Unit 2 test

AP Statistics Unit 2 Test

15 multiple-choice questions and 1 free-response question, every one from Unit 2: Probability, Random Variables, and Probability Distributions, written in the format of the AP Statistics Exam.

Take the multiple choice on its own, timed at the exam's pace or in practice mode with each answer checked as you give it. Or sit the whole test, free response included, with each section on its own clock; the free-response scoring guide opens when that section ends.

AP Statistics Unit 2 Test

15 multiple-choice and 1 free-response question, a shorter set laid out like the AP Statistics Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time
Section I: Multiple Choice	15 of the exam's 42	32 minutes 9 seconds
Section II: Free Response	1 of the exam's 4	22 minutes 30 seconds

A graphing calculator with statistical capabilities is expected on both sections.

One section at a time. When a section's clock reaches zero, the section ends and is scored.
Multiple choice is scored when its section ends, with an explanation for every choice.
Free response is never scored by the page. When the section ends, you see the scoring guide and a model response for each part and tick the points your answer earns.
At the end you get your score and the time you used for each section.
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Time for each section Standard time Time and a half Double time Untimed Study mode

Choose before Section I starts. Time and a half and double time match common AP testing accommodations; untimed shows the time you use without ending a section. Study mode is untimed and shows each answer as you go.

Section I: Multiple Choice

15 questions · 32 minutes 9 seconds

A regional transit agency checked a random sample of 400 trips from last month. Of the 300 bus trips in the sample, 75 arrived late, and of the 100 light rail trips, 25 arrived late. An analyst makes a mosaic plot of the data with one bar for each type of vehicle, divided into a late segment and a not-late segment. Which describes the mosaic plot?
Answer: A.
1. A
  Correct. In a mosaic plot each bar's width shows how many trips are in that group, and the bus trips are $300 / 400 = 0.75$ of the sample against $100 / 400 = 0.25$ for light rail. Within each bar the segments show proportions: $75 / 300 = 0.25$ of bus trips and $25 / 100 = 0.25$ of light rail trips were late, so the late segments are the same height.
2. B
  This draws the counts, 75 and 25, as segment heights. The heights in a mosaic plot are proportions within each bar, and both are 0.25. The widths are where the group sizes show up, so the bars are not equally wide.
3. C
  That describes a segmented bar chart, where every bar has the same width. A mosaic plot makes each bar's width match its share of the data, so the bus bar is three times as wide as the light rail bar.
4. D
  This counts the group sizes twice. The wider bus bar already shows that there are three times as many bus trips. The heights are the proportions late, 0.25 for both, so the late segments are equally tall.
A factory that makes ceramic floor tiles inspected all 8,000 tiles it made on one day. The table shows the result of each inspection by the shift that made the tile.
$\begin{matrix} Passed & Sold as seconds & Discarded & Total \\ Day shift & 4,300 & 500 & 200 & 5,000 \\ Night shift & 2,250 & 450 & 300 & 3,000 \\ Total & 6,550 & 950 & 500 & 8,000 \end{matrix}$
What proportion of the tiles made by the night shift were discarded?
Answer: B.
1. A
  0.038 is $300 / 8,000$ , a joint relative frequency: the share of all the day's tiles that were both made at night and discarded. The question restricts to night-shift tiles, so divide by the night-shift total.
2. B
  Correct. Restrict to the night-shift row. Of its 3,000 tiles, 300 were discarded, so the conditional relative frequency is $300 / 3,000 = 0.100$ .
3. C
  0.375 is $3,000 / 8,000$ , the share of all tiles that the night shift made. That is a marginal relative frequency, and it ignores whether the tiles were discarded.
4. D
  0.600 is $300 / 500$ , the share of discarded tiles that came from the night shift. That conditions on the wrong variable: it divides by the column total instead of the row total.
A darts player hits the inner ring she aims at on 70% of her throws. To estimate the probability that she hits it on at least 4 of her next 5 throws, her coach ran a simulation of 200 trials, each one a set of 5 simulated throws. The table shows how many trials gave each number of hits.
$\begin{matrix} Hits in 5 throws & 0 & 1 & 2 & 3 & 4 & 5 & Total \\ Number of trials & 1 & 6 & 25 & 62 & 71 & 35 & 200 \end{matrix}$
Based on the simulation, what is the estimated probability that she hits the inner ring on at least 4 of her next 5 throws?
Answer: D.
1. A
  0.175 is $35 / 200$ , the trials with all 5 hits. That is more than 4 hits. At least 4 also counts the trials with exactly 4.
2. B
  0.355 is $71 / 200$ , the trials with exactly 4 hits. At least 4 means 4 or 5, so the 35 trials with 5 hits belong in the count too.
3. C
  0.470 is $(1 + 6 + 25 + 62) / 200$ , the trials with 3 or fewer hits. That is the complement of the event asked about.
4. D
  Correct. The trials with at least 4 hits are the 71 with 4 and the 35 with 5. The relative frequency $(71 + 35) / 200 = 106 / 200 = 0.530$ estimates the probability.
In a board game, a player rolls two fair four-sided dice, each numbered 1 to 4, and moves forward the sum of the two numbers. What is the probability that on one roll the player moves forward at least 4 spaces?
Answer: C.
1. A
  0.188 is $3 / 16$ , the probability of a sum of 2 or 3. That is the complement of the event asked about, so subtract it from 1.
2. B
  0.714 is $5 / 7$ , which treats the seven possible sums, 2 through 8, as equally likely. They are not: a sum of 5 can happen four ways and a sum of 2 only one way. The equally likely outcomes are the 16 ordered pairs of rolls.
3. C
  Correct. The two dice give $4 \times 4 = 16$ equally likely ordered outcomes. Only (1, 1), (1, 2) and (2, 1) have a sum less than 4, so $P (at least 4) = 1 - 3 / 16 = 13 / 16 \approx 0.813$ .
4. D
  0.875 is $1 - 2 / 16$ , which counts (1, 2) and (2, 1) as one outcome. They are different outcomes, since either die can show the 1, so 3 of the 16 outcomes have a sum less than 4.
A car rental company finds that for a randomly selected rental, the probability that the car is returned late is 0.12, the probability that it is returned with new damage is 0.05, and the probability that it is returned both late and with new damage is 0.01. Are the events "returned late" and "returned with new damage" mutually exclusive?
Answer: A.
1. A
  Correct. Two events are mutually exclusive when they cannot happen together, which means their joint probability is 0. Here $P (late \cap damaged) = 0.01$ , so some rentals are both late and damaged, and the events are not mutually exclusive.
2. B
  Comparing the joint probability with $0.12 \times 0.05 = 0.006$ is a check for independence, not for mutual exclusivity. Mutually exclusive events have a joint probability of exactly 0.
3. C
  A sum below 1 says nothing about overlap. Mutual exclusivity is decided by the joint probability, and here it is 0.01, not 0.
4. D
  Small is not the same as zero. Even a joint probability of 0.01 means that about 1 rental in 100 is both late and damaged, so the two events can happen together.
On a commuter rail line, the probability that a randomly selected morning train leaves its first station late is 0.25, and the probability that it arrives downtown late is 0.30. The probability that it both leaves late and arrives late is 0.20. A morning train has just left its first station late. What is the probability that it arrives downtown late?
Answer: D.
1. A
  0.20 is the probability that a train both leaves late and arrives late, out of all morning trains. The train is already known to have left late, so the probability has to be found among late-leaving trains only.
2. B
  0.35 is $0.25 + 0.30 - 0.20$ , the probability that a train leaves late or arrives late. That is a union, not a conditional probability.
3. C
  0.67 is $0.20 / 0.30$ , the probability that a train left late given that it arrived late. The condition is reversed: here the train is known to have left late.
4. D
  Correct. $P (arrives late | leaves late) = \frac{P (both)}{P (leaves late)} = \frac{0.20}{0.25} = 0.80$ .
A highway over a mountain pass is closed on 80% of winter mornings that follow a night with snow and on 10% of winter mornings that follow a night without snow. Snow falls on 35% of winter nights at the pass. What is the probability that the highway is closed on a randomly selected winter morning?
Answer: B.
1. A
  0.280 is $0.35 (0.80)$ , the probability of snow and a closure. The highway can also be closed after a night without snow, and that path has to be added.
2. B
  Correct. By the multiplication rule, $P (snow \cap closed) = 0.35 (0.80) = 0.280$ and $P (no snow \cap closed) = 0.65 (0.10) = 0.065$ . The two paths cannot both happen, so $P (closed) = 0.280 + 0.065 = 0.345$ .
3. C
  0.450 is the plain average of 0.80 and 0.10. That would be right only if half of the nights had snow. Each conditional probability has to be weighted by how often its condition happens: 0.35 and 0.65.
4. D
  0.900 adds the two conditional probabilities. They describe different groups of mornings, so they cannot be added; each has to be multiplied by the probability of its condition first.
At a hardware store's registers, 30% of customers scan a loyalty card and 45% pay with a phone. Among the customers who pay with a phone, 30% scan a loyalty card. What is the probability that a randomly selected customer scans a loyalty card or pays with a phone, or both?
Answer: C.
1. A
  0.135 is $0.30 (0.45)$ , the probability that a customer does both. The question asks for either one or both, which is the union.
2. B
  0.450 is $0.30 + 0.45 - 0.30$ , which subtracts the conditional probability 0.30 in place of the joint probability. The overlap to subtract is $P (card \cap phone) = 0.45 (0.30) = 0.135$ .
3. C
  Correct. Knowing that a customer pays with a phone does not change the chance of scanning a card (0.30 either way), so the events are independent and $P (card \cap phone) = 0.30 (0.45) = 0.135$ . Then $P (card \cup phone) = 0.30 + 0.45 - 0.135 = 0.615$ .
4. D
  0.750 is $0.30 + 0.45$ , which counts the customers who do both twice. Subtract the overlap, 0.135, once.
Questions 9 to 11 refer to the following.
An online garden supply store ships orders of seeds, tools and plants. For a randomly selected order, let $X$ be the number of items in the order that turn out to be out of stock after the order is placed. The probability distribution of $X$ is shown. Orders are independent of one another.
$\begin{matrix} x & 0 & 1 & 2 & 3 \\ P (X = x) & 0.55 & 0.25 & 0.15 & 0.05 \end{matrix}$

Which list gives the cumulative probability distribution of $X$ for $x = 0, 1, 2$ and 3, in that order?
Answer: C.
1. A
  These are $P (X < x)$ , which leave out the value itself. A cumulative distribution gives $P (X \leq x)$ , so its first entry is $P (X = 0) = 0.55$ , not 0.
2. B
  These are the probabilities of the single values, the distribution as given. A cumulative distribution adds them up as $x$ increases.
3. C
  Correct. A cumulative distribution gives $P (X \leq x)$ : $0.55$ , then $0.55 + 0.25 = 0.80$ , then $0.80 + 0.15 = 0.95$ , then $0.95 + 0.05 = 1.00$ . The last entry is 1 because every order has at most 3 out-of-stock items.
4. D
  These are $P (X \geq x)$ , which add from the top down. A cumulative distribution adds from the smallest value up, $P (X \leq x)$ .
Uses the material above question 9.

For these orders, $μ_{X} = 0.70$ and $σ_{X} = 0.90$ . What does the value 0.90 tell the store?
Answer: A.
1. A
  Correct. The standard deviation of a random variable is the typical distance of its values from the mean over the long run. Over many orders, the number of out-of-stock items is typically about 0.90 away from 0.70.
2. B
  The 68% figure comes from the empirical rule, which describes normal distributions. This distribution is discrete and skewed right. The values within 0.90 of 0.70 are 0 and 1, and $P (X = 0) + P (X = 1) = 0.80$ , not 0.68.
3. C
  0.90 is a parameter of the whole distribution. The standard deviation of a particular sample of orders is a statistic, and it varies from sample to sample around a value near 0.90.
4. D
  This reads the standard deviation as a percent. The orders with no more than $0.70 + 0.90 = 1.6$ out-of-stock items are those with 0 or 1, and they make up $0.55 + 0.25 = 0.80$ of orders.
Uses the material above question 9.

The store's next 6 orders are filled. What is the probability that exactly 2 of these 6 orders have at least one out-of-stock item?
Answer: B.
1. A
  0.019 is $(0.45)^{2} (0.55)^{4}$ , the probability of one particular order of results, such as the first two orders. Exactly 2 of 6 can happen in $(\binom{6}{2}) = 15$ different orders, so multiply by 15.
2. B
  Correct. An order has at least one out-of-stock item with probability $p = 1 - 0.55 = 0.45$ , and orders are independent, so the count is binomial with $n = 6$ and $p = 0.45$ . Then $P (exactly 2) = (\binom{6}{2}) (0.45)^{2} (0.55)^{4} \approx 0.278$ .
3. C
  0.297 uses $p = 0.25$ , the probability of exactly one out-of-stock item. An order counts if it has one or more, so $p = 1 - P (X = 0) = 0.45$ .
4. D
  0.442 is $P (at most 2)$ , which adds the probabilities of 0, 1 and 2 such orders. The question asks for exactly 2.
A store receives a crate of 20 car batteries, and 4 of the batteries are dead. A clerk selects 5 batteries from the crate at random, without replacement, and tests each one. Let $X$ be the number of dead batteries among the 5 tested. Is $X$ a binomial random variable?
Answer: D.
1. A
  Two outcomes and a fixed number of trials are only part of the setting. The trials must also be independent with the same probability of success, and here they are not: after a dead battery is drawn, only 3 of the remaining 19 are dead.
2. B
  Random selection does not make the trials independent. Drawing without replacement from a small crate changes the chance of a dead battery after every draw.
3. C
  Ten successes and ten failures is a condition for using a normal model, not part of the definition of a binomial random variable. A binomial count can have any expected number of successes.
4. D
  Correct. The first battery is dead with probability $4 / 20 = 0.20$ , but the second is dead with probability $3 / 19 \approx 0.158$ if the first was dead and $4 / 19 \approx 0.211$ if it was not. The trials are not independent, so $X$ is not binomial. The 5 batteries are 25% of the crate, far more than 10%, so the dependence cannot be ignored.
A machine cuts wooden dowels whose lengths are approximately normal with mean 30.00 centimeters and standard deviation 0.08 centimeter. A dowel is rejected if it is shorter than 29.90 centimeters or longer than 30.10 centimeters. What proportion of the dowels the machine cuts are rejected?
Answer: B.
1. A
  0.106 is the area in one tail only, the dowels longer than 30.10 centimeters. Dowels shorter than 29.90 centimeters are rejected too, and by symmetry that tail has the same area.
2. B
  Correct. The limits are $z = \frac{30.10 - 30.00}{0.08} = 1.25$ and $z = - 1.25$ . Each tail beyond them has area about 0.106, so the proportion rejected is about $2 (0.106) \approx 0.211$ .
3. C
  0.789 is the area between 29.90 and 30.10 centimeters, the proportion of dowels that are accepted. The question asks for the dowels outside those limits.
4. D
  0.894 is the area to the left of 30.10 centimeters. It includes the short dowels, which are rejected, and leaves out the long ones, which are also rejected.
At a state high school track meet, the girls' 100-meter dash times were approximately normal with mean 13.1 seconds and standard deviation 0.5 second. The girls' long jump distances were approximately normal with mean 4.9 meters and standard deviation 0.6 meter. Maya ran the 100 meters in 12.2 seconds, and Ana long jumped 5.9 meters. Which athlete had the better performance relative to the other girls in her event?
Answer: A.
1. A
  Correct. Maya's z-score is $\frac{12.2 - 13.1}{0.5} = - 1.80$ , and in a race a lower time is better, so she ran faster than about 96.4% of the girls. Ana's z-score is $\frac{5.9 - 4.9}{0.6} \approx 1.67$ , so she jumped farther than about 95.2% of the girls. Maya's performance is the more unusual one.
2. B
  A z-score has to be read in the direction that is better for the event. In the 100 meters a negative z-score means a faster time, so Maya's $- 1.80$ is 1.80 standard deviations better than the mean, which beats Ana's 1.67.
3. C
  Seconds and meters cannot be compared directly. Each difference has to be measured in its own event's standard deviations: Maya's 0.9 second is 1.80 standard deviations, and Ana's 1.0 meter is about 1.67.
4. D
  Standardizing is what makes the comparison possible. Converting each result to a z-score, or a percentile, puts both events on the same scale.
A roadside assistance service records how long each driver who calls waits for a tow truck. Last year's wait times are strongly skewed to the right, with mean 38 minutes and standard deviation 22 minutes. Treating last year's calls as the population, an analyst simulates 1,000 random samples of 40 calls and records the mean wait time of each sample. Which best describes the distribution of the 1,000 sample means?
Answer: C.
1. A
  This describes the population of wait times, not the sample means. A mean of 40 waits averages out the occasional very long wait, so the means are neither strongly skewed nor as spread out as single waits.
2. B
  The shape is right but the spread is not. Each sample mean averages 40 waits, so the means vary much less from sample to sample than individual waits vary from driver to driver.
3. C
  Correct. By the central limit theorem, the means of random samples of 40 have a distribution that is approximately normal even though the waits are skewed. The means center near the population mean of 38 minutes, and because each one averages 40 waits, they are much less spread out than individual waits.
4. D
  Less spread is right, but the shape is not inherited from the population. The central limit theorem says that the means of samples this large are approximately normal even when the population is strongly skewed.

Section II: Free Response

1 question · 22 minutes 30 seconds

Question 2: Multi-Focus on Practices 3 and 4 10 points

At a family arcade, a player of a ball-rolling game rolls 3 balls up a ramp toward a target. Each ball lands in the small top ring with probability 0.2, independently of the other balls. The machine pays prize tickets based on how many of the 3 balls land in the top ring. Let $T$ be the number of tickets paid for a randomly selected play. One probability is missing from the table.

Tickets paid for one play of the ball-rolling game
Balls in the top ring 0 1 2 3

Tickets paid, $t$ 5 10 30 100
$P (T = t)$ 0.512 ? 0.096 0.008

Use the given information to respond to parts A, B, C, D, and E. Label any subparts (e.g., i and ii) that may be present.
1. A.
  
  i. Show that the probability that exactly 1 of the 3 balls lands in the top ring is 0.384.
  
  ii. Calculate the probability that a randomly selected play pays at least 30 tickets.
  Question 2, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Shows that exactly 1 ball in the top ring can happen in 3 ways (the first, second or third ball), each with probability $0.2 (0.8) (0.8) = 0.128$ by independence, so $P (T = 10) = 3 (0.128) = 0.384$ . The binomial formula $(\binom{3}{1}) (0.2)^{1} (0.8)^{2}$ also earns the point, as does $1 - (0.512 + 0.096 + 0.008) = 0.384$ with the reason that the probabilities add to 1.Earns it: "The one hit can be ball 1, 2 or 3. Each way has probability $0.2 \times 0.8 \times 0.8 = 0.128$ , so the total is $3 \times 0.128 = 0.384$ ."; "The probabilities must add to 1: $1 - 0.512 - 0.096 - 0.008 = 0.384$ ."Does not: " $0.2 (0.8) (0.8) = 0.128$ " (one particular order only; there are three orders in which exactly one ball lands in the ring); "0.384" with no work (the question asks the response to show where the value comes from)
  - 1 point Calculates $P (T \geq 30) = P (T = 30) + P (T = 100) = 0.096 + 0.008 = 0.104$ .Earns it: " $P (T \geq 30) = 0.096 + 0.008 = 0.104$ "Does not: "0.008" (only the 100-ticket outcome; at least 30 includes 30); " $1 - 0.096 = 0.904$ " (subtracts the wrong probability from 1)
  Model response
  i. Exactly one ball lands in the top ring if the first ball does and the other two do not, or the second does and the others do not, or the third does and the others do not. By independence each of these has probability $0.2 (0.8) (0.8) = 0.128$ , and they cannot happen together, so $P (T = 10) = 3 (0.128) = 0.384$ .
  
  ii. $P (T \geq 30) = P (T = 30) + P (T = 100) = 0.096 + 0.008 = 0.104$ .
2. B.
  
  i. Calculate the mean of $T$ . Show your work.
  
  ii. Interpret the mean of $T$ in context.
  Question 2, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Calculates $μ_{T} = 5 (0.512) + 10 (0.384) + 30 (0.096) + 100 (0.008) = 10.08$ tickets, with work shown. A mean consistent with an incorrect probability from part A also earns the point.Earns it: " $μ_{T} = 2.56 + 3.84 + 2.88 + 0.80 = 10.08$ tickets"Does not: " $\frac{5 + 10 + 30 + 100}{4} = 36.25$ " (treats the four ticket amounts as equally likely); "10.08" with no work shown
  - 1 point Interprets the mean as a long-run average in context: over many plays, the game pays an average of about 10.08 tickets per play. The response must refer to the long run (or to many plays).Earns it: "If the game is played many, many times, the average payout will be about 10.08 tickets per play."Does not: "A player usually gets 10.08 tickets." (no play pays 10.08 tickets, and most plays pay 5; the mean is a long-run average); "The expected value is 10.08 tickets." (restates the value without interpreting it)
  Model response
  i. $μ_{T} = 5 (0.512) + 10 (0.384) + 30 (0.096) + 100 (0.008) = 2.56 + 3.84 + 2.88 + 0.80 = 10.08$ tickets.
  
  ii. Over a large number of plays, the game pays an average of about 10.08 tickets per play. No single play pays 10.08 tickets; that is the long-run average.
3. C.
  
  Calculate the standard deviation of $T$ . Show your work.
  Question 2, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Calculates $σ_{T} = \sqrt{\sum (t_{i} - 10.08)^{2} P (t_{i})} = \sqrt{115.9936} \approx 10.77$ tickets, with work shown. The shortcut $\sqrt{E (T^{2}) - μ_{T}^{2}} = \sqrt{217.6 - {10.08}^{2}}$ also earns the point, as does technology output with the values and probabilities identified. A value consistent with an incorrect mean from part B earns the point if the work is shown.Earns it: " $σ_{T} = \sqrt{(5 - 10.08)^{2} (0.512) + (10 - 10.08)^{2} (0.384) + (30 - 10.08)^{2} (0.096) + (100 - 10.08)^{2} (0.008)} = \sqrt{116.0} \approx 10.77$ "Does not: "116.0" (that is the variance; the standard deviation is its square root); "About 43.85, the standard deviation of the four numbers 5, 10, 30 and 100" (does not weight each value by its probability)
  Model response
  $σ_{T} = \sqrt{(5 - 10.08)^{2} (0.512) + (10 - 10.08)^{2} (0.384) + (30 - 10.08)^{2} (0.096) + (100 - 10.08)^{2} (0.008)}$
  
  $= \sqrt{115.9936} \approx 10.77$ tickets.
4. D.
  
  Call a play a big win if it pays at least 30 tickets. A regular player says she had 4 big wins in her last 10 plays, and she thinks the machine has become easier than the arcade describes. Assume that plays are independent and that the probabilities in the table are correct. Let $Y$ be the number of big wins in 10 plays.
  
  i. Explain why $Y$ has a binomial distribution, and identify the values of $n$ and $p$ .
  
  ii. Calculate $P (Y \geq 4)$ .
  
  iii. Based on your answer to (ii), does the player's result give a reason to doubt that the machine works as the arcade describes? Explain.
  Question 2, part D: your response. It stays in this browser tab and is never sent anywhere.
  Score part D
  - 1 point Explains that $Y$ counts the successes (big wins) in a fixed number of independent trials (10 plays), each with two outcomes (big win or not) and the same probability of success, and identifies $n = 10$ and $p = 0.104$ (or the value from part A (ii)).Earns it: "Each of the 10 plays is a big win or not, the plays are independent, and each has the same probability 0.104 of being a big win. $Y$ counts big wins, so it is binomial with $n = 10$ and $p = 0.104$ ."Does not: "Two outcomes, $n = 10$ , $p = 0.104$ ." (does not mention independence or the same probability on every play); " $n = 10$ and $p = 0.2$ " (0.2 is the probability for one ball, not for a big win on a play)
  - 1 point Calculates $P (Y \geq 4) = \sum_{k = 4}^{10} (\binom{10}{k}) (0.104)^{k} (0.896)^{10 - k} = 1 - P (Y \leq 3) \approx 0.0147$ , with the distribution and values identified (for example, binomial with $n = 10$ , $p = 0.104$ , at least 4).Earns it: " $P (Y \geq 4) = 1 - P (Y \leq 3) = 1 - 0.9853 = 0.0147$ , binomial with $n = 10$ , $p = 0.104$ "Does not: " $P (Y = 4) = 0.0127$ " (the probability of exactly 4, not 4 or more); " $1 - P (Y \leq 4) = 0.0020$ " (leaves out $Y = 4$ ); "0.0147" with no indication of the distribution or values used
  - 1 point Answers yes and explains using the probability: if the machine works as described, 4 or more big wins in 10 plays would happen only about 1.5% of the time, which is unusual, so the result gives a reason to doubt the arcade's description. A conclusion consistent with an incorrect probability in (ii) earns the point if it is justified by that probability.Earns it: "Yes. If the probabilities are right, 4 or more big wins in 10 plays happens only about 1.5% of the time. That is rare, so her result makes me doubt the machine works as described."Does not: "Yes, because 4 is more than the mean of 1.04." (being above the mean is not enough; the response must use the probability); "No, because 0.0147 is the probability that the machine works as described." (the probability was found by assuming the machine works as described; it is not the probability of that)
  Model response
  i. Each of the 10 plays is either a big win or not. The plays are independent, and each has the same probability, $P (T \geq 30) = 0.104$ , of being a big win. $Y$ counts the big wins in a fixed number of plays, so $Y$ is binomial with $n = 10$ and $p = 0.104$ .
  
  ii. $P (Y \geq 4) = \sum_{k = 4}^{10} (\binom{10}{k}) (0.104)^{k} (0.896)^{10 - k} = 1 - P (Y \leq 3) \approx 1 - 0.9853 = 0.0147$ .
  
  iii. Yes. If the machine works as the arcade describes, a player gets 4 or more big wins in 10 plays only about 1.5% of the time. That is unusual enough to give a reason to doubt the description: big wins may be more likely now than 0.104.
5. E.
  
  The time one play takes, from the first roll until the last ball stops, is approximately normal with mean 40 seconds and standard deviation 6 seconds.
  
  i. Calculate the probability that a randomly selected play takes more than 50 seconds.
  
  ii. Calculate the time that separates the quickest 10% of plays from the rest.
  Question 2, part E: your response. It stays in this browser tab and is never sent anywhere.
  Score part E
  - 1 point Calculates $P (time > 50)$ for a normal distribution with mean 40 and standard deviation 6: $z = \frac{50 - 40}{6} \approx 1.67$ , so the probability is about 0.048. The distribution, its parameters and the boundary must be identified, for example in a labeled sketch or in calculator notation with labeled values. A table value from $z = 1.67$ , 0.0475, also earns the point.Earns it: " $z = (50 - 40) / 6 = 1.67$ , so $P (Z > 1.67) = 0.048$ ."; "normal, mean 40, SD 6, area above 50 is 0.048"Does not: "0.952" (the probability that a play takes less than 50 seconds); "0.048" with no indication of the distribution, the parameters or the boundary
  - 1 point Calculates the 10th percentile of the distribution: $40 - 1.28 (6) \approx 32.3$ seconds, using $z \approx - 1.28$ for an area of 0.10 to the left. The work must show the z-score or identify the distribution and the area.Earns it: " $z = - 1.28$ , so the time is $40 - 1.28 (6) = 32.3$ seconds."Does not: " $40 + 1.28 (6) = 47.7$ seconds" (that is the 90th percentile, which separates the slowest 10% of plays); " $40 - 2 (6) = 28$ seconds" (the empirical rule gives the bottom 2.5%, not the bottom 10%)
  Model response
  i. The time for a play is normal with mean 40 seconds and standard deviation 6 seconds. $z = \frac{50 - 40}{6} \approx 1.67$ , so $P (time > 50) = P (Z > 1.67) \approx 0.048$ .
  
  ii. The quickest 10% of plays lie below the 10th percentile. The z-score with an area of 0.10 to its left is about $- 1.28$ , so the time is $40 - 1.28 (6) \approx 32.3$ seconds. About 10% of plays take less than 32.3 seconds.
Your score: 0 of 10, from the boxes you ticked.

Tickets paid for one play of the ball-rolling game
Balls in the top ring	0	1	2	3
Tickets paid, $t$	5	10	30	100
$P (T = t)$	0.512	?	0.096	0.008

AP Statistics Unit 2 Test

Section I: Multiple Choice

Section II: Free Response

Question 2: Multi-Focus on Practices 3 and 4 10 points

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