Histogram
In this article, we will discuss:
Here is one more link to practice a few extra questions: Maths Genie Histogram Questions
Frequency density = frequency/width of bin
Solved Example:
Question 1: Suppose we have collected data on the heights of 50 students in a class, and we want to create a histogram to represent this data. We group the heights into bins, with each bin representing a range of heights. Let’s say we have five bins with the following ranges:
Solution:
We then count the number of students whose height falls within each bin.
Let’s say we get the following frequencies:
To calculate the frequency density for each bin, we use the formula:
Frequency density = frequency/width of bin
For example, for the first bin, the width is 10 cm, so the frequency density is:
Frequency density for 150-160 cm
bin = 8 / 10
= 0.8
Similarly, we can calculate the frequency density for each bin:
• Frequency density for 150-160 cm bin = 0.8
• Frequency density for 160-170 cm bin = 1.2
• Frequency density for 170-180 cm bin = 1.6
• Frequency density for 180-190 cm bin = 0.9
• Frequency density for 190-200 cm bin = 0.5
By using frequency density instead of frequency, we can compare the distribution of data across bins with different widths.
In this example, we can see that the distribution is highest in the 170-180 cm bin, which has the highest frequency density of 1.6.
To draw a histogram, follow these steps:
Solved Example
Question: Let’s say we have the following data set representing the scores of 30 students in a math test: 80, 78, 90, 85, 92, 84, 77, 85, 88, 80, 75, 72, 85, 90, 88, 92, 78, 82, 86, 90, 82, 83, 85, 88, 92, 89, 86, 80, 78, 85. We aim to construct a histogram to visually represent the score distribution. Here’s how we can do it:
Solution:
70-74: 1/5 = 0.2
75-79: 5/5 = 1
80-84: 7/5 = 1.4
85-89: 11/5 = 2.2
90-94: 6/5 = 1.2
To find the frequency for each bin, you can use the formula frequency equals area.
Solved Example
Question: The table below and its corresponding histogram show the mass, in kg, of some newborn bottlenose dolphins.
(a) Use the table and histogram to find the value of k in the formula
(b) Estimate the number of dolphins whose weight is greater than 13 kg.
Solution:
(a)
Start by finding the frequency density in terms of k – add two columns to the table, one for class width, one for frequency density
We can use either of the two intervals that feature both in the table and on the histogram to find the value of k
Using the first bar
Check using the other (2nd) bar to check: 7.5k = 3.75, k = 0.5
(b) Estimate the number of dolphins whose weight is greater than 13 kg.
Solution:
We can see from the table that there are 6 dolphins in the interval 15 ≤ m <30.
So we need to estimate the number of dolphins that are in the interval 13≤ m <15.
For 13≤ m <15, the histogram shows the frequency density is 1.5 and we found the value of k in part(a).
Using the formula given in the question
So the total number of can be estimated by
6 + 6 = 12
There are approximately 12 dolphins with a weight greater than 13kg
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