# Cumulative Frequency Histogram

How to make and solve a cumulative frequency histogram: 4 examples and their solutions.

## Example 1

### Example

### Solution

First, make a cumulative frequency table.

Draw a 3 column table like this.

Name the titles

Interval, Frequency, and Cumulative Frequency.

Frequency Table

Fill in the Cumulative Frequency column.

For the first row,

the frequency is 2.

So the cumulative frequency is

2.

The previous cumulative frequency is 2.

The frequency of this row is 3.

Then the cumulative frequency is

2 + 3 = 5.

The previous cumulative frequency is 7.

The frequency of this row is 5.

Then the cumulative frequency is

7 + 5 = 12.

The previous cumulative frequency is 12.

The frequency of this row is 6.

Then the cumulative frequency is

12 + 6 = 18.

The previous cumulative frequency is 18.

The frequency of this row is 2.

Then the cumulative frequency is

18 + 2 = 20.

Use the Interval column

and the Cumulative Frequency column

to make a cumulative frequency histogram.

Frequency Histogram

Draw a rectangular form like this.

Write Test Scores

at the bottom of the form.

Write Cumulative Frequency

on the left side of the form.

Write the intervals of the test scores

at the bottom of the form.

The cumulative frequencies are from 0 to 20.

So write 0 to 20

on the left side of the form.

And inside the rectangle,

lightly draw horizontal lines

that show each frequency.

The cumulative frequency of 50-59 is 2.

So, for 50-59,

draw a vertical bar

whose height is 2.

The cumulative frequency of 60-69 is 5.

So, for 60-69,

draw a vertical bar

whose height is 5.

(in the middle of 4 and 6)

Draw the bar right next to the previous bar

so that

there would be no space between the bars.

The cumulative frequency of 70-79 is 12.

So, for 70-79,

draw a vertical bar

whose height is 12.

The cumulative frequency of 80-89 is 18.

So, for 80-89,

draw a vertical bar

whose height is 18.

The cumulative frequency of 90-99 is 20.

So, for 90-99,

draw a vertical bar

whose height is 20.

So this is the cumulative frequency histogram

of the given test scores.

## Example 2

### Example

### Solution

This bar shows

the number of students

whose scores are less than 80 points.

Its height is 12.

So the percent of students

whose scores are less than 80 points is,

the height of this bar,

12

over,

the total number of students,

20

times 100.

So the percent of students

whose scores are less than 80 points is

[12/20]⋅100.

Cancel the denominator 20

and reduce 100 to, 100/20, 5.

12⋅5 = 60

This is the percent of students.

So write %.

So 60% is the answer.

## Example 3

### Example

### Solution

This bar shows

the number of students

whose scores are less than 70 points.

Its height is 6.

Then the number of students

whose scores are

more than or equal to 70 points

is this blue height:

20 - 6.

So the percent of students

whose scores are

more than or equal to 70 points is,

the length of the blue height,

20 - 6

over,

the total number of students,

20

times 100.

So the percent of students

whose scores are

more than or equal to 70 points is

[(20 - 6)/20]⋅100.

20 - 6 = 14

14/20 = 7/10

Cancel the denominator 10

and reduce 100 to, 100/10, 10.

7⋅10 = 70

This is the percent of students.

So write %.

So 70% is the answer.

## Example 4

### Example

### Solution

To find the quartiles Q_{1}, Q_{2}, and Q_{3},

find the related cumulaitve frequencies

n_{1}, n_{2}, and n_{3}.

n_{1} is the related cumulative frequency of Q_{1}.

So n_{1} is equal to,

the total number of the values, 20

times,

Q_{1} is the 1/4 of the data, 1/4.

20⋅[1/4] = 5

So n_{1} = 5.

This means

there are n_{1} = 5 values

between 0 and Q_{1}.

n_{2} is the related cumulative frequency of Q_{2}.

So n_{2} is equal to,

the total number of the values, 20

times,

Q_{2} is the 1/2 of the data, 1/2.

20⋅[1/2] = 10

So n_{2} = 10.

This means

there are n_{2} = 10 values

between 0 and Q_{2}.

n_{3} is the related cumulative frequency of Q_{3}.

So n_{3} is equal to,

the total number of the values, 20

times,

Q_{3} is the 3/4 of the data, 3/4.

Cancel the denominator 4

and reduce 20 to, 20/4, 5.

3⋅5 = 15

So n_{3} = 15.

This means

there are n_{3} = 15 values

between 0 and Q_{3}.

n_{1} = 5

n_{2} = 10

n_{3} = 15

Write these values

on the Cumulative Frequency axis.

Find the interval that have Q_{1}.

Find the nearest bar

that covers n_{1} = 5:

the interval 60-69.

So the interval 60-69 has Q_{1}.

By the same way,

find the interval that have Q_{2}.

Find the nearest bar

that covers n_{2} = 10:

the interval 70-79.

So the interval 70-79 has Q_{2}.

Find the interval that have Q_{3}.

Find the nearest bar

that covers n_{3} = 15:

the interval 90-99.

So the interval 90-99 has Q_{3}.

So

Q_{1}: 60-69

Q_{2}: 70-79

Q_{3}: 90-99

is the answer.