Histograms and Frequency Distributions

1. What a Histogram Actually Shows

A histogram summarizes numerical data by grouping values into intervals, sometimes called bins. Each bar represents the observations that fall inside one interval.

The Four Things to Read First

1. Horizontal axis

What numerical quantity is being grouped?

2. Vertical axis

What does the height represent: frequency, relative frequency, or percent?

3. Interval

Which numerical values belong to this bar?

4. Bar height

How many observations, or what proportion of observations, are represented?

Original SATMath800 example: the bar over 60–70 has frequency 11.

Core idea: A histogram does not show individual values one by one. It shows how many observations are grouped into each numerical interval.

2. Bins and Intervals — The Most Important Reading Skill

A histogram groups numerical values into intervals. The exact meaning of an interval depends on how the problem defines it.

Example of an Interval Definition

Suppose a problem states that the first interval contains integers greater than or equal to 10 but less than 20.

\[ 10 \le x < 20 \]

Therefore, 10 is included, while 20 is not.

Do not assume the endpoints. If the question explicitly defines the intervals, use that definition.

Worked Example • Reading an Interval

A histogram has an interval labeled 30–40. The problem states that this interval represents integers greater than or equal to 30 but less than 40. Which value belongs to this interval?

A) 29

B) 30

C) 40

D) 41

Solution

1

The interval is \(30 \le x < 40\).

2

30 satisfies both conditions: \(30 \ge 30\) and \(30 < 40\).

3

40 does not qualify because the interval stops before 40.

Answer: B

3. Frequency vs. Relative Frequency

Frequency is a count. It tells you how many observations fall in an interval.

Relative frequency is a proportion or percentage. It tells you what fraction of all observations fall in an interval.

\[ \text{Relative frequency} = \frac{\text{frequency}}{\text{total number of observations}} \]

Frequency

If 18 students scored from 70 to 80, the frequency is \(18\).

Relative Frequency

If 18 of 60 students scored from 70 to 80, then

\[ \frac{18}{60}=0.30=30\% \]

If the vertical axis says Frequency, read heights as counts. If it says Relative Frequency or Percent, read the height as a proportion or percentage instead.

4. How to Read a Histogram Precisely

Use this routine whenever you see a histogram.

Step 1

Read the horizontal axis.

Step 2

Read the vertical axis.

Step 3

Read the interval definition.

Step 4

Read the relevant bar height.

Step 5: Translate the graph into words before doing arithmetic. For example: “There are 11 observations from 60 up to but not including 70.”

Worked Example • Direct Reading

In the histogram from Section 1, how many observations have scores from 60 up to but not including 70?

A) 6

B) 7

C) 11

D) 12

Solution

1

Locate the interval \(60 \le x < 70\).

2

The bar for that interval has height \(11\).

3

Because the vertical axis is frequency, the height means 11 observations.

Answer: C

5. Finding the Most Common Interval

The interval with the tallest bar has the greatest frequency. This interval is sometimes called the modal interval.

The tallest bar identifies the interval with the greatest frequency.

Worked Example • Modal Interval

Which interval contains the greatest number of observations?

A) 10–20

B) 20–30

C) 30–40

D) 40–50

Solution

1

Compare the bar heights: 4, 9, 13, and 8.

2

The largest frequency is \(13\).

3

That frequency belongs to 30–40.

Answer: C

6. Finding the Total Number of Observations

If a histogram shows frequency and all intervals are included, add the bar heights to find the total number of observations.

\[ \text{Total} = f_1+f_2+f_3+\cdots+f_n \]

Worked Example • Total Frequency

A histogram has five bars with frequencies \(4,7,9,6,4\). How many observations are in the data set?

Solution

1

Add every bar height.

2

\(4+7+9+6+4=30\).

Answer: 30 observations

Important: This works when the vertical axis is frequency. If the vertical axis is a percentage, add percentages to get 100%, not the number of observations.

7. What a Histogram Can Tell You About Center

A histogram can show where observations are concentrated, but grouping the data into intervals may prevent you from finding an exact mean or median.

Mean

The exact average requires the individual values or enough information to reconstruct their total.

Median

The exact middle value may be hidden inside an interval.

Typical location

The histogram can still show where the data are concentrated.

A histogram can make the general location of a distribution visible without revealing every value.

SAT rule: If the graph groups values into intervals, do not invent an exact mean or median unless the question gives enough additional information to determine it.

8. Range and Spread from a Histogram

Spread describes how widely the data are distributed. A histogram can show the overall spread, but the exact minimum or maximum may not be visible.

Range

\(\text{Range}=\text{maximum}-\text{minimum}\).

Histogram limitation

If the first occupied interval is 40–50, the exact minimum could be many values within that interval.

Worked Example • What Can Be Determined?

A histogram’s first occupied interval is 40–50 and its last occupied interval is 80–90. Can the exact range be determined from the histogram alone?

A) Yes, it is exactly 50.

B) Yes, it is exactly 40.

C) No, because the exact minimum and maximum are not known.

D) No, because histograms never show spread.

Solution

1

The first occupied interval only tells us that the minimum is somewhere in 40–50.

2

The last occupied interval only tells us that the maximum is somewhere in 80–90.

3

Therefore, the exact range cannot be determined from the grouped information alone.

Answer: C

9. Comparing Two Histograms

When two histograms are compared, first make sure the horizontal scales and interval definitions are comparable. Then compare concentration, center, spread, and shape.

Center

Where is the data concentrated?

Spread

How wide is the distribution?

Shape

Is it symmetric, skewed, clustered, or separated by gaps?

Frequency

How many observations are in important intervals?

The first question is not “Which looks taller?” It is “What does the shared scale show?”

Comparison routine: 1) confirm the scales, 2) locate the concentration, 3) compare spread, 4) compare shape, and 5) make only a claim supported by the graphs.

10. Distribution Shape

A histogram’s shape is the overall pattern formed by its bars. Common descriptions include approximately symmetric, skewed, clustered, and separated by gaps.

Approximately symmetric

The left and right sides have a similar overall pattern around a central region.

Right-skewed

Most observations are relatively lower, with a longer tail extending toward larger values.

Left-skewed

Most observations are relatively higher, with a longer tail extending toward smaller values.

Shape is about the overall pattern, not one individual bar.

11. Histograms and Unusual Values

A histogram may suggest an unusual region when one observation or a small group lies far from the main concentration. But a grouped graph may hide the exact value of that observation.

A separated bar can suggest an unusual region, but grouped data still limit exact conclusions.

Be precise: “This bar is separated from the main cluster” is a graph-supported statement. “The value is exactly 92” is not supported unless the graph or problem gives that exact value.

12. How to Construct a Histogram from a Set of Numbers

A histogram starts with the raw data. The basic process is simple: choose intervals, count how many values fall in each interval, and draw bars whose heights match those counts.

Original SATMath800 Example

Suppose the data are:

\[ 12,\ 14,\ 17,\ 21,\ 22,\ 24,\ 27,\ 31,\ 34,\ 36,\ 38,\ 42 \]

We will use four simple intervals: 10–19, 20–29, 30–39, and 40–49.

Step 1 — Choose the intervals

Interval	Values that belong	Frequency
10–19	12, 14, 17	3
20–29	21, 22, 24, 27	4
30–39	31, 34, 36, 38	4
40–49	42	1

Step 2 — Draw one bar for each interval

The frequencies are 3, 4, 4, and 1, so the histogram contains 12 observations in total.

Why this construction works

1

Every original number is assigned to exactly one interval.

2

Each interval is counted: 3, 4, 4, and 1.

3

The bar heights are those frequencies.

4

The frequencies add to \(3+4+4+1=12\), matching the 12 original observations.

Construction shortcut: Raw data → choose intervals → count → draw bars. If you cannot explain where a bar height came from, stop and recount the data.

13. Frequency Tables → Histograms

A frequency table and a frequency histogram contain the same basic information: the intervals and their counts. The histogram turns the table into a visual pattern.

Interval	Frequency
0–9	3
10–19	6
20–29	8
30–39	5

Worked Example • Table to Graph

Which interval should have the tallest bar when the table above is represented as a frequency histogram?

A) 0–9

B) 10–19

C) 20–29

D) 30–39

Solution

1

The tallest bar corresponds to the greatest frequency.

2

The frequencies are 3, 6, 8, and 5.

3

The largest is 8, which belongs to 20–29.

Answer: C

14. Histograms vs. Bar Graphs

Both displays use rectangular bars, but they summarize different kinds of information.

Histogram

Used for numerical data.
Values are grouped into intervals.
Adjacent numerical intervals normally touch.
Bar height represents frequency or another numerical measure.

Bar Graph

Often used for categories.
Each bar represents a category.
Categories are distinct rather than continuous intervals.
Gaps commonly separate categories.

Do not use “bars touch” as your only test. First identify what the horizontal axis represents. Numerical intervals are the key feature of a histogram.

15. Changing Bin Width — Why the Picture Can Change

The same raw data can produce different-looking histograms if the interval width changes. The data did not change; the grouping changed.

Different bin widths can make the same distribution look smoother or more detailed.

Key idea: A visual difference does not automatically mean the underlying data changed. Check the intervals first.

16. Histograms in SAT Word Problems

In a context question, translate the graph into the language of the situation. The arithmetic is often simple; the reading is the real challenge.

Worked Example • Context + Histogram

A study records the number of minutes 40 students spend exercising each week. A frequency histogram shows that 14 students exercise from 60 up to but not including 90 minutes. How many students are represented by that interval?

A) 14

B) 26

C) 40

D) 54

Solution

1

The vertical axis is frequency.

2

The bar for 60–90 has height 14.

3

Therefore, 14 students are in that interval.

Answer: A

17. What a Histogram Cannot Tell You Exactly

Grouping information is useful, but it removes some detail. A strong SAT student knows when the graph supports an exact answer and when it does not.

Question	Usually determined exactly?
Which interval has the highest frequency?	Yes
Total frequency, if all bars are shown	Yes
Exact minimum inside the first occupied interval	Not necessarily
Exact maximum inside the last occupied interval	Not necessarily
Exact mean from grouped intervals alone	Not necessarily
Exact median from grouped intervals alone	Not necessarily

Precision principle: Never extract more precision from a graph than the graph contains.

18. SAT Traps You Should Recognize Instantly

Trap 1: Reading a Bar as a Value

The bar height is usually a frequency or relative frequency, not the numerical value on the horizontal axis.

Trap 2: Ignoring Interval Definitions

Do not automatically include both endpoints. Follow the interval definition given by the problem.

Trap 3: Assuming the Tallest Bar Gives the Mean

The tallest bar identifies the most frequent interval, not the exact mean.

Trap 4: Assuming a Gap Gives an Exact Outlier

A gap shows separation in grouped data; it does not reveal an exact value.

Trap 5: Adding Percentages as Counts

If the axis is percent, convert using the total number of observations before finding a count.

Trap 6: Comparing Heights Without Checking Scales

Before comparing two histograms, verify that their horizontal intervals and vertical scales are comparable.

19. Your SAT Decision Framework

OBSERVE

Read both axes and the interval labels.

IDENTIFY

Decide whether the height is frequency, relative frequency, or percent.

READ

Locate the relevant interval or intervals.

REASON

Add, compare, convert, or interpret only what the graph supports.

VERIFY

Check units, endpoints, totals, and whether your conclusion is exact or approximate.

ANSWER

Choose the statement that matches the graph—not the statement that sounds most sophisticated.

Computer Scientist’s Approach: Observe → Identify → Decompose → Reason → Solve → Verify. Histograms reward this approach because the visual itself contains structured information.

20. Quick Concept Check

Quick Check

1. What does the height of a frequency histogram bar represent?

2. If an interval is \(20 \le x < 30\), is 30 included?

3. What does the tallest bar tell you?

4. Can a histogram always give the exact mean?

5. Why can changing bin width change the appearance of a histogram?

Answers

1

The number of observations in that interval.

2

No. The interval stops before 30.

3

The interval with the greatest frequency.

4

No. Grouping may prevent the exact mean from being determined.

5

Different bin widths group the same observations differently, changing the visual summary.

20. Advanced Practice — SAT-Style Histogram Problems

What makes these questions “advanced”?

The hardest histogram questions are rarely about reading one bar. They combine a histogram with another idea: mean or median, a change to the data set, interval endpoints, possible values, comparison of two distributions, or a statement that must be true.

The questions below are original SATMath800 questions. Their structures are modeled on the kinds of reasoning used in official SAT practice, but the numbers, contexts, data sets, answer choices, and visuals are original.

Question 1 • Reading Frequency and Percentage

The histogram summarizes the number of books read by 40 students during a summer. The interval from 4 up to but not including 6 contains 10 students. What percentage of the students read from 4 up to but not including 6 books?

All 40 students are represented by the five bars.

A) 10%

B) 20%

C) 25%

D) 40%

Solution

1

The interval from 4 up to but not including 6 has frequency 10.

2

There are 40 students in all.

3

Convert the fraction to a percentage: \(\dfrac{10}{40}=0.25=25\%\).

Answer: C

Question 2 • Locating the Median from a Histogram

The histogram represents 41 measurements. Which interval contains the median of the data set?

A) 10–20

B) 20–30

C) 30–40

D) 40–50

Solution

1

There are 41 observations, so the median is the 21st value when the data are ordered.

2

Cumulative frequencies are 5, then 13, then 27.

3

The 21st value falls after the first 13 observations but before the first 28 observations. Therefore it is in the 30–40 interval.

Answer: C

Question 3 • Adding a Value: Mean vs. Median

The histogram summarizes data set A, which contains 40 values. A new value of 18 is added to data set A to create data set B with 41 values. Which of the following must be true?

Frequencies: 4, 6, 18, and 12. Total = 40.

I. The mean of data set B is less than the mean of data set A.

II. The median of data set B is less than the median of data set A.

A) I only

B) II only

C) I and II

D) Neither I nor II

Solution

1

All 40 original values are greater than 18, so the added value is below every original value. Adding a value below the original mean makes the new mean smaller. Thus I must be true.

2

For 41 values, the median is the 21st value. After inserting 18 at the beginning, that 21st value is the 20th value of data set A.

3

In data set A, the first 10 values are below 40, and the next 18 values are in the 40–50 interval. Therefore both the 20th and 21st values of A are in the 40–50 interval.

4

The 20th and 21st values could be equal. If they are equal, the two medians are equal. Therefore II does not have to be true.

Answer: A

Question 4 • Smallest Possible Difference Between Means

Two data sets of 20 integers each are summarized in the histograms shown. For each histogram, an interval such as 40–50 represents integers greater than or equal to 40 but less than 50. What is the smallest possible difference between the mean of data set A and the mean of data set B?

In both histograms, the four frequencies are 2, 5, 8, and 5.

A) 0

B) 1

C) 4

D) 5

Solution

1

To make the difference as small as possible, make A as small as possible and B as large as possible.

2

Because the values are integers, the smallest values allowed in A’s intervals are 40, 50, 60, and 70.

3

The largest values allowed in B’s intervals are 44, 54, 64, and 74.

4

Thus the smallest possible mean of A is \(\dfrac{2(40)+5(50)+8(60)+5(70)}{20}=58\).

5

The largest possible mean of B is \(\dfrac{2(44)+5(54)+8(64)+5(74)}{20}=54\).

6

Smallest possible difference = \(58-54=4\).

Answer: C

Question 5 • Which Mean Could Be Possible?

A data set contains 30 integers. The histogram shows the distribution of the values. Which of the following could be the mean of the data set?

A) 22

B) 25

C) 33

D) 36

Solution

1

The smallest possible sum occurs when the values are 10, 20, and 30 within their respective intervals: \(5(10)+10(20)+15(30)=700\).

2

The largest possible sum is \(5(19)+10(29)+15(39)=970\).

3

So the mean must be between \(700/30\approx23.33\) and \(970/30\approx32.33\).

4

Among the choices, only 25 lies in this possible range. It is achievable; for example, five values can be 19, one value can be 25, nine values can be 20, and fifteen values can be 30. Their sum is \(750\), so the mean is 25.

Answer: B

Question 6 • Comparing Two Distributions

The two histograms summarize data sets A and B. Both data sets contain 20 values. Which statement must be true?

A) The mean of A is greater than the mean of B.

B) The median of A is greater than the median of B.

C) The ranges of A and B must be equal.

D) The standard deviations of A and B must be different.

Solution

1

For A, the cumulative frequencies are 2, 7, 15, 20. The 10th and 11th values are therefore in the 30–40 interval.

2

For B, the cumulative frequencies are 5, 13, 18, 20. The 10th and 11th values are in the 20–30 interval.

3

Therefore the median of A must be greater than the median of B.

4

The histogram does not give the exact endpoints inside the occupied intervals, so the exact ranges and standard deviations are not forced to have the relationships stated in the other choices.

Answer: B

Question 7 • Student-Produced Response • Cumulative Frequency

A histogram summarizes 50 observations. The frequencies of the five intervals are 6, 9, 15, 12, and 8, respectively. The intervals are 0–10, 10–20, 20–30, 30–40, and 40–50, where the lower endpoint is included and the upper endpoint is not included.

What is the smallest possible number of observations that are less than 30?

Solution

1

“Less than 30” includes every observation in the first three intervals.

2

Add their frequencies: \(6+9+15=30\).

3

Because the intervals end at 30 and the upper endpoint is not included, every value in the 20–30 interval is still less than 30.

Answer: 30

Question 8 • Mean vs. Tallest Bar

A student says, “The 30–40 interval has the tallest bar, so the mean of the data set must be between 30 and 40.” The histogram shows 50 observations. Which statement best explains why the student’s conclusion is not necessarily true?

A) The tallest bar identifies the median, not the mean.

B) The tallest bar identifies the most frequent interval, but the mean depends on all the values.

C) The mean is always in the interval with the greatest frequency.

D) A histogram cannot be used to make any statement about the mean.

Solution

1

The tallest bar tells us where the greatest number of observations are concentrated: here, in the 30–40 interval.

2

The mean is computed from the values of all observations, not just the most common interval.

3

Values in all the other intervals also contribute to the mean, so the tallest bar alone does not determine the mean.

Answer: B

Question 9 • Must Be True • Median After Adding a Value

A histogram summarizes a data set of 51 values. The frequencies in the four intervals 0–10, 10–20, 20–30, and 30–40 are 7, 12, 18, and 14, respectively. A new value of 5 is added to the data set. Which statement must be true about the median of the new data set?

A) The new median is less than 20.

B) The new median is between 20 and 30.

C) The new median is between 30 and 40.

D) The new median must equal 25.

Solution

1

The original data set has 51 values. After adding one value, the new data set has 52 values.

2

For 52 values, the median is the average of the 26th and 27th values.

3

The added value 5 is inserted at the beginning, so the 26th and 27th values of the new set correspond to the 25th and 26th values of the original set.

4

In the original set, the first 19 values are below 20, and the next 18 values are in the 20–30 interval. Therefore the 25th and 26th values are both between 20 and 30.

5

The average of two numbers that are both between 20 and 30 is also between 20 and 30.

Answer: B

Question 10 • Advanced Comparison • Possible Mean

Data set A and data set B each contain 24 integers. The histograms have the same frequencies in corresponding bars, but every interval for B is 10 units lower than the corresponding interval for A. For example, A’s 50–60 interval corresponds to B’s 40–50 interval. What is the smallest possible difference between the mean of A and the mean of B?

The four frequencies are 3, 7, 9, and 5 in both data sets.

A) 0

B) 1

C) 10

D) 24

Solution

1

Every interval in A is 10 units higher than the corresponding interval in B.

2

To minimize the difference, use the smallest allowed integer in every A interval and the largest allowed integer in every B interval.

3

The smallest possible mean of A is \(\dfrac{3(50)+7(60)+9(70)+5(80)}{24}=\dfrac{1600}{24}\approx66.67\).

4

The largest possible mean of B is \(\dfrac{3(49)+7(59)+9(69)+5(79)}{24}=\dfrac{1576}{24}\approx65.67\).

5

The smallest possible difference is therefore \(\dfrac{1600}{24}-\dfrac{1576}{24}=\dfrac{24}{24}=1\).

Answer: B

Question 11 • Individual-Value Histogram • Arithmetic Mean

The histogram shows the number of books read by each of 30 students during a summer program. Each bar represents an individual numerical value rather than an interval. What is the arithmetic mean number of books read by the students?

The frequencies for 2, 3, 4, 5, 6, and 7 books are 2, 4, 7, 8, 6, and 3, respectively.

A) 4.2

B) 4.5

C) 4.7

D) 5.1

Solution

1

Use each value multiplied by its frequency. This is a weighted mean.

2

The total number of students is 2+4+7+8+6+3=30.

3

The total number of books is 2*2+3*4+4*7+5*8+6*6+7*3=141.

4

Therefore, the mean is 141/30=4.7.

Answer: C

Question 12 • Advanced • Which Mean and Median Could Be Possible?

A data set contains 30 integers. Its histogram has 5 values in the interval 0–10, 8 values in 10–20, 10 values in 20–30, and 7 values in 30–40. The lower endpoint is included and the upper endpoint is excluded in each interval. Which pair could be the mean and median of the data set, respectively?

Frequencies: 5, 8, 10, and 7. There are 30 integers in all.

A) Mean = 26, median = 24

B) Mean = 24, median = 22

C) Mean = 24, median = 31

D) Mean = 18, median = 29

Solution

1

Because there are 30 values, the median is the average of the 15th and 16th values. The first 13 values are below 20, and the next 10 values are in 20–30. Therefore, the median must be between 20 and 29.

2

The smallest possible total is 5*0+8*10+10*20+7*30=450, so the mean is at least 450/30=15.

3

The largest possible total is 5*9+8*19+10*29+7*39=760, so the mean is at most 760/30≈25.33. Thus A is impossible because a mean of 26 is too large, and C is impossible because a median of 31 is outside the 20–29 interval.

4

For D, a median of 29 forces the 15th and 16th values to be 29. Even using the smallest possible values elsewhere, the total is at least 5*0+8*10+(20+29+8*29)+7*30=571, giving a mean greater than 18.

5

B is achievable. For example, use five 9s, eight 19s, the ten middle values 20,20,24,26,26,26,26,27,27,28, and seven 39s. The total is 720, so the mean is 24. The 15th and 16th values are 20 and 24, so the median is 22. Thus B is achievable.

Answer: B

Advanced Histogram Checklist: Read the interval definition first. Then count or locate cumulative frequency. For means, remember that a grouped interval gives a range of possible values, not one exact value. For medians, locate the middle position using cumulative frequency. When a value is added, track how the ordered positions change. When comparing two histograms, compare what must be true—not what merely could be true.

Latest Updates

Categories

1. What a Histogram Actually Shows

The Four Things to Read First

1. Horizontal axis

2. Vertical axis

3. Interval

4. Bar height

2. Bins and Intervals — The Most Important Reading Skill

Example of an Interval Definition

Solution

3. Frequency vs. Relative Frequency

Frequency

Relative Frequency

4. How to Read a Histogram Precisely

Step 1

Step 2

Step 3

Step 4

Solution

5. Finding the Most Common Interval

Solution

6. Finding the Total Number of Observations

Solution

7. What a Histogram Can Tell You About Center

Mean

Median

Typical location

8. Range and Spread from a Histogram

Range

Histogram limitation

Solution

9. Comparing Two Histograms

Center

Spread

Shape

Frequency

10. Distribution Shape

Approximately symmetric

Right-skewed

Left-skewed

11. Histograms and Unusual Values

12. How to Construct a Histogram from a Set of Numbers

Original SATMath800 Example

Step 1 — Choose the intervals

Step 2 — Draw one bar for each interval

Why this construction works

13. Frequency Tables → Histograms

Solution

14. Histograms vs. Bar Graphs

Histogram

Bar Graph

15. Changing Bin Width — Why the Picture Can Change

16. Histograms in SAT Word Problems

Solution

17. What a Histogram Cannot Tell You Exactly

18. SAT Traps You Should Recognize Instantly

Trap 1: Reading a Bar as a Value

Trap 2: Ignoring Interval Definitions

Trap 3: Assuming the Tallest Bar Gives the Mean

Trap 4: Assuming a Gap Gives an Exact Outlier

Trap 5: Adding Percentages as Counts

Trap 6: Comparing Heights Without Checking Scales

19. Your SAT Decision Framework

OBSERVE

IDENTIFY

READ

REASON

VERIFY

ANSWER

20. Quick Concept Check

Answers

20. Advanced Practice — SAT-Style Histogram Problems

What makes these questions “advanced”?

Solution

Solution

Solution

Solution

Solution

Solution