SAT Data Analysis: Master Every Question Type

SAT Data Analysis Questions | 40 SAT-Style Questions with Solutions | SATMath800
SAT Math · Data Analysis

Read the data.
Beat the question.

40 original SAT-style questions designed to take you from basic statistical skills to the questions where the real challenge is deciding what the data actually tell you.

40 SAT-STYLE QUESTIONS
DATA PATTERN
Your SAT Data Analysis Path 5 levels · 40 questions

How to use this page

Don’t rush through the answer choices. For every question, first identify what information is actually given. Then decide which statistical idea applies.

The first 24 questions build the core toolkit. The final 16 questions deliberately become more conceptual and test your ability to reason from statistical information.

The SAT Data Mindset

READ IDENTIFY CALCULATE INTERPRET INFER
Part I · Questions 1–24

Build the Data Analysis Toolkit

Start with the fundamental statistical ideas and learn how to read the major SAT data displays: tables, box plots, histograms, dot plots, scatterplots, and two-way tables.

01

Mean

Find the average—and use the average to work backward.

01 Finding the Mean
Core Skill

A teacher records the numbers of minutes that 5 students spent completing a practice assignment:

\[ 42,\quad35,\quad38,\quad45,\quad40 \]

What is the mean number of minutes?

A) 38
B) 40
C) 42
D) 45
Solution

Add the values:

\[ 42+35+38+45+40=200 \]

There are 5 values, so:

\[ \text{Mean}=\frac{200}{5}=\boxed{40} \]
Answer: B
💡 SAT Strategy: Mean means total ÷ number of values.
02 Finding an Unknown from the Mean
Reverse Thinking

The mean of the five numbers

\[ 18,\quad24,\quad27,\quad31,\quad x \]

is 26. What is the value of \(x\)?

A) 24
B) 27
C) 30
D) 32
Solution

If the mean is 26 and there are 5 values, the total must be:

\[ 5*(26)=130 \]

The known values total:

\[ 18+24+27+31=100 \]

Therefore:

\[ x=130-100=\boxed{30} \]
Answer: C
🧠 SAT Shortcut: When the mean is given, first calculate the required total: \[ \text{Required total}=(\text{mean})*(\text{number of values}) \]
02

Median

Find the middle—but only after putting the data in order.

03 Finding the Median
Trap Question

The numbers below represent the number of books read by 7 students during the summer:

\[ 3,\quad8,\quad5,\quad11,\quad6,\quad4,\quad9 \]

What is the median?

A) 5
B) 6
C) 8
D) 11
Solution

Step 1 — Sort the data.

\[ 3,\quad4,\quad5,\quad6,\quad8,\quad9,\quad11 \]

Step 2 — Find the middle value.

There are 7 values, so the median is the 4th value:

\[ \boxed{6} \]
Answer: B
⚠️ SAT Trap: Do not simply take the middle number in the order given. The data must be ordered first. Here, 11 is in the middle of the original list, but it is not the median.
04 Median with an Unknown
Position

The five numbers below are arranged in increasing order:

\[ 12,\quad18,\quad x,\quad27,\quad35 \]

The median is 23. What is \(x\)?

A) 21
B) 23
C) 24
D) 25
Solution

There are five ordered values, so the median is the third value.

\[ x=\boxed{23} \]
Answer: B
💡 SAT Strategy: If the data are already ordered, look immediately at the middle position.
03

Range

Measure the distance from the minimum to the maximum.

05 Finding the Range
Core Skill

The daily high temperatures, in degrees Fahrenheit, during a five-day period were:

\[ 71,\quad76,\quad68,\quad80,\quad74 \]

What was the range?

A) 8
B) 10
C) 12
D) 14
Solution

The maximum is \(80\), and the minimum is \(68\).

\[ \text{Range}=80-68=\boxed{12} \]
Answer: C
06 Comparing Ranges
Compare

The numbers of visitors to two museums over five days are shown below.

Museum A Museum B
120, 135, 140, 150, 155 105, 125, 140, 160, 180

How much greater is the range for Museum B than the range for Museum A?

A) 30
B) 35
C) 40
D) 45
Solution

For Museum A:

\[ 155-120=35 \]

For Museum B:

\[ 180-105=75 \]

Difference:

\[ 75-35=\boxed{40} \]
Answer: C
🧠 Remember: Range uses only two values—the maximum and the minimum.
04

Percentiles

Understand where a value stands relative to a group.

07 Interpreting a Percentile
Interpret

A student scored at the 82nd percentile on a mathematics assessment. Which statement is the best interpretation?

A) The student answered 82% of the questions correctly.
B) The student’s score was 82 points.
C) The student’s score was higher than approximately 82% of the scores in the comparison group.
D) The student scored exactly 18 points above the mean.
Solution

A percentile describes a student’s relative position compared with other scores.

Answer: C
⚠️ Common Trap: The 82nd percentile does not mean 82% correct.
08 Comparing Percentiles
Interpret

Student A scored at the 74th percentile on a test, while Student B scored at the 91st percentile.

Which statement must be true?

A) Student B answered 17% more questions correctly.
B) Student B’s numerical score was 17 points higher.
C) Student B performed better relative to the comparison group.
D) Student A’s score was below the mean.
Solution

A percentile tells us about relative standing. The 91st-percentile student had a higher position within the comparison group.

Answer: C
💡 Remember: Percentile rank is about position, not the number of questions answered correctly.
05

Quartiles & IQR (Interquartile Range)

Measure the spread of the middle 50% of the data.

09 Finding the IQR
Core Skill

Consider the ordered data set:

\[ 4,\quad7,\quad9,\quad12,\quad15,\quad18,\quad21,\quad25 \]

The first quartile is \(Q_1=7\), and the third quartile is \(Q_3=21\). What is the interquartile range?

A) 12
B) 14
C) 16
D) 18
Solution
\[ \mathrm{IQR}=Q_3-Q_1 \] \[ 21-7=\boxed{14} \]
Answer: B
10 Comparing IQR
Compare Spread

Two groups have the following quartiles:

Group \(Q_1\) \(Q_3\)
A 18 30
B 12 32

Which group has the greater interquartile range?

A) Group A
B) Group B
C) The two groups have the same IQR
D) It cannot be determined
Solution

Group A:

\[ 30-18=12 \]

Group B:

\[ 32-12=20 \]

Therefore, Group B has the greater IQR.

Answer: B
💡 SAT Strategy: For IQR, look only at \(Q_3-Q_1\).
06

Box Plots

Read the five-number summary directly from a visual.

THE FIVE-NUMBER SUMMARY
Minimum \(Q_1\) Median \(Q_3\) Maximum
11 Reading a Box Plot
Visual Reading

The box plot above represents the distribution of the number of minutes that students spent exercising each day.

The five-number summary is:

\[ 20,\quad30,\quad42,\quad55,\quad70 \]

What is the median?

A) 30
B) 42
C) 55
D) 70
Solution

The median is represented by the line inside the box.

\[ \boxed{42} \]
Answer: B
💡 SAT Strategy: In a box plot, the line inside the box is the median.
12 Comparing Two Box Plots
Compare Spread

Two box plots represent the distributions of monthly transportation costs for two groups of students.

Group \(Q_1\) \(Q_3\)
A 40 70
B 35 80

Which group has the greater interquartile range?

A) Group A
B) Group B
C) The two groups have the same IQR
D) It cannot be determined
Solution
\[ \mathrm{IQR}_A=70-40=30 \] \[ \mathrm{IQR}_B=80-35=45 \]

Therefore, Group B has the greater IQR.

Answer: B
07

Histograms

Read grouped numerical data and interpret frequency.

A HISTOGRAM GROUPS DATA INTO INTERVALS
4
9
12
7
3
0–2 2–4 4–6 6–8 8–10
13 Reading Frequency
Visual Reading

A histogram shows the number of students who spent the following amounts of time studying for a test.

Study time (hours) Number of students
0–24
2–49
4–612
6–87
8–103

How many students studied between 4 and 6 hours?

A) 7
B) 9
C) 12
D) 15
Solution

The interval from 4 to 6 hours has a frequency of 12.

Answer: C
💡 SAT Strategy: In a histogram, the height of a bar represents frequency.
14 Comparing Intervals
Compare

The histogram below represents the commute times, in minutes, of employees at a company.

Commute time Frequency
0–105
10–2014
20–3018
30–4011
40–504

How many more employees have commute times between 20 and 30 minutes than between 40 and 50 minutes?

A) 7
B) 11
C) 14
D) 22
Solution
\[ 18-4=\boxed{14} \]
Answer: C
⚠️ Common Trap: The question asks how many more, so subtract.
08

Dot Plots

Every dot represents an individual observation.

DOT PLOT — NUMBER OF GOALS
1 2 3 4 5
15 Reading a Dot Plot
Median

A dot plot represents the number of goals scored by a soccer team in 9 games:

\[ 1,\quad2,\quad2,\quad2,\quad3,\quad3,\quad4,\quad5,\quad5 \]

What is the median number of goals?

A) 2
B) 3
C) 4
D) 5
Solution

There are 9 observations, so the median is the 5th value.

\[ 1,\quad2,\quad2,\quad2,\quad\boxed{3},\quad3,\quad4,\quad5,\quad5 \]
Answer: B
16 Comparing Dot Plots
Range

Class A

\[ 6,\quad7,\quad7,\quad8,\quad8,\quad8,\quad9,\quad9 \]

Class B

\[ 4,\quad6,\quad7,\quad8,\quad8,\quad9,\quad10,\quad12 \]

Which statement is true?

A) Class A has a greater range.
B) Class B has a greater range.
C) The two classes have the same range.
D) The range cannot be determined.
Solution
\[ \text{Range}_A=9-6=3 \] \[ \text{Range}_B=12-4=8 \]

Therefore, Class B has the greater range.

Answer: B
🧠 SAT Strategy: For range, find the minimum and maximum first.
09

Frequency Tables

Use frequency to organize repeated observations.

17 Mean from a Frequency Table
Weighted Thinking

The table shows the number of books read by students during a month.

Books read Frequency
12
25
36
43

What is the mean number of books read?

A) 2.25
B) 2.50
C) 2.625
D) 3.00
Solution

Multiply each value by its frequency:

\[ 1*2=2 \] \[ 2*5=10 \] \[ 3*6=18 \] \[ 4*3=12 \]

Total books:

\[ 2+10+18+12=42 \]

Total students:

\[ 2+5+6+3=16 \]

Therefore:

\[ \text{Mean}=\frac{42}{16}=\boxed{2.625} \]
Answer: C
18 Missing Frequency
Algebra + Data

The table shows the number of pets owned by students.

Number of pets Number of students
04
17
2\(x\)
33

There are 20 students in total. What is \(x\)?

A) 4
B) 5
C) 6
D) 7
Solution
\[ 4+7+x+3=20 \] \[ 14+x=20 \] \[ x=\boxed{6} \]
Answer: C
10

Scatterplots & Association

Recognize direction and strength before doing any arithmetic.

POSITIVE ASSOCIATION
19 Identifying Association
Scatterplot

A researcher creates a scatterplot comparing the number of hours students spend studying with their scores on a mathematics assessment. The points generally rise from left to right.

Which statement best describes the association?

A) Negative association
B) Positive association
C) No association
D) The variables have the same numerical values.
Solution

As study time increases, assessment scores generally increase. Therefore, the variables have a positive association.

Answer: B
💡 SAT Strategy: Look at the overall direction of the point cloud—not whether every individual point follows the pattern.
20 Strength of Association
Scatterplot

Two scatterplots show the relationship between \(x\) and \(y\).

  • Plot A: The points lie very close to a downward-sloping line.
  • Plot B: The points generally slope downward but are widely scattered.

Which statement is true?

A) Plot A shows a stronger negative association.
B) Plot B shows a stronger negative association.
C) Both show positive associations.
D) Neither plot shows an association.
Solution

Both plots have a downward trend, so both have negative associations. However, Plot A’s points are much closer to a clear line.

Therefore, Plot A has the stronger negative association.

Answer: A
🧠 Remember: Direction tells you positive or negative. Strength tells you how closely the points follow the pattern.
11

Line of Best Fit

Use a model to make predictions and interpret slope in context.

A MODEL FOR THE GENERAL TREND
\[ y=m*x+b \]
21 Interpreting the Slope
Model

A line of best fit for a data set relating the number of hours a student studies, \(x\), to the student’s test score, \(y\), is

\[ y=6.5*x+42 \]

What does the slope of 6.5 represent?

A) A student who studies 0 hours is predicted to score 6.5 points.
B) Each additional hour of studying is associated with a predicted increase of 6.5 points in the test score.
C) The average test score is 6.5 points.
D) The test score increases by 42 points for every additional hour of studying.
Solution

The slope tells us how much \(y\) changes for a 1-unit increase in \(x\).

Here, \(x\) represents hours studied and \(y\) represents test score. Therefore, an additional hour corresponds to a predicted increase of 6.5 points.

Answer: B
💡 SAT Strategy: Always attach the slope to its units: \[ \frac{\text{change in test score}}{\text{change in hours}} \]
22 Making a Prediction
Model

The line of best fit for the relationship between the age of a car, \(x\), in years and its value, \(y\), in thousands of dollars is

\[ y=28-2.4*x \]

According to the model, what is the predicted value of a 5-year-old car?

A) \$12,000
B) \$16,000
C) \$18,000
D) \$25,600
Solution

A 5-year-old car means \(x=5\).

\[ y=28-2.4*(5) \] \[ y=28-12 \] \[ y=16 \]

Because \(y\) is measured in thousands of dollars:

\[ \boxed{\$16,000} \]
Answer: B
⚠️ Common Trap: The equation gives \(y=16\), but the units are thousands of dollars.
12

Two-Way Tables & Relative Frequency

Compare groups carefully—especially the denominator.

23 Reading a Two-Way Table
Relative Frequency
Activity No Activity Total
Freshmen 48 32 80
Sophomores 54 26 80
Juniors 24 16 40
Total 126 74 200

What percentage of the surveyed juniors participate in an after-school activity?

A) 12%
B) 20%
C) 60%
D) 63%
Solution

The question asks about juniors, so the denominator is the total number of juniors:

\[ 40 \]

Of those 40 students, 24 participate.

\[ \frac{24}{40}*100\%=60\% \]
Answer: C
🧠 SAT Strategy: Circle the group named in the question. That group usually determines your denominator.
24 Comparing Relative Frequencies
Compare
Morning Evening Total
Grade 11 72 48 120
Grade 12 81 99 180

Which statement is true?

A) A greater number of Grade 11 students prefer evening study than Grade 12 students.
B) A greater percentage of Grade 11 students prefer morning study than Grade 12 students.
C) The same percentage of students in both grades prefer morning study.
D) Exactly 27% of Grade 12 students prefer morning study.
Solution

Grade 11:

\[ \frac{72}{120}*100\%=60\% \]

Grade 12:

\[ \frac{81}{180}*100\%=45\% \]

Therefore, a greater percentage of Grade 11 students prefer morning study.

Answer: B
⚠️ Common Trap: Grade 12 has 81 morning students versus 72 for Grade 11, but Grade 12 also has many more students overall. Percentages require the correct denominator.
Part II · Questions 25–40

🧠 The Data Detective Challenge

The SAT doesn’t always ask you to calculate. Sometimes the real challenge is deciding what the numbers actually mean—and what they do not prove.

Calculate less. Reason more.

Before choosing an answer, ask yourself: What is guaranteed by the information given?

Given What information do I actually have?
Meaning What does that information tell me?
Limit What does it NOT tell me?
13

Standard Deviation

Understand spread around the mean.

25 What Does Standard Deviation Tell Us?
🧠 Challenge

Two neighborhoods have the same mean home price.

Mean Standard Deviation
Neighborhood A \$400,000 \$15,000
Neighborhood B \$400,000 \$60,000

Which statement is best supported by the information?

A) Homes in Neighborhood A are generally closer in price to \$400,000 than homes in Neighborhood B.
B) Every home in Neighborhood A costs between \$385,000 and \$415,000.
C) The least expensive home in Neighborhood A costs more than the least expensive home in Neighborhood B.
D) Neighborhood A has a higher median home price.
Solution

Both neighborhoods have the same mean:

\[ \$400,000 \]

But Neighborhood A has the smaller standard deviation:

\[ \$15,000<\$60,000 \]

A smaller standard deviation indicates that the values tend to be less spread out around the mean.

Answer: A
⚠️ Important: Standard deviation does not tell us the exact minimum, maximum, or range. It also does not mean every value lies within one standard deviation of the mean.
26 Comparing Standard Deviations
🧠 Challenge

Two classes receive scores on the same assessment.

Class Mean Standard Deviation
A 80 3
B 80 9

Which statement must be true?

A) Class A has a higher median score.
B) Class B has a greater range.
C) Scores in Class A are less spread out around the mean than scores in Class B.
D) Every score in Class A is between 77 and 83.
Solution

The means are identical, but:

\[ 3<9 \]

Therefore, Class A’s scores have less variability around the mean.

Answer: C
🧠 Remember: Standard deviation describes spread. It does not determine the exact minimum, maximum, median, or range.
14

Comparing Distributions

Read center and spread together.

27 Same Mean, Different Spread
🧠 Challenge

The distributions of delivery times for two restaurants are summarized below.

Restaurant Mean Standard Deviation
A 32 min 4 min
B 32 min 11 min

Which statement is best supported?

A) Restaurant A generally has more consistent delivery times.
B) Restaurant B generally has more consistent delivery times.
C) Restaurant A has a higher average delivery time.
D) Restaurant B has a higher median delivery time.
Solution

Both means are 32 minutes. Restaurant A has the smaller standard deviation:

\[ 4<11 \]

Therefore, Restaurant A’s delivery times tend to be less spread out and more consistent.

Answer: A
28 Same Standard Deviation, Different Mean
🧠 Challenge
Group Mean Standard Deviation
X 68 5
Y 82 5

Which statement is best supported?

A) Group X has greater variability.
B) Group Y has greater variability.
C) The two groups have the same standard deviation.
D) The two groups have the same median.
Solution

The standard deviations are both:

\[ 5 \]

Therefore, the two groups have the same measured amount of spread according to standard deviation.

Answer: C
💡 Think of it this way: Different centers do not necessarily mean different spreads.
15

Transforming a Data Set

What changes when the same number is added to every value?

A DISTRIBUTION SHIFTS — BUT DOES NOT STRETCH
\[ 4,\quad7,\quad10 \] \[ \Downarrow\quad +10 \] \[ 14,\quad17,\quad20 \]

Every value moves by the same amount, so the distances between the values remain unchanged.

29 Adding a Constant
🧠 Challenge

The mean of a data set is 48 and the standard deviation is 6. A new data set is created by adding 10 to every value.

What are the mean and standard deviation of the new data set?

A) Mean = 48, SD = 16
B) Mean = 58, SD = 6
C) Mean = 58, SD = 16
D) Mean = 58, SD = 60
Solution

Adding 10 to every value shifts the entire distribution by 10.

\[ \text{New mean}=48+10=58 \]

But the distances between the data values do not change, so the standard deviation remains 6.

\[ \boxed{\text{Mean}=58,\quad\text{SD}=6} \]
Answer: B
30 What Changes?
🧠 Challenge

A data set has:

  • mean = 72
  • median = 70
  • range = 24
  • standard deviation = 5

A new data set is created by subtracting 8 from every value. Which quantity remains unchanged?

A) Mean
B) Median
C) Range
D) Both mean and median
Solution

Subtracting 8 from every value shifts the distribution but does not change the distances between values.

Therefore, the range remains unchanged.

Answer: C
🧠 SAT Shortcut: Add or subtract the same constant: mean and median change; range and standard deviation do not.
16

Scaling a Data Set

Multiplying every value stretches the distribution.

31 Multiplying Every Value
🧠 Challenge

The mean of a data set is 15, and its standard deviation is 4. Every value in the data set is multiplied by 3.

What are the new mean and standard deviation?

A) Mean = 18, SD = 7
B) Mean = 45, SD = 12
C) Mean = 45, SD = 4
D) Mean = 15, SD = 12
Solution

The mean is multiplied by 3:

\[ 3*15=45 \]

The standard deviation is also multiplied by 3:

\[ 3*4=12 \]
\[ \boxed{\text{Mean}=45,\quad\text{SD}=12} \]
Answer: B
32 Scaling and Range
🧠 Challenge

A data set has a range of 18. A new data set is created by multiplying every value by 5.

What is the range of the new data set?

A) 18
B) 23
C) 90
D) 100
Solution

Multiplying every value by 5 multiplies the distance between the minimum and maximum by 5.

\[ 5*18=\boxed{90} \]
Answer: C
💡 Key distinction: Adding a constant leaves spread unchanged. Multiplying by a constant scales the spread.
17

Outliers & Their Impact

One unusual value can dramatically change some statistics.

ONE EXTREME VALUE CHANGES THE STORY
\[ 300,\quad310,\quad315,\quad325,\quad330 \] \[ \hspace{20px}\downarrow \] \[ 1,500 \] Extreme outlier
33 Effect of an Outlier
🧠 Challenge

A data set contains the following home prices, in thousands of dollars:

\[ 300,\quad310,\quad315,\quad325,\quad330 \]

A home priced at \$1,500,000 is added to the data set. Which statistic is most likely to increase substantially?

A) Median
B) Mean
C) Minimum
D) First quartile
Solution

The new value is extremely large compared with the other values. The mean uses every value, so an extreme value can pull the mean substantially upward.

The median is much less affected by a single extreme value.

Answer: B
34 Which Measure Is More Resistant?
🧠 Challenge

A researcher records the following annual incomes, in thousands of dollars:

\[ 42,\quad45,\quad47,\quad49,\quad52,\quad55,\quad500 \]

Which measure of center would generally be more representative of the income of a typical person in this group?

A) Mean
B) Median
C) Range
D) Standard deviation
Solution

The value 500 is an extreme outlier. Because the mean uses every value, the \$500,000 income pulls the mean upward.

The median is much less affected by the extreme value.

Answer: B
💡 Big Idea: For distributions with strong outliers, the median often gives a better picture of a typical value.
18

What Must Be True?

Separate guaranteed conclusions from merely possible ones.

35 Comparing Two Distributions
🔥 Advanced
Group A Group B
Mean 75 75
Standard deviation 4 10

Which statement must be true?

A) The median score for Group A is 75.
B) Group A has a smaller range than Group B.
C) Scores in Group A have less variability around the mean than scores in Group B.
D) Every score in Group A is between 71 and 79.
Solution

Both groups have the same mean:

\[ 75 \]

But:

\[ 4<10 \]

Therefore, Group A has less variability around the mean.

Answer: C
⚠️ Why the other choices fail: Standard deviation does not determine the median, range, or exact limits of every score.
36 A Transformation
🔥 Advanced

A data set has a mean of 52 and a range of 18. A new data set is created by adding 7 to every value.

Which statement must be true?

A) The new mean is 59 and the new range is 25.
B) The new mean is 52 and the new range is 18.
C) The new mean is 59 and the new range is 18.
D) The new mean is 52 and the new range is 25.
Solution

Adding 7 to every value shifts the entire distribution by 7.

\[ 52+7=59 \]

But the range measures the distance between the minimum and maximum. Both values increase by 7, so their difference remains unchanged:

\[ (\text{maximum}+7)-(\text{minimum}+7) \] \[ =\text{maximum}-\text{minimum} \] \[ =18 \]
Answer: C
💡 SAT Shortcut: Adding a constant moves the distribution but does not stretch it.
19

What Can Be Inferred?

Don’t claim more than the data actually support.

37 Standard Deviation Does Not Give Exact Limits
🔥 Advanced

A data set has a mean of 100 and a standard deviation of 12. Which statement is best supported?

A) Every value in the data set is between 88 and 112.
B) Most values must be exactly 12 units from the mean.
C) The data values tend to vary around a center of 100, with a standard deviation of 12.
D) The range of the data set is 24.
Solution

Standard deviation describes the amount of variability around the mean. It does not tell us the exact minimum or maximum.

Therefore, we cannot conclude that all values lie between \(88\) and \(112\).

Answer: C
⚠️ Major SAT Trap: Mean ± standard deviation is not automatically the minimum and maximum.
38 What Does the Data Actually Support?
🔥 Advanced
Group A Group B
Mean 78 84
Median 80 83
Standard deviation 5 5

Which statement can be inferred from the table?

A) Every student in Group B scored higher than every student in Group A.
B) Group B had a higher mean score than Group A.
C) Group B had twice the variability of Group A.
D) The highest score in Group B was 6 points higher than the highest score in Group A.
Solution

The table directly gives:

\[ \text{Mean}_A=78 \] \[ \text{Mean}_B=84 \]

Since:

\[ 84>78 \]

Group B had a higher mean score.

Answer: B
⚠️ Data Detective Rule: A statistical summary describes a group. It usually does not tell you what happened to every individual observation.
20

Real-World Statistical Reasoning

The final challenge: combine several statistical ideas.

39 House Prices
🏆 Final Stretch

Two neighborhoods have the following statistics for home prices.

Neighborhood A Neighborhood B
Mean price \$420,000 \$420,000
Median price \$415,000 \$410,000
Standard deviation \$18,000 \$52,000

Which statement is best supported?

A) Neighborhood A has less variation in home prices than Neighborhood B.
B) Every home in Neighborhood A costs between \$402,000 and \$438,000.
C) The least expensive home in Neighborhood A costs more than the least expensive home in Neighborhood B.
D) Neighborhood A has no homes priced below \$400,000.
Solution

The two neighborhoods have the same mean:

\[ \$420,000 \]

But:

\[ \$18,000<\$52,000 \]

Therefore, Neighborhood A has a smaller standard deviation and less variability around the mean.

Answer: A
⚠️ Data Detective Move: The table does not give the minimum or maximum home price. Therefore, B, C, and D make claims that cannot be guaranteed.
40 The Final Data Detective Challenge
🏆 FINAL BOSS

A researcher compares the weekly exercise times of two groups.

Group X Group Y
Mean 150 min 150 min
Median 148 min 149 min
Standard deviation 12 min 30 min

The researcher then adds 20 minutes to every observation in both groups.

Which statement is true about the resulting data sets?

A) Both means increase by 20 minutes, and both standard deviations increase by 20 minutes.
B) Both means increase by 20 minutes, while both standard deviations remain unchanged.
C) Both means remain unchanged, while both standard deviations increase by 20 minutes.
D) The means and standard deviations of both groups remain unchanged.
Solution

Adding 20 minutes to every observation shifts each distribution by 20 minutes.

For both groups:

\[ 150+20=170 \]

So both means increase by 20 minutes.

However, the distance between observations does not change. Therefore, the standard deviations remain:

\[ 12\text{ min} \qquad\text{and}\qquad 30\text{ min} \]
Answer: B
🏆 Final SAT Shortcut: When the same number is added to every value:

Mean changes.    Median changes.    Range stays the same.    Standard deviation stays the same.
🏆

Data Detective Complete.

You’ve gone from calculating basic statistics to reasoning about what distributions, standard deviations, transformations, and real-world data actually tell you.

✓ Calculate the mean and median
✓ Read a box plot
✓ Interpret a histogram
✓ Analyze a dot plot
✓ Interpret standard deviation
✓ Recognize the effect of an outlier
✓ Understand data transformations
✓ Determine what must be true
✓ Distinguish inference from assumption
✓ Reason from real-world statistics

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