed{P(L) = 0.11385}$
Think of this as a weighted average of the two left-handedness rates — you weight each group’s rate by how large that group is. Since single births make up the overwhelming majority (96.5%), they drive the overall rate, which lands very close to 11%. The small multiple-birth group nudges it up just slightly to about 11.4%.

(b)
We use the definition of conditional probability to find $P(M \mid L)$.
$P(M \mid L) = \dfrac{P(M \cap L)}{P(L)}$
The joint probability $P(M \cap L)$ is found using the multiplication rule:
$P(M \cap L) = P(M) \cdot P(L \mid M) = (0.035)(0.22) = 0.0077$
Using $P(L) = 0.11385$ from part (a):
$P(M \mid L) = \dfrac{0.0077}{0.11385}$
$\boxed{P(M \mid L) \approx 0.0676}$
This is a classic Bayes-style reversal — we flipped from “given multiple birth, what’s the chance of being left-handed?” to “given left-handed, what’s the chance of multiple birth?” Even though multiple-birth children are twice as likely to be left-handed (22% vs. 11%), they are so rare (only 3.5% of all births) that among all left-handed children, only about 6.8% actually come from multiple births. Rarity wins.

(c)
Let $X$ represent the number of left-handed children in a random sample of $20$. Since each child is selected independently with the same probability of being left-handed, $X$ follows a binomial distribution:
$X \sim \operatorname{Binomial}(n = 20,\; p = 0.11385)$
We want $P(X \geq 3)$. Using the complement:
$P(X \geq 3) = 1 – P(X \leq 2) = 1 – \bigl[P(X = 0) + P(X = 1) + P(X = 2)\bigr]$
Using the binomial formula $\displaystyle P(X = k) = \binom{20}{k}(0.11385)^{k}(0.88615)^{20-k}$:
$P(X = 0) = \dbinom{20}{0}(0.11385)^{0}(0.88615)^{20} \approx 0.0891$
$P(X = 1) = \dbinom{20}{1}(0.11385)^{1}(0.88615)^{19} \approx 0.2289$
$P(X = 2) = \dbinom{20}{2}(0.11385)^{2}(0.88615)^{18} \approx 0.2797$
Summing the complement terms:
$P(X \leq 2) \approx 0.0891 + 0.2289 + 0.2797 = 0.5977$
$P(X \geq 3) = 1 – 0.5977$
$\boxed{P(X \geq 3) \approx 0.402}$
The complement trick is your best friend here — instead of adding up $P(X=3) + P(X=4) + \cdots + P(X=20)$, which would take forever, you subtract the small pile of easy cases (0, 1, or 2 left-handed kids) from 1. With only about an 11.4% chance per child, there’s roughly a 40% chance that 3 or more kids in a group of 20 will be left-handed — higher than you might expect, because “at least 3” covers most of the distribution.

Question

A grocery store purchases melons from two distributors, J and K. Distributor J provides melons from organic farms. The distribution of the diameters of the melons from Distributor J is approximately normal with mean $133$ millimeters (mm) and standard deviation $5$ mm.

(a) For a melon selected at random from Distributor J, what is the probability that the melon will have a diameter greater than $137$ mm?

Distributor K provides melons from nonorganic farms. The probability is $0.8413$ that a melon selected at random from Distributor K will have a diameter greater than $137$ mm. For all the melons at the grocery store, $70$ percent of the melons are provided by Distributor J and $30$ percent are provided by Distributor K.

(b) For a melon selected at random from the grocery store, what is the probability that the melon will have a diameter greater than $137$ mm?

(c) Given that a melon selected at random from the grocery store has a diameter greater than $137$ mm, what is the probability that the melon will be from Distributor J?

Most-appropriate topic codes (AP Statistics):

• Topic $2.6$ — Conditional Probability (Parts $ \mathrm{b} $, $ \mathrm{c} $)
• Topic $2.7$ — Independent Events and Mutually Exclusive Events (Parts $ \mathrm{b} $, $ \mathrm{c} $)
• Topic $2.11$ — The Normal Distribution (Part $ \mathrm{a} $)

▶️ Answer/Explanation

(a)
Let $X$ denote the diameter (in mm) of a randomly selected melon from Distributor J. We are told that $X$ follows an approximately normal distribution with mean $\mu = 133$ mm and standard deviation $\sigma = 5$ mm.
First, convert the boundary value to a $z$-score:
$ z = \dfrac{137 – 133}{5} = \dfrac{4}{5} = 0.8 $
Now find the probability to the right of $z = 0.8$ using the standard normal table:
$ P(X > 137) = P(Z > 0.8) = 1 – P(Z < 0.8) = 1 – 0.7881 $
$ \boxed{P(X > 137) = 0.2119} $

(b)
Define the following events:
$J$: the melon is from Distributor J
$K$: the melon is from Distributor K
$G$: the melon has a diameter greater than $137$ mm
We are given $P(J) = 0.70$, $P(K) = 0.30$, $P(G \mid J) = 0.2119$, and $P(G \mid K) = 0.8413$.
Using the Law of Total Probability:
$ P(G) = P(G \mid J)\cdot P(J) + P(G \mid K)\cdot P(K) $
$ P(G) = (0.2119)(0.70) + (0.8413)(0.30) $
$ P(G) = 0.14833 + 0.25239 $
$ \boxed{P(G) = 0.4007} $

(c)
We want the conditional probability that the melon is from Distributor J, given that its diameter is greater than $137$ mm. Using the definition of conditional probability:
$ P(J \mid G) = \dfrac{P(J \cap G)}{P(G)} $
The joint probability $P(J \cap G)$ was found in part (b):
$ P(J \cap G) = P(G \mid J)\cdot P(J) = (0.2119)(0.70) = 0.14833 $
Substituting into the formula:
$ P(J \mid G) = \dfrac{0.14833}{0.40072} $
$ \boxed{P(J \mid G) \approx 0.3701} $
Even though Distributor J supplies $70\%$ of the melons in the store, only about $37\%$ of the large-diameter melons (over $137$ mm) come from Distributor J — this makes sense because Distributor K’s melons are much more likely to be large, so they dominate that group despite being the smaller supplier.

Question

Consider an experiment in which two men and two women will be randomly assigned to either a treatment group or a control group in such a way that each group has two people. The people are identified as Man 1, Man 2, Woman 1, and Woman 2. The six possible arrangements are shown below.

Two possible methods of assignment are being considered: the sequential coin flip method, as described in part (a), and the chip method, as described in part (b). For each method, the order of the assignment will be Man 1, Man 2, Woman 1, Woman 2.

(a) For the sequential coin flip method, a fair coin is flipped until one group has two people. An outcome of tails assigns the person to the treatment group, and an outcome of heads assigns the person to the control group. As soon as one group has two people, the remaining people are automatically assigned to the other group.

(i) Complete the table below by calculating the probability of each arrangement occurring if the sequential coin flip method is used.

(ii) For the sequential coin flip method, what is the probability that Man 1 and Man 2 are assigned to the same group?

(b) For the chip method, two chips are marked “treatment” and two chips are marked “control.” Each person selects one chip at random without replacement.

(i) Complete the table below by calculating the probability of each arrangement occurring if the chip method is used.

(ii) For the chip method, what is the probability that Man 1 and Man 2 are assigned to the same group?

(c) Sixteen participants consisting of 10 students and 6 teachers at an elementary school will be used for an experiment to determine lunch preference for the school population of students and teachers. As the participants enter the school cafeteria for lunch, they will be randomly assigned to receive one of two lunches so that 8 will receive a salad, and 8 will receive a grilled cheese sandwich. The students will enter the cafeteria first, and the teachers will enter next. Which method, the sequential coin flip method or the chip method, should be used to assign the treatments? Justify your choice.

Most-appropriate topic codes (AP Statistics):

• Topic $1.13$ — Experimental Design (Part $ \mathrm{c} $)
• Topic $2.6$ — Conditional Probability (Parts $ \mathrm{a} $, $ \mathrm{b} $)
• Topic $2.7$ — Independent Events and Mutually Exclusive Events (Parts $ \mathrm{a} $, $ \mathrm{b} $)

▶️ Answer/Explanation

(a)(i)
Let T (tail) represent being assigned to the treatment group and H (head) represent being assigned to the control group. The process stops as soon as one group fills up. We trace each possible sequence of flips:

(a)(ii)
Man 1 and Man 2 are assigned to the same group only in Arrangements A (both in treatment) and D (both in control). So the probability is:

$P(A) + P(D) = \dfrac{1}{4} + \dfrac{1}{4} = \boxed{\dfrac{1}{2}}$

(b)(i)
Let T represent being assigned to the treatment group and C represent being assigned to the control group. Since chips are drawn without replacement from a pool of 2 T chips and 2 C chips, the probabilities change at each draw. Working through each arrangement:

(b)(ii)
Man 1 and Man 2 are in the same group only in Arrangements A and D. Therefore:

$P(A) + P(D) = \dfrac{1}{6} + \dfrac{1}{6} = \boxed{\dfrac{1}{3}}$

(c)
The chip method should be used. Here is the reasoning:

From parts (a)(i) and (b)(i), the chip method gives every arrangement an equal probability of $\dfrac{1}{6}$, while the coin flip method assigns unequal probabilities — arrangements A and D each have probability $\dfrac{1}{4}$, while B, C, E, and F each have probability $\dfrac{1}{8}$.

From parts (a)(ii) and (b)(ii), the probability that both men end up in the same group is $\dfrac{1}{2}$ under the coin method but only $\dfrac{1}{3}$ under the chip method. Since students enter first and teachers enter next, the coin flip method is more likely to place all students together in one group — if teachers and students have different food preferences, this imbalance would make it impossible to tell whether any observed difference in lunch preference is due to the treatment (type of lunch) or the role of the participant (teacher vs. student).

The chip method, by giving all arrangements an equal chance, is therefore more appropriate for this experiment.

Question

A company manufactures model rockets that require igniters to launch. Once an igniter is used to launch a rocket, the igniter cannot be reused. Sometimes an igniter fails to operate correctly, and the rocket does not launch. The company estimates that the overall failure rate, defined as the percent of all igniters that fail to operate correctly, is 15 percent.

A company engineer develops a new igniter, called the super igniter, with the intent of lowering the failure rate. To test the performance of the super igniters, the engineer uses the following process.

Step 1: One super igniter is selected at random and used in a rocket.

Step 2: If the rocket launches, another super igniter is selected at random and used in a rocket.

Step 2 is repeated until the process stops. The process stops when a super igniter fails to operate correctly or 32 super igniters have successfully launched rockets, whichever comes first. Assume that super igniter failures are independent.

(a) If the failure rate of the super igniters is 15 percent, what is the probability that the first 30 super igniters selected using the testing process successfully launch rockets?

(b) Given that the first 30 super igniters successfully launch rockets, what is the probability that the first failure occurs on the thirty-first or the thirty-second super igniter tested if the failure rate of the super igniters is 15 percent?

(c) Given that the first 30 super igniters successfully launch rockets, is it reasonable to believe that the failure rate of the super igniters is less than 15 percent? Explain.

Most-appropriate topic codes (AP Statistics):

• Topic $2.6$ — Conditional Probability (Part $ \mathrm{b} $)
• Topic $2.7$ — Independent Events and Unions of Events (Parts $ \mathrm{a} $, $ \mathrm{b} $)
• Topic $3.6$ — $p$-Values (Part $ \mathrm{c} $)

▶️ Answer/Explanation

(a)

Since failures are independent and the failure rate is $p = 0.15$, the probability that a single super igniter successfully launches a rocket is:
$P(\text{success}) = 1 – 0.15 = 0.85$
For the first 30 super igniters to all successfully launch, each of the 30 independent trials must be a success:
$P(\text{first 30 all succeed}) = (0.85)^{30}$
$P(\text{first 30 all succeed}) = (0.85)^{30} \approx 0.0076$
$\boxed{P \approx 0.0076}$

(b)

We are told the first 30 igniters all succeeded. We want the probability that the first failure occurs on either the 31st or the 32nd igniter tested.
For the first failure to occur on the 31st igniter, the 31st must fail. Since failures are independent and we are conditioning on the 30 prior successes:
$P(\text{failure on 31st}) = 0.15$
For the first failure to occur on the 32nd igniter, the 31st must succeed AND the 32nd must fail:
$P(\text{failure on 32nd}) = (0.85)(0.15) = 0.1275$
These two events are mutually exclusive, so:
$P(\text{first failure on 31st or 32nd}) = 0.15 + (0.85)(0.15)$
$= 0.15 + 0.1275 = 0.2775$
$\boxed{P \approx 0.2775}$

(c)

Yes, it is reasonable to believe that the failure rate of the super igniters is less than 15 percent.
If the failure rate were truly 15 percent (same as the original igniters), the probability of the first 30 super igniters all successfully launching rockets would be only about $0.0076$, or less than 1 percent. This is an extremely unlikely outcome under the assumption that the failure rate is 15 percent.
Because observing 30 consecutive successes would be so rare if the failure rate were really 15 percent, getting this result in the test provides strong evidence that the true failure rate of the super igniters is lower than 15 percent. The data are inconsistent with a 15 percent failure rate, so it is indeed reasonable to believe the super igniters have a lower failure rate than the original igniters.

Question

Schools in a certain state receive funding based on the number of students who attend the school. To determine the number of students who attend a school, one school day is selected at random and the number of students in attendance that day is counted and used for funding purposes. The daily number of absences at High School A in the state is approximately normally distributed with mean of 120 students and standard deviation of 10.5 students.

(a) If more than 140 students are absent on the day the attendance count is taken for funding purposes, the school will lose some of its state funding in the subsequent year. Approximately what is the probability that High School A will lose some state funding?

(b) The principals’ association in the state suggests that instead of choosing one day at random, the state should choose 3 days at random. With the suggested plan, High School A would lose some of its state funding in the subsequent year if the mean number of students absent for the 3 days is greater than 140. Would High School A be more likely, less likely, or equally likely to lose funding using the suggested plan compared to the plan described in part (a)? Justify your choice.

(c) A typical school week consists of the days Monday, Tuesday, Wednesday, Thursday, and Friday. The principal at High School A believes that the number of absences tends to be greater on Mondays and Fridays, and there is concern that the school will lose state funding if the attendance count occurs on a Monday or Friday. If one school day is chosen at random from each of 3 typical school weeks, what is the probability that none of the 3 days chosen is a Tuesday, Wednesday, or Thursday?

Most-appropriate topic codes (AP Statistics):

• Topic $2.7$ — Normal Probability Distributions (Part $ \mathrm{a} $)
• Topic $4.1$ — Sampling Distributions for Sample Means (Part $ \mathrm{b} $)
• Topic $2.6$ — Probability Rules and Calculations of Probability (Part $ \mathrm{c} $)

▶️ Answer/Explanation

(a)
The daily number of absences follows an approximately normal distribution with $\mu = 120$ and $\sigma = 10.5$. We need $P(X > 140)$.
First, compute the $z$-score for $x = 140$:
$z = \dfrac{x – \mu}{\sigma} = \dfrac{140 – 120}{10.5} \approx 1.90$
From the standard normal table, $P(Z \leq 1.90) = 0.9713$, so:
$P(X > 140) = 1 – P(Z \leq 1.90) = 1 – 0.9713 = 0.0287$
$\boxed{P(\text{lose funding}) \approx 0.0287}$

(b)
High School A would be less likely to lose funding under the suggested plan.
Under the suggested plan, the relevant quantity is the sample mean $\bar{x}$ of absences over 3 days. By the Central Limit Theorem, $\bar{x}$ is approximately normally distributed with the same mean $\mu_{\bar{x}} = 120$ but a smaller standard deviation:
$\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{10.5}{\sqrt{3}} \approx 6.062$
Now compute the $z$-score for $\bar{x} = 140$:
$z = \dfrac{140 – 120}{6.062} \approx 3.30$
From the standard normal table, $P(Z \leq 3.30) = 0.9995$, so:
$P(\bar{x} > 140) = 1 – 0.9995 = 0.0005$
Since $0.0005 < 0.0287$, the school is less likely to lose funding under the 3-day plan. Taking the average over multiple days reduces variability, making it much harder for the mean to stray as far as 140 from the center of 120.

(c)
A typical school week has 5 days: Monday, Tuesday, Wednesday, Thursday, and Friday. The “bad” days (Monday or Friday) number 2 out of 5, while the “safe” days (Tuesday, Wednesday, or Thursday) number 3 out of 5.
We want the probability that none of the 3 days chosen (one from each of 3 weeks) is a Tuesday, Wednesday, or Thursday — meaning all 3 days must be Monday or Friday.
For any one week, the probability of choosing a Monday or Friday is:
$P(\text{Mon or Fri}) = \dfrac{2}{5} = 0.4$
Since the day chosen each week is independent of the other weeks:
$P(\text{none of the 3 days is Tue, Wed, or Thu}) = (0.4)^3 = 0.064$
$\boxed{P = 0.064}$

Question

The table below shows the political party registration by gender of all $500$ registered voters in Franklin Township.

(a) Given that a randomly selected registered voter is a male, what is the probability that he is registered for Party Y?

(b) Among the registered voters of Franklin Township, are the events “is a male” and “is registered for Party Y” independent? Justify your answer based on probabilities calculated from the table above.

(c) One way to display the data in the table is to use a segmented bar graph. The following segmented bar graph, constructed from the data in the party registration-Franklin Township table, shows party-registration distributions for males and females in Franklin Township.

In Lawrence Township, the proportions of all registered voters for Parties W, X, and Y are the same as for Franklin Township, and party registration is independent of gender. Complete the graph below to show the distributions of party registration by gender in Lawrence Township.

Most-appropriate topic codes (AP Statistics):

• Topic 2.6 — Conditional Probability (Part a)
• Topic 2.7 — Independent Events and Unions of Events (Part b)
• Topic 2.1 — Tabular and Graphical Representations for the Distributions of Two Categorical Variables (Part c)

▶️ Answer/Explanation

(a)
We need to find the conditional probability that a voter is registered for Party Y, given they are male. Looking only at the “Male” row, there are $200$ total males, and $48$ of them are registered for Party Y.
$P(\text{Party Y} \mid \text{Male}) = \frac{48}{200} = 0.24$

(b)
No, the events “is a male” and “is registered for Party Y” are not independent.
Two events $A$ and $B$ are independent if $P(A \mid B) = P(A)$. Let’s compare the conditional probability from part (a) to the overall marginal probability of being registered for Party Y.
$P(\text{Party Y}) = \frac{168}{500} = 0.336$
Since $P(\text{Party Y} \mid \text{Male}) = 0.24$ and $P(\text{Party Y}) = 0.336$, the probabilities are not equal ($0.24 \neq 0.336$). Knowing that a randomly selected voter is male changes the probability that they are registered for Party Y, so the events are dependent.

(c)
Because party registration is independent of gender in Lawrence Township, the distribution of party registration for both males and females must be identical to the overall marginal distribution of the town.
We first calculate the overall proportions for each party (which are the same as Franklin Township’s overall proportions):
Party W: $\frac{88}{500} = 0.176$
Party X: $\frac{244}{500} = 0.488$
Party Y: $\frac{168}{500} = 0.336$
To complete the segmented bar graph, you would draw identical bars for both the Male and Female categories with the following dividing lines:
• The segment for Party W starts at $0.0$ and ends at $0.176$.
• The segment for Party X starts at $0.176$ and ends at $0.176 + 0.488 = 0.664$.
• The segment for Party Y starts at $0.664$ and extends to $1.0$.

Question

Every year, each student in a nationally representative sample is given tests in various subjects. Recently, a random sample of $9,600$ twelfth-grade students from the United States were administered a multiple-choice United States history exam. One of the multiple-choice questions is below.

Of the $9,600$ students, $28$ percent answered the multiple-choice question correctly.

(a) Let $p$ represent the proportion of all United States twelfth-grade students who would answer the question correctly. Construct and interpret a 99 percent confidence interval for $p$.

Assume that students who actually know the answer correctly answer the question. For students who do not know the answer, assume that the question can be answered correctly by guessing completely at random from among the four multiple-choice options.

Let $k$ represent the proportion of all United States twelfth-grade students who actually know the answer to the question.

(b) A tree diagram of the possible outcomes for a randomly selected twelfth-grade student is provided below. Write the correct probability in each of the five empty boxes. Some of the probabilities may be expressions in terms of k.

(c) Based on the expectation from the probability tree diagram, express the proportion of all United States twelfth-grade students who would answer the question correctly, $p$, in terms of $k$.

(d) Using the formula from part (c) and the lower and upper endpoints of the confidence interval from part (a), construct a 99 percent confidence interval for $k$, the proportion of all United States twelfth-grade students who actually know the answer to the question.

Most-appropriate topic codes (AP Statistics):

• Topic 3.3 — Constructing a Confidence Interval for a Population Proportion (Part a)
• Topic 2.6 — Conditional Probability (Part b)
• Topic 2.7 — Independent Events and Unions of Events (Part c)
• Topic 3.4 — Justifying a Claim Based on a Confidence Interval for a Population Proportion (Part d)

▶️ Answer/Explanation

(a)
We want to construct a 99% confidence interval for the true proportion $p$.
First, calculate our point estimate and confirm design guidelines:
$\hat{p} = \frac{2,688}{9,600} = 0.28$
$n\hat{p} = 2,688 \ge 10 \quad \text{and} \quad n(1-\hat{p}) = 9,600 – 2,688 = 6,912 \ge 10$
The sample is random, and success/failure counts are well above 10, satisfying normality criteria. Using critical value $z^* = 2.576$:
$\text{Margin of Error} = z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{9,600}} = 2.576 \sqrt{\frac{0.28 \times 0.72}{9,600}} = 2.576 \times 0.004583 = 0.0118$
$\text{Interval} = 0.28 \pm 0.0118 \rightarrow (0.2682, 0.2918)$
Interpretation: We are 99% confident that the true proportion of all United States twelfth-grade students who would answer this history question correctly is between $0.2682$ and $0.2918$.

(b)

The branches are filled based on the structural logic given in the problem prompt:
• The top initial branch represents the probability a student Knows the answer, which is: $k$
• The bottom initial branch represents the probability a student Does not know the answer, which is: $1 – k$
• Given they know the answer, they choose the correct option with absolute certainty, making the conditional probability of a Correct response: $1$
• Given they do not know the answer, they choose by random guessing out of 4 options, making the conditional probability of a Correct response: $\frac{1}{4} = 0.25$

(c)
To find the overall proportion $p$ of correct answers, accumulate the final correct path terminals from your tree model:
$p = P(\text{Knows}) \times P(\text{Correct} \mid \text{Knows}) + P(\text{Doesn’t Know}) \times P(\text{Correct} \mid \text{Doesn’t Know})$
$p = k(1) + (1-k)(0.25)$
$p = k + 0.25 – 0.25k = 0.75k + 0.25$

(d)
To construct a 99% interval for $k$, isolate $k$ from our part (c) linear transformation model:
$p = 0.75k + 0.25 \implies p – 0.25 = 0.75k \implies k = \frac{p – 0.25}{0.75}$
Now, substitute the lower and upper bounds of $p$ from part (a) into this equation:
$\text{Lower Bound for } k = \frac{0.2682 – 0.25}{0.75} = \frac{0.0182}{0.75} = 0.0243$
$\text{Upper Bound for } k = \frac{0.2918 – 0.25}{0.75} = \frac{0.0418}{0.75} = 0.0557$
The resulting 99% confidence interval for the proportion of students who actually know the answer is $(0.0243, 0.0557)$.

Question

An airline claims that there is a $0.10$ probability that a coach-class ticket holder who flies frequently will be upgraded to first class on any flight. This outcome is independent from flight to flight. Sam is a frequent flier who always purchases coach-class tickets.

(a) What is the probability that Sam’s first upgrade will occur after the third flight?

(b) What is the probability that Sam will be upgraded exactly $2$ times in his next $20$ flights?

(c) Sam will take $104$ flights next year. Would you be surprised if Sam receives more than $20$ upgrades to first class during the year? Justify your answer.

Most-appropriate topic codes (AP Statistics):

• Topic 2.10 — The Binomial Distribution (Parts b & c)
• Topic 2.10 — The Binomial Distribution (Part a)
• Topic 2.7 — Independent Events and Unions of Events (Part a)

▶️ Answer/Explanation

(a)
Let $Y$ denote the number of flights Sam must make until he receives his first upgrade. The random variable $Y$ follows a geometric distribution with $p = 0.1$.
The probability that Sam’s upgrade will occur after his third flight is equivalent to the probability that he receives no upgrade on his first three flights.
$P(Y \ge 4) = 1 – P(Y \le 3)$
$= 1 – [P(Y=1) + P(Y=2) + P(Y=3)]$
$= 1 – [0.1 + 0.9(0.1) + (0.9)^2(0.1)]$
$= 1 – [0.1 + 0.09 + 0.081]$
$= 0.729$
$\boxed{0.729}$

(b)
Let $X$ denote the number of upgrades Sam will receive in $20$ flights. The random variable $X$ follows a binomial distribution with $n = 20$ independent trials and $p = 0.1$.
The probability that Sam will be upgraded exactly $2$ times is calculated as follows:
$P(X = 2) = \binom{20}{2}(0.1)^2(0.9)^{18}$
$\approx 0.2852$
$\boxed{P(X = 2) \approx 0.2852}$

(c)
Let $X$ denote the number of upgrades Sam will receive in $104$ flights. The random variable $X$ follows a binomial distribution with $n = 104$ independent trials and $p = 0.1$.
We need to find the probability of receiving more than $20$ upgrades:
$P(X > 20) = 1 – P(X \le 20)$
$\approx 1 – 0.9986$
$\approx 0.0014$
Because this probability is so small (less than $1\%$), it is very unlikely that Sam would receive more than $20$ upgrades in $104$ flights if the airline’s claim is correct. This would be expected to happen less than $1$ percent of the time.
Therefore, I would be surprised if Sam receives more than $20$ upgrades during the year.

Question

An advertising agency in a large city is conducting a survey of adults to investigate whether there is an association between highest level of educational achievement and primary source for news. The company takes a random sample of 2,500 adults in the city. The results are shown in the table below.

(a) If an adult is to be selected at random from this sample, what is the probability that the selected adult is a college graduate or obtains news primarily from the internet?

(b) If an adult who is a college graduate is to be selected at random from this sample, what is the probability that the selected adult obtains news primarily from the internet?

(c) When selecting an adult at random from the sample of 2,500 adults, are the events “is a college graduate” and “obtains news primarily from the internet” independent? Justify your answer.

(d) The company wants to conduct a statistical test to investigate whether there is an association between educational achievement and primary source for news for adults in the city. What is the name of the statistical test that should be used?

What are the appropriate degrees of freedom for this test?

Most-appropriate topic codes (AP Statistics):

• Topic $2.1$ — Tabular and Graphical Representations for the Distributions of Two Categorical Variables (Parts $\mathrm{a}$, $\mathrm{b}$, $\mathrm{c}$)
• Topic $2.6$ — Conditional Probability (Part $\mathrm{b}$)
• Topic $2.7$ — Independent Events and Unions of Events (Parts $\mathrm{a}$, $\mathrm{c}$)
• Topic $3.14$ — Setting Up a Chi-Square Test for Homogeneity or Independence (Part $\mathrm{d}$)

▶️ Answer/Explanation

(a)
Let $C$ = event that the adult is a college graduate, and $I$ = event that the adult obtains news primarily from the internet.
Using the Addition Rule:
$P(C \cup I) = P(C) + P(I) – P(C \cap I)$
Reading the values directly from the table:
$P(C) = \frac{693}{2500}, \qquad P(I) = \frac{687}{2500}, \qquad P(C \cap I) = \frac{245}{2500}$
$P(C \cup I) = \frac{693}{2500} + \frac{687}{2500} – \frac{245}{2500} = \frac{693 + 687 – 245}{2500} = \frac{1135}{2500}$
$\boxed{P(C \cup I) = \frac{1135}{2500} = 0.454}$
Don’t forget to subtract the overlap — college graduates who use the internet get counted in both the college graduate total and the internet total, so we subtract them once to avoid double-counting.

(b)
We want the conditional probability that an adult obtains news from the internet, given that the adult is a college graduate. From the table, among the 693 college graduates, 245 primarily use the internet:
$P(I \mid C) = \frac{P(C \cap I)}{P(C)} = \frac{\dfrac{245}{2500}}{\dfrac{693}{2500}} = \frac{245}{693}$
$\boxed{P(I \mid C) = \frac{245}{693} \approx 0.354}$
This is a conditional probability — we’ve already restricted our pool to only the 693 college graduates, so 693 becomes the new denominator. The 2,500 total cancels out entirely.

(c)
Two events are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$, which is equivalent to checking whether $P(I \mid C) = P(I)$.
From the table:
$P(I) = \frac{687}{2500} = 0.275$
$P(I \mid C) = \frac{245}{693} \approx 0.354$
Since $P(I \mid C) \approx 0.354 \neq 0.275 = P(I)$, the two events are not independent.
We can also verify using the multiplication rule directly:
$P(C) \cdot P(I) = \frac{693}{2500} \times \frac{687}{2500} = \frac{476{,}091}{6{,}250{,}000} \approx 0.0762$
$P(C \cap I) = \frac{245}{2500} = 0.098$
Since $0.098 \neq 0.0762$, the events are confirmed to be not independent. In real terms, college graduates are noticeably more likely to get their news from the internet than the general adult population — that difference in rates is exactly what “not independent” means here.

(d)
The appropriate test is the Chi-Square Test of Association (or Independence).
This test is used when we want to determine whether there is an association between two categorical variables — here, educational achievement (3 categories) and primary news source (5 categories).
The degrees of freedom are calculated as:
$\text{df} = (\text{number of rows} – 1) \times (\text{number of columns} – 1)$
$\text{df} = (5 – 1) \times (3 – 1) = 4 \times 2 = \boxed{8}$
There are 5 rows (news source categories) and 3 columns (education levels), not counting the totals row and column. The degrees of freedom formula captures how many cells in the table are “free to vary” once the row and column totals are fixed.

Question

The ELISA tests whether a patient has contracted HIV. The ELISA is said to be positive if it indicates that HIV is present in a blood sample, and the ELISA is said to be negative if it does not indicate that HIV is present in a blood sample. Instead of directly measuring the presence of HIV, the ELISA measures levels of antibodies in the blood that should be elevated if HIV is present. Because of variability in antibody levels among human patients, the ELISA does not always indicate the correct result.

As part of a training program, staff at a testing lab applied the ELISA to 500 blood samples known to contain HIV. The ELISA was positive for 489 of those blood samples and negative for the other 11 samples. As part of the same training program, the staff also applied the ELISA to 500 other blood samples known to not contain HIV. The ELISA was positive for 37 of those blood samples and negative for the other 463 samples.

(a) When a new blood sample arrives at the lab, it will be tested to determine whether HIV is present. Using the data from the training program, estimate the probability that the ELISA would be positive when it is applied to a blood sample that does not contain HIV.

(b) Among the blood samples examined in the training program that provided positive ELISA results for HIV, what proportion actually contained HIV?

(c) When a blood sample yields a positive ELISA result, two more ELISAs are performed on the same blood sample. If at least one of the two additional ELISAs is positive, the blood sample is subjected to a more expensive and more accurate test to make a definitive determination of whether HIV is present in the sample. Repeated ELISAs on the same sample are generally assumed to be independent. Under the assumption of independence, what is the probability that a new blood sample that comes into the lab will be subjected to the more expensive test if that sample does not contain HIV?

Most-appropriate topic codes (AP Statistics):

• Topic $2.6$ — Conditional Probability (Parts $\mathrm{a}$, $\mathrm{b}$)
• Topic $2.7$ — Independent Events and Unions of Events (Part $\mathrm{c}$)
• Topic $2.5$ — Mutually Exclusive Events (Part $\mathrm{c}$, complement rule)

▶️ Answer/Explanation

(a)

Out of the 500 blood samples known to not contain HIV, the ELISA returned a positive result for 37 of them.
So the estimated probability that the ELISA is positive given no HIV is:
$P(\text{positive} \mid \text{no HIV}) = \frac{37}{500} = 0.074$
$\boxed{P(\text{positive} \mid \text{no HIV}) \approx 0.074}$

(b)

First, find the total number of blood samples that returned a positive ELISA result across both groups:
$489 + 37 = 526 \text{ positive results in total}$
Of those 526 positive results, 489 actually came from samples that truly contained HIV.
So the proportion of positive ELISA results that actually contained HIV is:
$\frac{489}{526} \approx 0.9297$
$\boxed{\frac{489}{526} \approx 0.9297}$

(c)

From part (a), the probability the ELISA is positive for a sample with no HIV is $0.074$, and therefore the probability it is negative is $1 – 0.074 = 0.926$.
A sample with no HIV will be sent for the expensive test if: the 1st ELISA is positive AND at least one of the 2 follow-up ELISAs is also positive.
Find the probability that at least one of the two follow-up tests is positive (using the complement):
$P(\text{at least one positive in 2 follow-ups}) = 1 – P(\text{both negative}) = 1 – (0.926)^2$
$= 1 – 0.857476 = 0.142524$
Now multiply by the probability the first ELISA is positive:
$P(\text{subjected to expensive test}) = (0.074)(0.142524)$
$= 0.010547$
$\boxed{P(\text{subjected to expensive test}) \approx 0.0105}$

Question

A local arcade is hosting a tournament in which contestants play an arcade game with possible scores ranging from 0 to 20. The arcade has set up multiple game tables so that all contestants can play the game at the same time; thus contestant scores are independent. Each contestant’s score will be recorded as he or she finishes, and the contestant with the highest score is the winner.

After practicing the game many times, Josephine, one of the contestants, has established the probability distribution of her scores, shown in the table below.

Crystal, another contestant, has also practiced many times. The probability distribution for her scores is shown in the table below.

(a) Calculate the expected score for each player.

(b) Suppose that Josephine scores 16 and Crystal scores 17. The difference (Josephine minus Crystal) of their scores is $-1$. List all combinations of possible scores for Josephine and Crystal that will produce a difference (Josephine minus Crystal) of $-1$, and calculate the probability for each combination.

(c) Find the probability that the difference (Josephine minus Crystal) in their scores is $-1$.

(d) The table below lists all the possible differences in the scores between Josephine and Crystal and some associated probabilities.

Complete the table and calculate the probability that Crystal’s score will be higher than Josephine’s score.

Most-appropriate topic codes (AP Statistics):

• Topic $2.8$ — Introduction to Random Variables and Probability Distributions (Parts $\mathrm{a}$, $\mathrm{b}$, $\mathrm{c}$, $\mathrm{d}$)
• Topic $2.9$ — Parameters of Random Variables (Part $\mathrm{a}$)
• Topic $2.7$ — Independent Events and Unions of Events (Parts $\mathrm{b}$, $\mathrm{c}$, $\mathrm{d}$)

▶️ Answer/Explanation

(a)

The expected value (mean) of a discrete random variable is $\mu = \sum x_i \cdot P(x_i)$. We apply this to each player’s distribution.
For Josephine:
$\mu_J = 16(0.10) + 17(0.30) + 18(0.40) + 19(0.20)$
$\mu_J = 1.6 + 5.1 + 7.2 + 3.8 = \boxed{17.7}$
For Crystal:
$\mu_C = 17(0.45) + 18(0.40) + 19(0.15)$
$\mu_C = 7.65 + 7.20 + 2.85 = \boxed{17.7}$
Both players have the same expected score of $17.7$.

(b)

We need all pairs $(J, C)$ such that $J – C = -1$, i.e., Josephine’s score is exactly 1 less than Crystal’s. Since scores are independent, the probability of each pair is the product of the individual probabilities.
$J = 16,\ C = 17$: $\quad P = (0.10)(0.45) = 0.045$
$J = 17,\ C = 18$: $\quad P = (0.30)(0.40) = 0.120$
$J = 18,\ C = 19$: $\quad P = (0.40)(0.15) = 0.060$
These are the only three combinations that produce a difference of $-1$.

(c)

Since the three combinations in part (b) are mutually exclusive, we simply add their probabilities:
$P(J – C = -1) = 0.045 + 0.120 + 0.060$
$\boxed{P(J – C = -1) = 0.225}$

(d)

First, we find the missing probability for a difference of $-2$. Since all probabilities in the distribution must sum to 1:
$P(J – C = -2) = 1 – 0.015 – 0.225 – 0.325 – 0.260 – 0.090$
$P(J – C = -2) = 1 – 0.915 = \boxed{0.085}$
The completed distribution table is:

Crystal’s score is higher than Josephine’s when the difference $J – C < 0$, i.e., when the difference is $-3$, $-2$, or $-1$.
$P(\text{Crystal} > \text{Josephine}) = P(J – C < 0) = 0.015 + 0.085 + 0.225$
$\boxed{P(\text{Crystal} > \text{Josephine}) = 0.325}$

Question

The depth from the surface of Earth to a refracting layer beneath the surface can be estimated using methods developed by seismologists. One method is based on the time required for vibrations to travel from a distant explosion to a receiving point. The depth measurement $(M)$ is the sum of the true depth $(D)$ and the random measurement error $(E)$. That is, $M = D + E$. The measurement error $(E)$ is assumed to be normally distributed with mean $0$ feet and standard deviation $1.5$ feet.

(a) If the true depth at a certain point is $2$ feet, what is the probability that the depth measurement will be negative?

(b) Suppose three independent depth measurements are taken at the point where the true depth is $2$ feet. What is the probability that at least one of these measurements will be negative?

(c) What is the probability that the mean of the three independent depth measurements taken at the point where the true depth is $2$ feet will be negative?

Most-appropriate topic codes (AP Statistics):

• Topic 2.11 — The Normal Distribution (Part $\mathrm{a}$)
• Topic 2.7 — Independent Events and Unions of Events (Part $\mathrm{b}$)
• Topic 2.12 — Sampling Distributions and the Central Limit Theorem (Part $\mathrm{c}$)

▶️ Answer/Explanation

(a)

Since $M = D + E$ is a normal random variable plus a constant, $M$ is also normally distributed.
With true depth $D = 2$ feet, the distribution of $M$ has:
$\mu_M = 2\,\text{feet}, \qquad \sigma_M = 1.5\,\text{feet}$
We want $P(M < 0)$. Standardizing:
$P(M < 0) = P\!\left(Z < \frac{0 – 2}{1.5}\right) = P(Z < -1.33)$
Using the standard normal table:
$\boxed{P(M < 0) \approx 0.0918}$

(b)

Let each individual measurement be negative with probability $p = 0.0918$ (from part (a)), and let the three measurements be independent.

Using the complement rule — it is easier to find the probability that none of the three measurements is negative, then subtract from 1:

$P(\text{at least one negative}) = 1 – P(\text{none negative})$

$= 1 – (1 – 0.0918)^3$

$= 1 – (0.9082)^3$

$= 1 – 0.7491$

$\boxed{P(\text{at least one negative}) \approx 0.2509}$

(c)

Let $\bar{X}$ denote the mean of three independent depth measurements where the true depth is $2$ feet.
Since each measurement is normally distributed, the sampling distribution of $\bar{X}$ is also normal with:
$\mu_{\bar{X}} = 2\,\text{feet}, \qquad \sigma_{\bar{X}} = \frac{1.5}{\sqrt{3}} = 0.8660\,\text{feet}$
We want $P(\bar{X} < 0)$. Standardizing:
$P(\bar{X} < 0) = P\!\left(Z < \frac{0 – 2}{1.5/\sqrt{3}}\right) = P\!\left(Z < \frac{-2}{0.8660}\right) = P(Z < -2.31)$
Using the standard normal table:
$\boxed{P(\bar{X} < 0) \approx 0.0104}$

Question

Golf balls must meet a set of five standards in order to be used in professional tournaments. One of these standards is distance traveled. When a ball is hit by a mechanical device, Iron Byron, with a 10-degree angle of launch, a backspin of 42 revolutions per second, and a ball velocity of 235 feet per second, the distance the ball travels may not exceed 291.2 yards. Manufacturers want to develop balls that will travel as close to the 291.2 yards as possible without exceeding that distance. A particular manufacturer has determined that the distances traveled for the balls it produces are normally distributed with a standard deviation of 2.8 yards. This manufacturer has a new process that allows it to set the mean distance the ball will travel.

(a) If the manufacturer sets the mean distance traveled to be equal to 288 yards, what is the probability that a ball that is randomly selected for testing will travel too far?

(b) Assume the mean distance traveled is 288 yards and that five balls are independently tested. What is the probability that at least one of the five balls will exceed the maximum distance of 291.2 yards?

(c) If the manufacturer wants to be 99 percent certain that a randomly selected ball will not exceed the maximum distance of 291.2 yards, what is the largest mean that can be used in the manufacturing process?

Most-appropriate topic codes (AP Statistics):

• Topic 2.11 — The Normal Distribution (Parts $\mathrm{a}$, $\mathrm{c}$)
• Topic 2.10 — The Binomial Distribution (Part $\mathrm{b}$)
• Topic 2.7 — Independent Events and Unions of Events (Part $\mathrm{b}$)

▶️ Answer/Explanation

(a)
Let $D$ represent the distance a randomly selected ball travels. We are given that $D$ is normally distributed with mean $\mu = 288$ yards and standard deviation $\sigma = 2.8$ yards.
A ball travels “too far” if it exceeds 291.2 yards, so we need:
$P(D > 291.2)$
Converting to a $z$-score:
$z = \frac{291.2 – 288}{2.8} = \frac{3.2}{2.8} = 1.14$
$P(D > 291.2) = P(Z > 1.14) = 1 – P(Z \leq 1.14) = 1 – 0.8729 = 0.1271$
$\boxed{P(D > 291.2) \approx 0.1271}$

(b)
Since five balls are independently tested and the probability that any one ball exceeds 291.2 yards is $p = 0.1271$ (from part a), the number of balls exceeding the limit follows a binomial distribution with $n = 5$ and $p = 0.1271$.
Using the complement rule:
$P(\text{at least one} > 291.2) = 1 – P(\text{none} > 291.2)$
$= 1 – P(\text{all five} \leq 291.2)$
$= 1 – (1 – 0.1271)^5$
$= 1 – (0.8729)^5$
$= 1 – 0.5068$
$= 0.4932$
$\boxed{P(\text{at least one ball exceeds 291.2 yards}) \approx 0.4932}$

(c)
We want the manufacturer to be 99 percent certain that a randomly selected ball will not exceed 291.2 yards, meaning:
$P(D \leq 291.2) = 0.99$
The 99th percentile of the standard normal distribution corresponds to $z^* = 2.33$.
Setting up the equation with the unknown mean $\mu$:
$\frac{291.2 – \mu}{2.8} = 2.33$
Solving for $\mu$:
$291.2 – \mu = 2.33 \times 2.8 = 6.524$
$\mu = 291.2 – 6.524 = 284.676$
$\boxed{\mu = 284.676 \text{ yards}}$
In order to be 99 percent certain that a randomly selected ball will not exceed the maximum distance of 291.2 yards, the largest mean that can be used in the manufacturing process is 284.676 yards.

Question

At an archaeological site that was an ancient swamp, the bones from 20 brontosaur skeletons have been unearthed. The bones do not show any sign of disease or malformation. It is thought that these animals wandered into a deep area of the swamp and became trapped in the swamp bottom. The 20 left femur bones (thigh bones) were located and 4 of these left femurs are to be randomly selected without replacement for DNA testing to determine gender.

(a) Let $X$ be the number out of the 4 selected left femurs that are from males. Based on how these bones were sampled, explain why the probability distribution of $X$ is not binomial.

(b) Suppose that the group of 20 brontosaurs whose remains were found in the swamp had been made up of 10 males and 10 females. What is the probability that all 4 in the sample to be tested are male?

(c) The DNA testing revealed that all 4 femurs tested were from males. Based on this result and your answer from part (b), do you think that males and females were equally represented in the group of 20 brontosaurs stuck in the swamp? Explain.

(d) Is it reasonable to generalize your conclusion in part (c) pertaining to the group of 20 brontosaurs to the population of all brontosaurs? Explain why or why not.

Most-appropriate topic codes (AP Statistics):

• Topic 2.10 — The Binomial Distribution (Part a)
• Topic 2.7 — Independent Events and Unions of Events (Part a,b)
• Topic 2.4 — Introduction to Probability (Part b)
• Topic 2.3 — Estimating Probabilities Using Simulation (Part c)
•Topic 1.12 — Potential Problems with Sampling (Part d)

▶️ Answer/Explanation

(a)
The probability distribution of $X$ is not binomial because the bones are selected without replacement from a finite population of only 20 femurs. For a binomial distribution to apply, each trial must be independent — that is, the probability of success (selecting a male femur) must remain constant from one draw to the next. However, when sampling without replacement, the composition of the remaining pool changes with each selection, so the probability of drawing a male femur on each successive draw depends on what was drawn before it. Since the trials are not independent and the probability of success is not fixed, the distribution of $X$ is hypergeometric, not binomial.
$\boxed{X \text{ is not binomial because sampling is without replacement, making trials dependent}}$

(b)
With 10 males and 10 females among the 20 brontosaurs, compute the probability that all 4 selected femurs are male using the multiplication rule for dependent events (without replacement):
$P(\text{1st is male}) = \dfrac{10}{20}$
$P(\text{2nd is male} \mid \text{1st is male}) = \dfrac{9}{19}$
$P(\text{3rd is male} \mid \text{first two are male}) = \dfrac{8}{18}$
$P(\text{4th is male} \mid \text{first three are male}) = \dfrac{7}{17}$
Therefore:
$P(\text{all 4 are male}) = \dfrac{10}{20} \times \dfrac{9}{19} \times \dfrac{8}{18} \times \dfrac{7}{17}$
$= \dfrac{10 \times 9 \times 8 \times 7}{20 \times 19 \times 18 \times 17} = \dfrac{5040}{116280} \approx 0.0433$
This can also be expressed using combinations:
$P(\text{all 4 are male}) = \dfrac{\dbinom{10}{4}}{\dbinom{20}{4}} = \dfrac{210}{4845} \approx 0.0433$
$\boxed{P(\text{all 4 male}) \approx 0.0433}$

(c)
No, it does not seem likely that males and females were equally represented in the group of 20 brontosaurs. From part (b), if the group had exactly 10 males and 10 females, the probability of randomly selecting 4 males in a row is only about $4.33\%$. Since this probability is quite small (less than 5%), observing all 4 selected femurs being male is an unusual result under the assumption of equal representation. It is therefore more reasonable to think that males outnumbered females in this particular group of brontosaurs trapped in the swamp, though equal representation is possible — just unlikely given the data.
$\boxed{\text{Equal representation is unlikely; evidence suggests more males than females in the group}}$

(d)
No, it is not reasonable to generalize the conclusion from part (c) to the entire population of brontosaurs. The 20 brontosaurs found at the site do not constitute a random sample from the population of all brontosaurs — they represent only those individuals that happened to wander into that particular swamp and become trapped. This is a highly specific and non-random group. It is plausible that behavioral differences between male and female brontosaurs (for example, males may have been more likely to venture into deep swamp areas while foraging) could explain why males are overrepresented in this particular site. Such a non-representative sample cannot be used to draw conclusions about the broader population of all brontosaurs.
$\boxed{\text{Cannot generalize; the 20 brontosaurs are not a random sample of all brontosaurs}}$

Question

Two antibiotics are available as treatment for a common ear infection in children.

• Antibiotic A is known to effectively cure the infection 60 percent of the time. Treatment with antibiotic A costs $50.

• Antibiotic B is known to effectively cure the infection 90 percent of the time. Treatment with antibiotic B costs $80.

The antibiotics work independently of one another. Both antibiotics can be safely administered to children. A health insurance company intends to recommend one of the following two plans of treatment for children with this ear infection.

• Plan I: Treat with antibiotic A first. If it is not effective, then treat with antibiotic B.

• Plan II: Treat with antibiotic B first. If it is not effective, then treat with antibiotic A.

(a) If a doctor treats a child with an ear infection using plan I, what is the probability that the child will be cured?

If a doctor treats a child with an ear infection using plan II, what is the probability that the child will be cured?

(b) Compute the expected cost per child when plan I is used for treatment.

Compute the expected cost per child when plan II is used for treatment.

(c) Based on the results in parts (a) and (b), which plan would you recommend? Explain your recommendation.

Most-appropriate topic codes (AP Statistics):

• Topic 2.7 — Independent Events and Unions of Events (Part a)
• Topic 2.8 — Introduction to Random Variables and Probability Distributions (Part b)
• Topic 2.9 — Parameters of Random Variables (Parts b, c)

▶️ Answer/Explanation

(a)
Under Plan I, a child is cured if antibiotic A works, or if A fails but antibiotic B then works. Since the antibiotics act independently:
$P(\text{cured} \mid \text{Plan I}) = P(A \text{ cures}) + P(A \text{ fails}) \times P(B \text{ cures})$
$= 0.60 + (0.40)(0.90)$
$= 0.60 + 0.36$
$\boxed{P(\text{cured} \mid \text{Plan I}) = 0.96}$
Under Plan II, a child is cured if antibiotic B works, or if B fails but antibiotic A then works:
$P(\text{cured} \mid \text{Plan II}) = P(B \text{ cures}) + P(B \text{ fails}) \times P(A \text{ cures})$
$= 0.90 + (0.10)(0.60)$
$= 0.90 + 0.06$
$\boxed{P(\text{cured} \mid \text{Plan II}) = 0.96}$
Both plans result in the same probability of curing the infection: $0.96$.

(b)
For Plan I, there are two possible treatment scenarios:
— Antibiotic A cures the child (probability $0.60$): cost $= \$50$
— Antibiotic A fails, then antibiotic B is administered (probability $0.40$): cost $= \$50 + \$80 = \$130$
$E(\text{cost} \mid \text{Plan I}) = (0.60)(\$50) + (0.40)(\$130)$
$= \$30 + \$52$
$\boxed{E(\text{cost} \mid \text{Plan I}) = \$82}$
For Plan II, there are two possible treatment scenarios:
— Antibiotic B cures the child (probability $0.90$): cost $= \$80$
— Antibiotic B fails, then antibiotic A is administered (probability $0.10$): cost $= \$80 + \$50 = \$130$
$E(\text{cost} \mid \text{Plan II}) = (0.90)(\$80) + (0.10)(\$130)$
$= \$72 + \$13$
$\boxed{E(\text{cost} \mid \text{Plan II}) = \$85}$

(c)
Plan I is the better recommendation. Both plans produce identical probabilities of curing the infection ($0.96$), so neither plan has an advantage in terms of treatment effectiveness. However, Plan I has a lower expected cost per child ($\$82$) compared to Plan II ($\$85$). Since the insurance company is equally likely to achieve a cure under either plan but will spend $\$3$ less per child on average under Plan I, recommending Plan I is the more cost-effective choice.
$\boxed{\text{Recommend Plan I: same cure probability (0.96), lower expected cost (\$82 vs. \$85)}}$

Question

A simple random sample of adults living in a suburb of a large city was selected. The age and annual income of each adult in the sample were recorded. The resulting data are summarized in the table below.

(a) What is the probability that a person chosen at random from those in this sample will be in the 31–45 age category?

(b) What is the probability that a person chosen at random from those in this sample whose incomes are over $\$50{,}000$ will be in the 31–45 age category? Show your work.

(c) Based on your answers to parts (a) and (b), is annual income independent of age category for those in this sample? Explain.

Most-appropriate topic codes (AP Statistics):

• Topic 2.1 — Tabular and Graphical Representations for the Distributions of Two Categorical Variables (All parts)
• Topic 2.6 — Conditional Probability (Part b)
• Topic 2.7 — Independent Events and Unions of Events (Part c)
• Topic 2.2 — Summary Statistics for Two Categorical Variables (Part a)

▶️ Answer/Explanation

(a)

We want the probability that a randomly selected person from the sample falls in the 31–45 age category.
The total number of people in the sample is 207, and the number in the 31–45 age group is 89.
$ P(\text{age } 31\text{–}45) = \frac{89}{207} \approx 0.42995 $
$\boxed{P(\text{age } 31\text{–}45) \approx 0.4300}$

(b)

We now want the conditional probability that a person is in the 31–45 age category, given that their income is over $\$50{,}000$.
From the table, the total number of people with income over $\$50{,}000$ is 96, and among those, 35 are in the 31–45 age group.
$ P(\text{age } 31\text{–}45 \mid \text{income over } \$50{,}000) = \frac{35}{96} \approx 0.36458 $
$\boxed{P(\text{age } 31\text{–}45 \mid \text{income over } \$50{,}000) \approx 0.3646}$

(c)

For two variables to be independent, knowing the value of one variable should not change the probability of the other — in other words, the marginal probability and the conditional probability must be equal.
From part (a), $P(\text{age } 31\text{–}45) \approx 0.4300$, and from part (b), $P(\text{age } 31\text{–}45 \mid \text{income over } \$50{,}000) \approx 0.3646$.
Since these two probabilities are not equal ($0.4300 \neq 0.3646$), knowing a person’s income category does change the probability of being in the 31–45 age group.
Therefore, annual income and age category are not independent for those in this sample.

Question

Contestants on a game show spin a wheel like the one shown in the figure above. Each of the four outcomes on this wheel is equally likely and outcomes are independent from one spin to the next.

The contestant spins the wheel.
If the result is a skunk, no money is won and the contestant’s turn is finished.
If the result is a number, the corresponding amount in dollars is won. The contestant can then stop with those winnings or can choose to spin again, and his or her turn continues.
If the contestant spins again and the result is a skunk, all of the money earned on that turn is lost and the turn ends.
The contestant may continue adding to his or her winnings until he or she chooses to stop or until a spin results in a skunk.

(a) What is the probability that the result will be a number on all of the first three spins of the wheel?

(b) Suppose a contestant has earned $\$800$ on his or her first three spins and chooses to spin the wheel again. What is the expected value of his or her total winnings for the four spins?

(c) A contestant who lost at this game alleges that the wheel is not fair. In order to check on the fairness of the wheel, the data in the table below were collected for 100 spins of this wheel.

Based on these data, can you conclude that the four outcomes on this wheel are not equally likely? Give appropriate statistical evidence to support your answer.

Most-appropriate topic codes (AP Statistics):

• Topic 2.7 — Independent Events and Unions of Events (Part a)
• Topic 2.9 — Parameters of Random Variables (Part b)
• Topic 3.14 — Setting Up a Chi-Square Test for Homogeneity or Independence (Part c)
• Topic 3.15 — Carrying Out a Chi-Square Test for Homogeneity or Independence (Part c)

▶️ Answer/Explanation

(a)

There are four equally likely outcomes on the wheel: Skunk, $\$100$, $\$200$, and $\$500$. So the probability of landing on a number (i.e., not a skunk) on any single spin is $\dfrac{3}{4}$.
Since spins are independent, the probability of getting a number on all three of the first three spins is:
$ P(\text{number on all 3 spins}) = \left(\frac{3}{4}\right)^3 = \frac{27}{64} \approx 0.4219 $
$\boxed{P \approx 0.4219}$

(b)

The contestant currently has $\$800$ and chooses to spin a fourth time. The four equally likely outcomes on the fourth spin lead to the following total winnings:

The expected value of total winnings is:
$ E(\text{total winnings}) = 0\left(\frac{1}{4}\right) + 900\left(\frac{1}{4}\right) + 1000\left(\frac{1}{4}\right) + 1300\left(\frac{1}{4}\right) $
$ = \frac{0 + 900 + 1000 + 1300}{4} = \frac{3200}{4} = \$800 $
Alternatively, the expected gain from the fourth spin alone is:
$ E(\text{4th spin gain}) = (-800)\left(\frac{1}{4}\right) + 100\left(\frac{1}{4}\right) + 200\left(\frac{1}{4}\right) + 500\left(\frac{1}{4}\right) = \frac{-800+100+200+500}{4} = 0 $
So the expected total winnings $= \$800 + \$0 = \boxed{\$800}$.
Interestingly, the expected value of spinning again is exactly equal to the amount already won — so on average the fourth spin neither helps nor hurts.

(c)

Hypotheses:
$ H_0: p_1 = p_2 = p_3 = p_4 = \frac{1}{4} \quad \text{(all four outcomes are equally likely)} $
$ H_a: \text{at least one } p_i \neq \frac{1}{4} \quad \text{(the four outcomes are not equally likely)} $
Test: Chi-square goodness-of-fit test.
Conditions: The spins are independent (stated in the problem), and the expected count for each outcome is $100 \times \frac{1}{4} = 25 > 5$, so the sample size is large enough to proceed.
Expected counts: 25 for each of the four outcomes.
Test statistic:
$ \chi^2 = \sum \frac{(\text{Observed} – \text{Expected})^2}{\text{Expected}} $
$ = \frac{(33-25)^2}{25} + \frac{(21-25)^2}{25} + \frac{(20-25)^2}{25} + \frac{(26-25)^2}{25} $
$ = \frac{64}{25} + \frac{16}{25} + \frac{25}{25} + \frac{1}{25} = \frac{106}{25} = 4.24 $
Degrees of freedom: $df = 4 – 1 = 3$
P-value: $p\text{-value} \approx 0.237$ (from chi-square table with $df = 3$, the test statistic of 4.24 falls well below the critical value of 7.81 at $\alpha = 0.05$).
Conclusion: Since the $p\text{-value} \approx 0.237 > 0.05$, we fail to reject $H_0$. There is not convincing statistical evidence that the four outcomes on the wheel are not equally likely — the data are consistent with a fair wheel.

AP Statistics 2.7 Independent Events and Unions of Events- Exam Style Questions - FRQs - New Syllabus

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