Histograms

A key idea in statistics is that a histogram communicates frequency using the area of each bar, not just the height. In other words:

$\text{Area of a bar}\propto \text{Frequency}$

Each histogram bar is a rectangle, so its area is found by:

$\text{Area}=\text{Width}\times \text{Height}$

For a histogram:

• The Width is the class width (how wide the interval is on the x-axis).
• The Height is the frequency density.

So we can write: $\text{Frequency}=\text{Class Width}\times \text{Frequency Density}$

When some class intervals are wider than others, using frequency as the height would distort the picture. A wide interval could look far more important simply because its bar covers more horizontal distance, even if it contains only a few values.

Using frequency density as the height fixes this, because it makes the bar’s total area accurately represent how many data points are in that interval.

Here’s why the other statements don’t explain frequency density:

• Option B: Whether you include an axis break is a scaling/formatting decision and doesn’t determine how bar heights are calculated.
• Option C: Histogram bars touch because the variable is continuous, not because of frequency density.
• Option D: Histograms are for continuous numerical data; categorical groups are shown with bar charts instead.

We need frequency density so that the histogram follows its defining rule: the area of each bar correctly represents the frequency, even when class widths are different.

In statistics, specifically in Topic: Statistical Measures and Data Representation, histograms display continuous data. If the class widths on the x-axis are not equal, then plotting frequency as height would give a misleading visual because wider bars would automatically look “bigger.” Using frequency density corrects for the different widths.

Option D talks about categorical information. But histograms are only for continuous numerical data. Categories (like types of pets or shoe brands) are shown with bar charts, not histograms. So this cannot be the reason.

Option C suggests frequency density is used so bars touch. In a histogram, bars touch because the intervals are consecutive parts of a continuous scale. This is true regardless of whether you use frequency or frequency density, and regardless of equal/unequal widths. So C is not the reason.

Option B mentions an axis break, which is used to skip part of a scale when there’s a large gap. That has nothing to do with adjusting heights for different class widths. Frequency density neither avoids nor forces an axis break.

After eliminating B, C, and D, only Option A remains. A histogram’s rule is that area represents frequency.

Frequency density is defined by: $\text{Frequency Density}=\frac{\text{Frequency}}{\text{Class Width}}$

Rearranging shows: $\text{Area}=\text{Width}\times \text{Frequency Density}=\text{Frequency}$

So even if one class interval is much wider than another, the bar’s area still matches the number of data points.

A histogram represents continuous data grouped into class intervals. The important rule is that the area of each bar (not just its height) must be proportional to the frequency.

We rely on the rectangle area formula: $\text{Area}=\text{Width}\times \text{Height}$

For histograms we want the area to correspond to frequency, and we name the height Frequency Density, giving: $\text{Frequency}=\text{Class Width}\times \text{Frequency Density}$

If class widths are unequal, using frequency as the height would overemphasize wider classes because their areas would become too large.

Rearranging the relationship gives the correct height to use: $\text{Frequency Density}=\frac{\text{Frequency}}{\text{Class Width}}$

This ensures that each bar’s total area stays faithful to the frequency, no matter the interval width.

The remaining options don’t describe why we adjust heights using density:

• Option B: Axis breaks are about skipping part of a scale, not about correcting for different class widths.
• Option C: Bars touch because the variable is continuous, not because of frequency density.
• Option D: Histograms are for numerical (quantitative) data; categorical data uses bar charts.

Because the mathematics of histograms ties frequency to area, frequency density is required whenever class widths are unequal so that the bar areas correctly represent the frequencies.

Relative frequency needs the count in the class you’re focusing on. The lightest class interval is $19.5<x\le 20.0$.

From the table, that interval has frequency $3$.

Relative frequency is based on the whole set. The question states there are $50$ chicks in total.

Relative frequency is found using:

$\text{Relative Frequency}=\frac{\text{Frequency of Interval}}{\text{Total Frequency}}$

Substitute the values:

$\text{Relative Frequency}=\frac{3}{50}$

As a decimal, $\frac{3}{50}=0.06$.

Compare with the options:

• Option A shows the count ($3$), not a relative frequency.
• Option C ($0.03$) could happen if the division/decimal is done incorrectly.
• Option D ($0.60$) is what you’d get if you shift the decimal too far (treating $6\%$ as $60\%$).

Since $\frac{3}{50}=0.06$, the correct choice is Option B.

To form a relative frequency, we need two numbers: the frequency in the target class and the total frequency. For the lightest class interval:

• Frequency of $19.5<x\le 20.0$: 3
• Total frequency ($n$): 50

Relative frequency is the part out of the whole:

$\text{Relative Frequency}=\frac{\text{Frequency of Interval}}{\text{Total Frequency}}=\frac{3}{50}$

This is a standard idea in Statistical Measures and Data Representation.

To convert easily to a decimal, make the denominator 100. Since $50\times 2=100$, multiply top and bottom by 2:

$\frac{3\times 2}{50\times 2}=\frac{6}{100}$

Now convert:

$\frac{6}{100}=0.06$

• Option A is 3, which is the frequency (not relative frequency).
• Option C is 0.03, which could come from using the wrong total or halving by mistake.
• Option D is 0.60, which is a decimal-point placement error.

The proportional reasoning gives $0.06$, so Option B is correct.

Start by reading the table for the lightest class, $19.5<x\le 20.0$.

That class has frequency 3, and the full dataset contains 50 chicks.

A relative frequency is a fraction/decimal of the total, so it must lie between 0 and 1.

Option A is 3, which is the raw number in the class, so it cannot be the relative frequency.

Use the definition:

$\text{Relative Frequency}=\frac{\text{Frequency of interval}}{\text{Total frequency}}$

Substitute:

$\text{Relative Frequency}=\frac{3}{50}$

Compare $\frac{3}{50}$ with the remaining decimals:

• Option C (0.03): equals $\frac{3}{100}$, which happens if someone incorrectly uses 100 as the total instead of 50.
• Option D (0.60): equals $\frac{30}{50}$ or $\frac{3}{5}$, which often comes from moving the decimal the wrong way.

Convert:

$\frac{3}{50}=\frac{6}{100}=0.06$

This matches Option B.

The problem points out that the two schools do not have similar totals: one dataset comes from 50 students and the other from 500 students.

If we try to compare travel-time patterns between these two groups, using plain counts will distort the comparison because the larger school will automatically produce larger bar heights in every interval.

Using Frequency means using the actual number of students in each time bin. With 500 students, the bars will typically be about 10 times as tall as those from a sample of 50.

That difference in totals prevents a fair comparison of the distribution’s shape (where most students are) and spread (how varied the times are).

Consider the remaining choices:

• Class width describes the size of each interval along the horizontal axis (the bins), so it isn’t the vertical measure we need for comparing heights.
• Cumulative frequency adds counts as you move across bins. Since one school totals 50 and the other totals 500, the cumulative graphs would still finish at very different values, so the data still isn’t put on a common scale.

A fair comparison uses the Normalization Principle: convert each class count into a fraction of the whole group so both schools are measured on the same scale (from 0 to 1, or 0% to 100%).

This is done with Relative Frequency:

$\text{Relative Frequency}=\frac{\text{Frequency}}{\text{Total Population}}$

With this, each bar shows the share of students in that travel-time interval, independent of how many students were surveyed.

Relative frequency makes the vertical axis proportional, so both distributions can be compared on equal footing (each representing a total of 1 overall).

We are comparing two histograms of travel times, but the groups are very different in size:

• School A: 50 students
• School B: 500 students

To make a fair comparison, the vertical axis must represent the proportion in each interval, not just the number of students.

Class width is about how wide each bin is on the x-axis (for example, 0–10 minutes, 10–20 minutes). It does not describe bar height and does not adjust for different sample sizes, so it cannot create a fair vertical scale.

Frequency is the direct count in each bin. Because School B has 10 times as many students, its bars would generally be much taller even if the travel-time pattern is similar.

For instance, 25 students in School A is 50% of that school, but 25 students in School B is only 5%. So frequency exaggerates the larger school and does not compare the distribution shape fairly.

Cumulative frequency keeps adding the counts as you move through the bins. Like frequency, it depends on the total number surveyed.

That means School A’s cumulative graph would end at 50, while School B’s would end at 500, so they still cannot be compared on the same vertical scale.

Relative frequency (also called percentage frequency) is found by dividing the class frequency by the total number of students, N:

$\text{Relative Frequency}=\frac{\text{Frequency}}{N}$

This puts both schools on a common 0 to 1 (or 0% to 100%) scale, letting you compare how the travel times are distributed regardless of the different group sizes.

Frequency density describes how many values fall in each single unit of an interval’s width. We calculate it by dividing the frequency by the class width:

$\text{Frequency Density}=\frac{\text{Frequency}}{\text{Class Width}}$

This is essential for drawing histograms correctly, especially when class widths are not all the same.

Find the class width by subtracting the lower boundary from the upper boundary.

For the interval $10<x\le 25$: $\text{Class Width}=25-10=15$

Use the frequency density formula with the given frequency and the width you found.

Given the frequency is $45$ and the width is $15$: $\text{Frequency Density}=\frac{45}{15}=3.0$

Here is why the other answers do not fit:

• Option A (6.0): This could happen if someone incorrectly used a class width of $7.5$ instead of $15$ ($45/7.5=6.0$).
• Option B (2.25): This would result from dividing by $20$ rather than $15$ ($45/20=2.25$).
• Option C (0.25): This is like mixing up the calculation (for example, doing something closer to $\frac{1}{4}$ instead of $\frac{45}{15}$), so it is far too small to be correct.

Since the calculation gives $3.0$, the correct option is the one that matches $3.0$.

Calculate the interval width by taking the upper boundary minus the lower boundary.

For $10<x\le 25$:

$\text{Class Width}=25-10=15$

We know the class frequency is $45$.

In a histogram, frequency density is linked to frequency and class width by: $\text{Frequency Density}=\frac{\text{Frequency}}{\text{Class Width}}$

To verify answers, rearrange to: $\text{Frequency}=\text{Frequency Density}\times \text{Class Width}$

Then test each option by multiplying it by $15$ and seeing if you get $45$.

Test Option A ($6.0$):

$6.0\times 15=90$

Because $90\ne 45$, Option A cannot be correct.

Test Option B ($2.25$):

$2.25\times 15=33.75$

Because $33.75\ne 45$, Option B is incorrect.

Test Option C ($0.25$):

$0.25\times 15=3.75$

Because $3.75\ne 45$, Option C is incorrect.

Now test Option D ($3.0$):

$3.0\times 15=45$

This exactly matches the given frequency of $45$.

Only a frequency density of $3.0$ produces a frequency of $45$ when multiplied by the class width $15$, so that option must be the correct answer.