What is Data & statistics on the Algebra 1 Regents?

Data & statistics is a Statistics topic on the Algebra 1 Regents. RegentsLab has 21 practice questions on it, each with a full worked solution.

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Data & statistics: Algebra 1 Regents Practice

Practice Data & statistics for the Algebra 1 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 1 · 2 pts

A ball is launched upward and its height (feet) is recorded each second: at $0$ s, $11$; at $1$ s, $59$; at $2$ s, $75$; at $3$ s, $59$; at $4$ s, $11$. Based on these data, which statement is a valid conclusion?

(A) The ball reaches a maximum height of $11$ feet.
(B) The ball reaches its greatest height at $2$ seconds. ✓
(C) The ball lands on the ground at $4$ seconds.
(D) The ball was launched from a height of $0$ feet.

Worked solution

$\text{heights: } 11,\,59,\,\mathbf{75},\,59,\,11$ The data are symmetric and peak in the middle.

$\text{max } 75 \text{ ft at } t = 2$ Greatest value occurs at 2 seconds.

Answer: \text{maximum height at } t = 2 \text{ s}

Read the table for what it actually says: the largest height (75 ft) occurs at t = 2 s, and the launch height is the value at t = 0 (11 ft).

Question 12 · 2 pts

What is the median of the data set $\{4,\ 7,\ 9,\ 12,\ 18\}$ ?

(A) $9$ ✓
(B) $10$
(C) $12$
(D) $7$

Worked solution

$\{4,\,7,\,\mathbf{9},\,12,\,18\}$ Already ordered; take the middle.

$\text{median} = 9$

Answer: 9

The median is the middle value of an ordered list (average the two middles if the count is even).

Question 37 · 2 pts

For the data set $3,\ 5,\ 7,\ 8,\ 10,\ 12,\ 14,\ 16$, what is the interquartile range (IQR)?

(A) $9$
(B) $19$
(C) $7$ ✓
(D) $13$

Worked solution

$\text{Order the 8 values: } 3,\ 5,\ 7,\ 8,\ 10,\ 12,\ 14,\ 16$ The list is already in order.

$\text{Lower half } \{3,5,7,8\}:\quad Q_1=\frac{5+7}{2}=6$

$\text{Upper half } \{10,12,14,16\}:\quad Q_3=\frac{12+14}{2}=13$

$\text{IQR}=Q_3-Q_1=13-6=7$

Answer: $\text{IQR}=7$

The interquartile range measures the spread of the middle 50% of the data and equals $Q_3-Q_1$.

Question 39 · 2 pts

Four scatter plots have the correlation coefficients listed below. Which value indicates the strongest linear relationship?

(A) $0.88$
(B) $-0.09$
(C) $0.42$
(D) $-0.90$ ✓

Worked solution

$\text{Strength is measured by } |r|,\ \text{how close } r \text{ is to } 1 \text{ or } -1.$

$|-0.90|=0.90,\quad |0.88|=0.88,\quad |0.42|=0.42,\quad |-0.09|=0.09$

$\text{Greatest } |r| = 0.90 \Rightarrow r=-0.90 \text{ is strongest.}$

Answer: $r=-0.90$

The closer $|r|$ is to $1$, the stronger the linear relationship; the sign shows direction only, not strength.

Question 13 · 2 pts

A box plot of winter snowfall has a lower quartile of $Q_1 = 30$ inches and an upper quartile of $Q_3 = 80$ inches. What is the interquartile range (IQR)?

(A) $110$
(B) $30$
(C) $50$ ✓
(D) $80$

Worked solution

$\text{IQR} = Q_3 - Q_1$ Definition (on the reference sheet).

$= 80 - 30 = 50$ Subtract.

Answer: \text{IQR} = 50

The interquartile range is the width of the middle 50% of the data: upper quartile minus lower quartile.

Question 25 · 2 pts · constructed response

Of 100 students surveyed, 44 are seniors. Of the 65 who prefer videos, 42 are juniors. Determine how many seniors prefer music.

Model answer: 21

Worked solution

$\text{video seniors}=65-42=23$ 65 prefer videos; 42 are juniors.

$\text{music seniors}=44-23=21$ 44 seniors total, minus those who prefer videos.

Answer: 21\ \text{seniors prefer music}

Fill the two-way table: subtract junior video-watchers from the video total to get senior video-watchers, then subtract from all seniors.

Question 38 · 2 pts

A tutor models the relationship between hours studied, $x$, and a student's test score, $y$, with the line of best fit $y=2.5x+60$. Which statement best interprets the slope of this line?

(A) For each additional point scored, study time increases by $2.5$ hours.
(B) For each additional hour studied, the predicted test score increases by $2.5$ points. ✓
(C) A student who does not study is predicted to score $2.5$ points.
(D) For each additional hour studied, the predicted test score increases by $60$ points.

Worked solution

$\text{In } y=mx+b,\ \text{the slope } m \text{ is the change in } y \text{ per one-unit increase in } x.$

$\text{Here } m=2.5,\quad x=\text{hours studied},\quad y=\text{test score}.$

$\Delta y = 2.5 \text{ points for each } \Delta x = 1 \text{ hour}.$

Answer: The predicted score increases by $2.5$ points per additional hour studied.

The slope of a regression line is the unit rate of change of the response variable $y$ for each one-unit increase in the explanatory variable $x$.

Question 40 · 2 pts

The two-way frequency table shows the pet preferences of 80 students. | | Dog | Cat | Total | |---|---|---|---| | Male | 26 | 24 | 50 | | Female | 9 | 21 | 30 | | Total | 35 | 45 | 80 | According to the table, what percent of the female students prefer cats?

(A) $\approx 46.7\%$
(B) $30\%$
(C) $26.25\%$
(D) $70\%$ ✓

Worked solution

$\text{Restrict to the female row: } 30 \text{ females total, } 21 \text{ prefer cats.}$ A conditional relative frequency uses the given group's total as the denominator.

$\text{Relative frequency}=\frac{21}{30}$

$\frac{21}{30}=0.70=70\%$

Answer: $70\%$

A conditional relative frequency divides by the row (or column) total of the given group, here the $30$ female students.

Question 10 · 2 pts

What is the mean of the data set $10, 12, 14, 16, 48$?

(A) $38$
(B) $20$ ✓
(C) $13$
(D) $14$

Worked solution

$10 + 12 + 14 + 16 + 48 = 100$ Add all five values

$\text{mean} = \frac{100}{5}$ Divide by the number of values

$\text{mean} = 20$ Simplify

Answer: $20$

The mean is the sum of all values divided by how many there are. The total $100$ divided by $5$ gives $20$. (The median is $14$ and the range is $38$, which are different measures.)

Question 30 · 4 pts · constructed response

Consider the data set $2, 4, 6, 8, 10$. Suppose the value $10$ is changed to $100$. Explain what happens to the mean and to the median of the data set. Support your answer with calculations.

Model answer: The mean increases from $6$ to $24$, while the median stays at $6$. The mean is pulled higher by the large value, but the median is unaffected because the middle value of the ordered data does not change.

Worked solution

$\text{mean} = \dfrac{2+4+6+8+10}{5} = \dfrac{30}{5} = 6$ Original mean.

$\text{median} = 6$ Middle value of the ordered set $2,4,6,8,10$.

$\text{new mean} = \dfrac{2+4+6+8+100}{5} = \dfrac{120}{5} = 24$ Mean after changing 10 to 100.

$\text{new median} = 6$ Ordered set $2,4,6,8,100$ still has middle value 6.

Answer: The mean increases from $6$ to $24$, while the median stays at $6$. The mean is pulled higher by the large value, but the median is unaffected because the middle value of the ordered data does not change.

The mean uses the actual value of every data point, so a single large value (an outlier) raises it substantially, here from 6 to 24. The median depends only on position, and since the middle value is still 6, the median is resistant to the change.

Question 21 · 2 pts

Four sets of data were each fit with a line of best fit. Which value of the correlation coefficient $r$ indicates the strongest linear relationship?

(A) $r = 0.87$
(B) $r = -0.95$ ✓
(C) $r = -0.12$
(D) $r = 0.34$

Worked solution

$\text{strength} \propto |r|$ Strength depends on how close $|r|$ is to 1.

$|-0.95| = 0.95, \; |0.87| = 0.87, \; |0.34| = 0.34, \; |-0.12| = 0.12$

$0.95 > 0.87 > 0.34 > 0.12$ Largest absolute value is strongest.

Answer: $r = -0.95$

The strength of a linear relationship is measured by how close $|r|$ is to 1, regardless of sign. Since $|-0.95| = 0.95$ is the largest absolute value, it represents the strongest relationship. The negative sign only tells us the line has a negative slope.

Question 22 · 2 pts

Emma's scores on five quizzes were $70, 75, 80, 85,$ and $100$. Which statement about this data set is true?

(A) The median is greater than the mean.
(B) The mean and the median are equal.
(C) The mean is greater than the median. ✓
(D) The median of the data set is $82$.

Worked solution

$\text{Mean} = \frac{70 + 75 + 80 + 85 + 100}{5} = \frac{410}{5} = 82$

$\text{Median} = 80$ The middle value of the five ordered scores.

$82 > 80, \text{ so the mean is greater than the median.}$

Answer: The mean is greater than the median.

The mean of the five scores is 82 and the median, the middle value, is 80. Because the high score of 100 pulls the data to the right, the mean exceeds the median, which is typical of a distribution skewed toward larger values. The value 82 is the mean, not the median.

Question 10 · 2 pts

The number of goals scored by a soccer team in six games was $2, 4, 4, 6, 9,$ and $11$. Which statement about this data set is true?

(A) The mean is $6$ and the median is $4$.
(B) The mean is $6$ and the median is $6$.
(C) The mean is $6$ and the median is $5$. ✓
(D) The mean is $5$ and the median is $6$.

Worked solution

$\text{Mean} = \frac{2 + 4 + 4 + 6 + 9 + 11}{6} = \frac{36}{6} = 6$

$\text{Ordered data: } 2, 4, 4, 6, 9, 11$ With 6 values, the median is the average of the 3rd and 4th values.

$\text{Median} = \frac{4 + 6}{2} = 5$

Answer: The mean is $6$ and the median is $5$.

The mean is the sum $36$ divided by $6$ values, which is $6$. Since there is an even number of data values, the median is the average of the two middle values, $4$ and $6$, giving $5$. Choosing just the value $4$ or $6$ as the median is a frequent error with even-sized data sets.

Question 24 · 2 pts

A study compared the number of hours students spent watching television each week and their scores on a math test. The correlation coefficient for the data was $r = -0.9$. Which statement best describes the relationship between the two variables?

(A) There is a weak negative linear relationship.
(B) There is a strong positive linear relationship.
(C) There is no linear relationship.
(D) There is a strong negative linear relationship. ✓

Worked solution

$\text{The correlation coefficient } r \text{ ranges from } -1 \text{ to } 1.$

$|{-0.9}| = 0.9, \text{ which is close to } 1$ Values near 1 or -1 indicate a strong linear relationship.

$r < 0 \text{ means the relationship is negative: as hours increase, scores decrease.}$

Answer: There is a strong negative linear relationship.

The strength of a linear relationship is measured by how close $|r|$ is to 1, and the sign of $r$ gives the direction. Since $r = -0.9$ is close to $-1$, the data show a strong negative linear relationship: more television hours are associated with lower test scores.

Question 32 · 4 pts · constructed response

The number of books read last year by each of nine students is listed below: $3,\ 5,\ 5,\ 6,\ 7,\ 8,\ 8,\ 8,\ 10$. Determine the median and the interquartile range of this data set. Show how you arrived at each value.

Model answer: Median $=7$; interquartile range $=8-5=3$.

Worked solution

$\text{Data in order: } 3,5,5,6,7,8,8,8,10$ There are 9 values, so the median is the 5th value.

$\text{Median}=7$

$Q_1=\frac{5+5}{2}=5$ Lower half is $3,5,5,6$; $Q_1$ is the middle of that half.

$Q_3=\frac{8+8}{2}=8$ Upper half is $8,8,8,10$; $Q_3$ is the middle of that half.

$\text{IQR}=Q_3-Q_1=8-5=3$

Answer: Median $=7$; interquartile range $=8-5=3$.

With nine ordered values, the median is the fifth value, 7. The first quartile is the median of the four values below the median and the third quartile is the median of the four values above it, giving $Q_1=5$ and $Q_3=8$, so the interquartile range is $8-5=3$.

Question 10 · 2 pts

The number of books read last month by each of seven students is listed below. $$4, \ 6, \ 6, \ 7, \ 9, \ 10, \ 14$$ Which statement about these data is true?

(A) The median is greater than the mean.
(B) The mean is equal to the median.
(C) The mode is greater than the mean.
(D) The mean is greater than the median. ✓

Worked solution

$\text{mean} = \frac{4+6+6+7+9+10+14}{7} = \frac{56}{7} = 8$

$\text{median} = 7$ The data are ordered; the 4th of 7 values is the median.

$\text{mode} = 6$ 6 appears twice, more than any other value.

$8 > 7, \text{ so the mean is greater than the median.}$

Answer: The mean is greater than the median.

The mean of the seven values is $56/7 = 8$, the median is the middle value 7, and the mode is 6. Since $8 > 7$, the mean exceeds the median, which is typical for data skewed to the right by the large value 14.

Question 23 · 2 pts

A student computes the correlation coefficient for a set of bivariate data and gets $r = -0.92$. Which statement best describes the relationship between the two variables?

(A) An increase in one variable causes a decrease in the other variable.
(B) There is a strong negative linear relationship between the variables. ✓
(C) There is a strong positive linear relationship between the variables.
(D) There is a weak negative linear relationship between the variables.

Worked solution

$-1 \le r \le 1$ The correlation coefficient always lies between $-1$ and $1$.

$|-0.92| = 0.92 \text{ is close to } 1$ Values near $1$ or $-1$ indicate a strong linear relationship.

$r < 0 \text{ means the relationship is negative}$ As one variable increases, the other tends to decrease, but correlation does not prove causation.

Answer: There is a strong negative linear relationship between the variables.

The sign of $r$ gives the direction of the linear relationship and its absolute value gives the strength. Since $-0.92$ is negative and close to $-1$, the data show a strong negative linear relationship. The causation choice is wrong because correlation alone never establishes that one variable causes changes in the other.

Question 32 · 4 pts · constructed response

A film's box-office earnings (millions of dollars) over its first six weeks are: $185,\ 150,\ 90,\ 50,\ 25,\ 5$ (weeks $1$ through $6$). Write the linear regression equation (round values to the nearest hundredth), state the correlation coefficient to the nearest hundredth, and describe the fit.

Model answer: y = −37.57x + 215.67; r ≈ −0.98 → a strong (negative) linear fit

Worked solution

$y = -37.57x + 215.67$ Linear regression from the calculator, rounded to hundredths.

$r \approx -0.98$ Correlation coefficient, nearest hundredth.

$|r| \approx 1 \Rightarrow \text{strong fit}$ Close to −1 means a strong negative linear relationship.

Answer: y = -37.57x + 215.67,\ \ r \approx -0.98

Enter the data, run linear regression for the equation and r, then judge strength by how close $|r|$ is to 1. A negative slope and r near −1 is a strong negative fit.

Question 4 · 2 pts

Andrea's quiz scores are $82,\,87,\,90,\,84$ and Joe's are $91,\,78,\,94,\,67$. Which statement is correct?

(A) Both Andrea's mean and standard deviation are higher.
(B) Both Joe's mean and standard deviation are higher.
(C) Andrea's mean is higher, but Joe's standard deviation is higher. ✓
(D) Joe's mean is higher, but Andrea's standard deviation is higher.

Worked solution

$\bar{x}_A=85.75,\ \bar{x}_J=82.5$ Andrea's mean is higher.

$\text{Joe's spread }67\!-\!94 \Rightarrow \text{larger SD}$

Answer: \text{Andrea higher mean, Joe higher SD}

Compare means by averaging; standard deviation grows with spread, and Joe's scores vary far more widely.

Question 41 · 2 pts

The line of best fit for a data set is $y=3x+2$. For the observed data point $(4,\ 15)$, what is the residual?

(A) $1$ ✓
(B) $3$
(C) $14$
(D) $-1$

Worked solution

$\hat{y}=3(4)+2=14$ Predicted value at x = 4.

$\text{residual}=\text{observed}-\text{predicted}=y-\hat{y}$

$\text{residual}=15-14=1$

Answer: $\text{residual}=1$

A residual is the observed value minus the predicted value; a positive residual means the point lies above the line of best fit.

Question 42 · 2 pts

Two data sets each have five values and a mean of $50$. Set A is $\{50,\ 50,\ 50,\ 50,\ 50\}$ and Set B is $\{30,\ 40,\ 50,\ 60,\ 70\}$. Which statement correctly compares their standard deviations?

(A) Set A has the greater standard deviation because all of its values are equal.
(B) The two sets have equal standard deviations because they have the same mean.
(C) The standard deviations cannot be compared without more information.
(D) Set B has the greater standard deviation because its values are more spread out from the mean. ✓

Worked solution

$\text{Standard deviation measures typical distance of values from the mean.}$

$\text{Set A: every value} = 50 = \text{mean} \Rightarrow \text{all deviations} = 0 \Rightarrow \text{SD}=0.$

$\text{Set B: deviations reach } |70-50|=20 \Rightarrow \text{SD}>0.$

$\text{SD}_B > \text{SD}_A$

Answer: Set B has the greater standard deviation.

Standard deviation reflects how far data values fall from the mean; more spread means a larger standard deviation, regardless of the mean's value.

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