What is Probability & stats on the Algebra 2 Regents?

Probability & stats is a Statistics & Probability topic on the Algebra 2 Regents. RegentsLab has 31 practice questions on it, each with a full worked solution.

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Probability & stats: Algebra 2 Regents Practice

Practice Probability & stats for the Algebra 2 Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 97 · 2 pts

A survey of 80 students recorded whether they play a school sport and whether they made the honor roll, as shown. | | Plays a sport | No sport | Total | | --- | --- | --- | --- | | Honor roll | 18 | 12 | 30 | | Not on honor roll | 32 | 18 | 50 | | Total | 50 | 30 | 80 | To the nearest hundredth, what is the probability that a randomly selected student made the honor roll, given that the student plays a sport?

(A) 0.23
(B) 0.60
(C) 0.36 ✓
(D) 0.63

Worked solution

$P(\text{honor roll} \mid \text{sport}) = \dfrac{\text{honor roll and sport}}{\text{total who play a sport}}$ A conditional probability restricts the sample space to the given event.

$= \dfrac{18}{50}$ 18 students play a sport and made honor roll; 50 students play a sport.

$= 0.36$

Answer: 0.36

A conditional probability $P(A\mid B)$ uses only the outcomes where $B$ occurs as the denominator, so here the sample space shrinks to the 50 students who play a sport.

Question 99 · 2 pts

A set of data is normally distributed with a mean of 64 and a standard deviation of 8. What is the $z$-score of a data value of 76?

(A) -1.5
(B) 1.5 ✓
(C) 0.19
(D) 12

Worked solution

$z = \dfrac{x - \mu}{\sigma}$ The z-score formula measures distance from the mean in standard deviations.

$= \dfrac{76 - 64}{8} = \dfrac{12}{8}$

$= 1.5$

Answer: 1.5

A $z$-score tells how many standard deviations a value lies from the mean, computed as $z=\frac{x-\mu}{\sigma}$.

Question 100 · 2 pts

For two events, $P(A)=0.45$, $P(B)=0.35$, and $P(A \text{ and } B)=0.15$. What is $P(A \text{ or } B)$?

(A) 0.15
(B) 0.80
(C) 0.65 ✓
(D) 0.95

Worked solution

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ The addition rule for the union of two events.

$= 0.45 + 0.35 - 0.15$

$= 0.65$

Answer: 0.65

The addition rule subtracts $P(A\text{ and }B)$ so the outcomes in the overlap are not counted twice.

Question 105 · 2 pts

A medical researcher wants to know whether a new drug lowers blood pressure. She randomly assigns $200$ volunteers to two groups: one group receives the drug and the other receives a placebo. After $8$ weeks she compares the average blood pressure of the two groups. Which best describes the study and the conclusions it allows?

(A) It is an observational study, so no cause-and-effect conclusion can be made.
(B) It is a controlled experiment, but a cause-and-effect conclusion can never be drawn from any study.
(C) It is a sample survey, and it can establish a cause-and-effect relationship.
(D) It is a controlled experiment, and because treatments were randomly assigned, a cause-and-effect conclusion is appropriate. ✓

Worked solution

$\text{Treatments (drug vs. placebo) are imposed by the researcher} \Rightarrow \text{this is an experiment.}$

$\text{Random assignment balances other variables across the two groups.}$ This is what separates an experiment from an observational study.

$\Rightarrow \text{any difference in outcome can be attributed to the treatment (cause and effect).}$

Answer: It is a controlled experiment, and because treatments were randomly assigned, a cause-and-effect conclusion is appropriate.

Only randomized experiments, in which treatments are imposed and randomly assigned, support cause-and-effect conclusions; observational studies and surveys do not.

Question 24 · 2 pts

The scores on a standardized test are approximately normally distributed with a mean of 75 and a standard deviation of 5. Approximately what percent of the scores are between 70 and 80?

(A) $95\%$
(B) $34\%$
(C) $50\%$
(D) $68\%$ ✓

Worked solution

$70 = 75 - 5 \text{ and } 80 = 75 + 5$ The interval is one standard deviation on each side of the mean

$\text{Empirical rule: about } 68\% \text{ of a normal distribution lies within 1 standard deviation of the mean}$

$\text{So approximately } 68\% \text{ of scores fall between 70 and 80}$

Answer: $68\%$

The interval from 70 to 80 spans exactly one standard deviation below and above the mean of 75. By the empirical rule for normal distributions, about 68% of the data falls within one standard deviation of the mean. The distractor 95% corresponds to two standard deviations, and 34% is only one side of the mean.

Question 31 · 4 pts · constructed response

In a survey of 120 students at a high school, 60 students play a sport, 48 students play a musical instrument, and 24 students do both. Determine whether the events "plays a sport" and "plays a musical instrument" are independent. Justify your answer.

Model answer: Yes, the events are independent: $P(S)=\tfrac{60}{120}=0.5$, $P(M)=\tfrac{48}{120}=0.4$, and $P(S)\cdot P(M)=0.2=\tfrac{24}{120}=P(S \text{ and } M)$.

Worked solution

$P(S)=\frac{60}{120}=0.5$ Probability a student plays a sport.

$P(M)=\frac{48}{120}=0.4$ Probability a student plays an instrument.

$P(S \text{ and } M)=\frac{24}{120}=0.2$

$P(S)\cdot P(M)=(0.5)(0.4)=0.2=P(S \text{ and } M)$ The independence test is satisfied, so the events are independent.

Answer: Yes, the events are independent: $P(S)=\tfrac{60}{120}=0.5$, $P(M)=\tfrac{48}{120}=0.4$, and $P(S)\cdot P(M)=0.2=\tfrac{24}{120}=P(S \text{ and } M)$.

Two events are independent exactly when $P(A \text{ and } B)=P(A)\cdot P(B)$. Here both sides equal $0.2$, so knowing that a student plays a sport does not change the probability that the student plays an instrument.

Question 31 · 2 pts

A fair coin is flipped $3$ times. What is the probability of getting exactly $2$ heads?

(A) $\tfrac{2}{3}$
(B) $\tfrac{1}{2}$
(C) $\tfrac{3}{8}$ ✓
(D) $\tfrac{1}{8}$

Worked solution

$\binom{3}{2}\left(\tfrac12\right)^3 = 3\cdot\tfrac18$ Binomial probability.

$= \tfrac{3}{8}$

Answer: \tfrac{3}{8}

Count the favorable outcomes (HHT, HTH, THH = 3) over the total (2³ = 8), or use the binomial formula.

Question 98 · 2 pts

On a standardized exam, scores are normally distributed with a mean of 80 and a standard deviation of 6. Using the empirical rule (68-95-99.7), approximately what percent of students scored above 92?

(A) 95%
(B) 47.5%
(C) 2.5% ✓
(D) 5%

Worked solution

$92 = 80 + 2(6)$ 92 is exactly 2 standard deviations above the mean.

$P(\text{within } 2\text{ SD}) = 95\% \;\Rightarrow\; \text{both tails} = 100\% - 95\% = 5\%$

$\text{upper tail} = \dfrac{5\%}{2} = 2.5\%$

Answer: 2.5%

Under the empirical rule about $95\%$ of data lies within $2$ standard deviations of the mean, leaving $5\%$ split equally between the two tails, so each tail holds $2.5\%$.

Question 101 · 2 pts

A polling organization simulated 500 random samples to estimate the proportion of voters who support a measure. The simulated sample proportions were approximately normal with a mean of 0.62 and a standard deviation of 0.05. Which interval captures the middle 95% of the simulated proportions?

(A) (0.42, 0.82)
(B) (0.52, 0.72) ✓
(C) (0.62, 0.72)
(D) (0.57, 0.67)

Worked solution

$\text{middle } 95\% \;\Rightarrow\; \text{mean} \pm 2(\text{SD})$ About 95% of an approximately normal distribution lies within 2 standard deviations.

$0.62 \pm 2(0.05) = 0.62 \pm 0.10$

$(0.62 - 0.10,\; 0.62 + 0.10) = (0.52,\; 0.72)$

Answer: (0.52, 0.72)

For an approximately normal sampling distribution, about $95\%$ of values fall within $2$ standard deviations of the mean, giving the symmetric interval $\text{mean}\pm 2\sigma$.

Question 103 · 2 pts

In a city, 60% of households own a car, 30% own a bicycle, and 18% own both a car and a bicycle. If a household is selected at random, what is the probability that it owns a bicycle, given that it owns a car?

(A) 0.50
(B) 0.30 ✓
(C) 0.18
(D) 0.60

Worked solution

$P(\text{bike} \mid \text{car}) = \dfrac{P(\text{bike and car})}{P(\text{car})}$ Definition of conditional probability.

$= \dfrac{0.18}{0.60}$

$= 0.30$

Answer: 0.30

Conditional probability divides the joint probability of both events by the probability of the given (conditioning) event.

Question 106 · 2 pts

To estimate the proportion of voters who favor a ballot measure, a pollster runs $500$ simulated samples of size $100$. The simulated sample proportions are approximately normal with mean $0.46$ and standard deviation $0.05$. Using a margin of two standard deviations about the mean, which interval gives the plausible values for the true proportion?

(A) $(0.41,\ 0.51)$
(B) $(0.36,\ 0.56)$ ✓
(C) $(0.26,\ 0.66)$
(D) $(0.46,\ 0.56)$

Worked solution

$\text{margin} = 2 \times 0.05 = 0.10$

$0.46 - 0.10 = 0.36 \quad \text{and} \quad 0.46 + 0.10 = 0.56$

$\text{Interval: } (0.36,\ 0.56)$

Answer: $(0.36,\ 0.56)$

A margin of error of two standard deviations of the sampling distribution about the sample mean captures the plausible values for the population proportion.

Question 107 · 2 pts

Scores on a standardized test are normally distributed with a mean of $500$ and a standard deviation of $100$. If $2000$ students took the test, approximately how many scored between $300$ and $600$?

(A) About $1630$ students ✓
(B) About $1360$ students
(C) About $1900$ students
(D) About $950$ students

Worked solution

$300 = 500 - 2(100), \quad 600 = 500 + 1(100)$ 300 is 2 SD below the mean; 600 is 1 SD above.

$P(-2\sigma \text{ to } 0) = 47.5\%, \quad P(0 \text{ to } +1\sigma) = 34\%$

$47.5\% + 34\% = 81.5\%$

$0.815 \times 2000 = 1630$

Answer: About $1630$ students.

The empirical rule splits the normal curve into half-standard-deviation regions (47.5% from the mean out to 2 SD, 34% within 1 SD), so any interval bounded by whole standard deviations can be totaled.

Question 109 · 2 pts

At a school, the probability that a student plays a sport is $0.6$, the probability that a student plays an instrument is $0.3$, and the probability that a student does both is $0.2$. What is the probability that a randomly chosen student plays a sport or plays an instrument?

(A) $0.9$
(B) $0.5$
(C) $0.2$
(D) $0.7$ ✓

Worked solution

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

$= 0.6 + 0.3 - 0.2 = 0.7$

Answer: $0.7$

For events that can occur together, the Addition Rule subtracts the overlap once so the shared students are not double counted.

Question 111 · 2 pts

On an English test the class mean is $75$ with a standard deviation of $5$, and Maria scored $88$. On a History test the class mean is $80$ with a standard deviation of $10$, and Jordan scored $96$. Relative to their own classes, who performed better?

(A) They performed equally well because both scored above their class means.
(B) Maria, whose z-score of $2.6$ exceeds Jordan's $1.6$. ✓
(C) Jordan, because his raw score of $96$ is higher than Maria's $88$.
(D) Jordan, because he scored $16$ points above his mean while Maria scored only $13$ above hers.

Worked solution

$z = \frac{x - \bar{x}}{\sigma}$

$z_{\text{Maria}} = \frac{88 - 75}{5} = \frac{13}{5} = 2.6$

$z_{\text{Jordan}} = \frac{96 - 80}{10} = \frac{16}{10} = 1.6$

$2.6 > 1.6 \Rightarrow \text{Maria did better relative to her class.}$

Answer: Maria, whose z-score of $2.6$ exceeds Jordan's $1.6$.

z-scores rescale values from different distributions onto a common scale (standard deviations from the mean), allowing a fair comparison across the two tests.

Question 8 · 2 pts

For two events $A$ and $B$, it is known that $P(A)=0.4$, $P(B)=0.3$, and $P(A\text{ and }B)=0.12$. Which statement is true about events $A$ and $B$?

(A) They are dependent because $P(A)+P(B)\ne P(A\text{ and }B)$.
(B) They are independent because $P(A)\cdot P(B)=P(A\text{ and }B)$. ✓
(C) They are independent because $P(A\text{ and }B)\ne 0$.
(D) They are dependent because $P(A)\ne P(B)$.

Worked solution

$P(A)\cdot P(B) = (0.4)(0.3) = 0.12$ Test the multiplication rule for independence.

$P(A\text{ and }B) = 0.12$ Given.

$0.12 = 0.12,\ \text{so } A \text{ and } B \text{ are independent.}$

Answer: They are independent because $P(A)\cdot P(B)=P(A\text{ and }B)$.

Two events are independent exactly when $P(A\text{ and }B)=P(A)\cdot P(B)$. Here $(0.4)(0.3)=0.12$ matches the given joint probability, so the events are independent. The other statements use incorrect tests such as adding probabilities or comparing $P(A)$ to $P(B)$.

Question 20 · 2 pts

For two events $A$ and $B$, it is known that $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \text{ and } B) = 0.2$. Which statement is true?

(A) $A$ and $B$ are dependent because $P(A) + P(B) \neq P(A \text{ and } B)$
(B) $A$ and $B$ are independent because $P(A) + P(B) - P(A \text{ and } B) = 0.7$
(C) $A$ and $B$ are dependent because $P(A) \neq P(B)$
(D) $A$ and $B$ are independent because $P(A) \cdot P(B) = P(A \text{ and } B)$ ✓

Worked solution

$\text{Independence test: } P(A) \cdot P(B) = P(A \text{ and } B)$

$P(A) \cdot P(B) = (0.4)(0.5) = 0.2$

$P(A \text{ and } B) = 0.2, \text{ so the products match}$

$\text{Therefore } A \text{ and } B \text{ are independent.}$

Answer: $A$ and $B$ are independent because $P(A) \cdot P(B) = P(A \text{ and } B)$

Two events are independent exactly when the product of their individual probabilities equals the probability of their intersection. Here $(0.4)(0.5) = 0.2 = P(A \text{ and } B)$, so the events are independent. Adding probabilities or comparing $P(A)$ with $P(B)$ does not test independence.

Question 31 · 4 pts · constructed response

In a survey of 400 students at a high school, 240 students play a sport, 180 students play a musical instrument, and 90 students do both. Determine whether the events "plays a sport" and "plays a musical instrument" are independent. Justify your answer.

Model answer: Not independent, because $P(S \cap M) = 0.225$ while $P(S) \cdot P(M) = (0.6)(0.45) = 0.27$, and $0.225 \ne 0.27$.

Worked solution

$P(S) = \frac{240}{400} = 0.6$ Probability a student plays a sport.

$P(M) = \frac{180}{400} = 0.45$ Probability a student plays an instrument.

$P(S \cap M) = \frac{90}{400} = 0.225$ Probability a student does both.

$P(S) \cdot P(M) = (0.6)(0.45) = 0.27$ Test for independence: compare $P(S \cap M)$ with $P(S)\cdot P(M)$.

$0.225 \ne 0.27, \ \text{so the events are not independent.}$

Answer: Not independent, because $P(S \cap M) = 0.225$ while $P(S) \cdot P(M) = (0.6)(0.45) = 0.27$, and $0.225 \ne 0.27$.

Two events are independent exactly when the probability of their intersection equals the product of their individual probabilities. Here $P(S)\cdot P(M)=0.27$ but the actual overlap is $P(S \cap M)=0.225$, so the events are not independent.

Question 23 · 2 pts

At a high school, the probability that a randomly selected student plays a sport is $0.4$, the probability that a student takes an art class is $0.3$, and the probability that a student does both is $0.12$. Which statement is true about the events "plays a sport" and "takes an art class"?

(A) The events are independent because $P(S) \cdot P(A) = P(S \text{ and } A)$. ✓
(B) The events are independent because $P(S \text{ and } A) \neq 0$.
(C) The events are dependent because $P(S) + P(A) \neq P(S \text{ and } A)$.
(D) The events are dependent because $P(S) \neq P(A)$.

Worked solution

$\text{Independence test: events } S \text{ and } A \text{ are independent if } P(S) \cdot P(A) = P(S \text{ and } A).$

$P(S) \cdot P(A) = (0.4)(0.3) = 0.12$

$P(S \text{ and } A) = 0.12$ This matches the product, so the events are independent.

$\text{Equivalently, } P(S \mid A) = \frac{0.12}{0.3} = 0.4 = P(S).$

Answer: The events are independent because $P(S) \cdot P(A) = P(S \text{ and } A)$.

Two events are independent exactly when the product of their individual probabilities equals the probability of their intersection. Here $(0.4)(0.3) = 0.12$, which equals $P(S \text{ and } A)$, so knowing a student takes art does not change the probability that the student plays a sport. Adding probabilities or comparing $P(S)$ to $P(A)$ are not valid tests for independence.

Question 37 · 4 pts · constructed response

A survey of 200 students at a high school found that 120 of the students play a school sport and 80 of the students are seniors. Of the students surveyed, 48 are seniors who play a school sport. Determine whether the events "a student plays a school sport" and "a student is a senior" are independent. Justify your answer.

Model answer: The events are independent because $P(\text{sport}) \cdot P(\text{senior}) = (0.6)(0.4) = 0.24$, which equals $P(\text{sport and senior}) = \frac{48}{200} = 0.24$.

Worked solution

$P(\text{sport}) = \frac{120}{200} = 0.6$

$P(\text{senior}) = \frac{80}{200} = 0.4$

$P(\text{sport and senior}) = \frac{48}{200} = 0.24$

$P(\text{sport}) \cdot P(\text{senior}) = (0.6)(0.4) = 0.24$ Compare the product of the individual probabilities to the joint probability.

$0.24 = 0.24, \text{ so the events are independent}$ Equivalently, $P(\text{sport} \mid \text{senior}) = \frac{48}{80} = 0.6 = P(\text{sport})$.

Answer: The events are independent because $P(\text{sport}) \cdot P(\text{senior}) = (0.6)(0.4) = 0.24$, which equals $P(\text{sport and senior}) = \frac{48}{200} = 0.24$.

Two events $A$ and $B$ are independent exactly when $P(A \text{ and } B) = P(A) \cdot P(B)$. Here the joint probability, $\frac{48}{200} = 0.24$, equals the product $(0.6)(0.4) = 0.24$, so knowing a student is a senior does not change the probability that the student plays a sport, and the events are independent.

Question 8 · 2 pts

The scores on a standardized test are approximately normally distributed with a mean of 75 and a standard deviation of 5. Approximately what percent of the scores are between 70 and 85?

(A) $68\%$
(B) $81.5\%$ ✓
(C) $47.5\%$
(D) $95\%$

Worked solution

$70=75-5 \text{ is 1 standard deviation below the mean}$

$85=75+2(5) \text{ is 2 standard deviations above the mean}$

$P(70<x<75)\approx 34\%,\quad P(75<x<85)\approx 47.5\%$ From the 68-95-99.7 rule, half of 68% and half of 95%.

$34\%+47.5\%=81.5\%$

Answer: $81.5\%$

The interval runs from 1 standard deviation below the mean to 2 standard deviations above it. By the empirical rule, about 34% of the data lies between the mean and 1 standard deviation below, and about 47.5% lies between the mean and 2 standard deviations above, giving roughly $34\%+47.5\%=81.5\%$.

Question 14 · 2 pts

For two events $A$ and $B$, it is known that $P(A)=0.5$, $P(B)=0.4$, and $P(A\text{ and }B)=0.2$. Which statement is true?

(A) $A$ and $B$ are dependent because $P(A)+P(B)\ne P(A\text{ and }B)$
(B) $A$ and $B$ are independent because $P(A)+P(B)$ is less than 1
(C) $A$ and $B$ are mutually exclusive because $P(A\text{ and }B)\ne 0$
(D) $A$ and $B$ are independent because $P(A)\cdot P(B)=P(A\text{ and }B)$ ✓

Worked solution

$P(A)\cdot P(B)=(0.5)(0.4)=0.2$ Independence test: compare the product to $P(A\text{ and }B)$.

$P(A\text{ and }B)=0.2=P(A)\cdot P(B)$

$\text{Therefore } A \text{ and } B \text{ are independent.}$

Answer: $A$ and $B$ are independent because $P(A)\cdot P(B)=P(A\text{ and }B)$

Two events are independent exactly when $P(A\text{ and }B)=P(A)\cdot P(B)$. Here $(0.5)(0.4)=0.2$, which matches the given joint probability, so the events are independent. Mutually exclusive would require $P(A\text{ and }B)=0$, and sums of probabilities play no role in the independence test.

Question 30 · 4 pts · constructed response

At a high school, the probability that a randomly selected student plays a sport is $0.5$, the probability that a student is in the band is $0.4$, and the probability that a student both plays a sport and is in the band is $0.2$. Determine whether the events "plays a sport" and "is in the band" are independent. Justify your answer.

Model answer: The events are independent because $P(S)\cdot P(B)=(0.5)(0.4)=0.2=P(S\text{ and }B)$.

Worked solution

$\text{Events are independent if } P(S\text{ and }B)=P(S)\cdot P(B)$ State the test for independence.

$P(S)\cdot P(B)=(0.5)(0.4)=0.2$

$P(S\text{ and }B)=0.2$ This equals the product, so the condition holds.

$\text{Therefore the events are independent.}$

Answer: The events are independent because $P(S)\cdot P(B)=(0.5)(0.4)=0.2=P(S\text{ and }B)$.

Two events are independent exactly when the probability of both occurring equals the product of their individual probabilities. Since $(0.5)(0.4)=0.2$ matches the given joint probability, playing a sport gives no information about band membership, so the events are independent.

Question 102 · 2 pts

A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random without replacement. What is the probability that both marbles are red?

(A) 25/64
(B) 5/16
(C) 5/14 ✓
(D) 5/8

Worked solution

$P(\text{1st red}) = \dfrac{5}{8}$ 5 red marbles out of 8 total.

$P(\text{2nd red} \mid \text{1st red}) = \dfrac{4}{7}$ One red and one marble have been removed.

$P(\text{both red}) = \dfrac{5}{8}\cdot\dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$

Answer: 5/14

Without replacement the draws are dependent, so the second probability is conditioned on the first by reducing both the favorable count and the total.

Question 104 · 2 pts

A machine produces bolts whose lengths are normally distributed with a mean of 50 mm and a standard deviation of 2 mm. In a batch of 1000 bolts, approximately how many bolts are longer than 54 mm?

(A) 160 bolts
(B) 25 bolts ✓
(C) 50 bolts
(D) 475 bolts

Worked solution

$54 = 50 + 2(2)$ 54 is 2 standard deviations above the mean.

$P(\text{length} > 54) = \dfrac{100\% - 95\%}{2} = 2.5\%$ The upper tail beyond 2 standard deviations.

$0.025 \times 1000 = 25$

Answer: 25 bolts

Beyond $2$ standard deviations on one side lies about $2.5\%$ of a normal distribution; multiplying this proportion by the batch size gives the expected count.

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