Create a histogram for the number of books bought by [latex]\scriptsize 50[/latex] part-time college learners.

[latex]\scriptsize \displaystyle \begin{align*}&1;\text{ }1;\text{ }1;\text{ }1;\text{ }1;\text{ }1;\text{ }1;\text{ }1;\text{ }1;\text{ }1;\text{ }1\\&2;\text{ }2;\text{ }2;\text{ }2;\text{ }2;\text{ }2;\text{ }2;\text{ }2;\text{ }2;\text{ }2\\&3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3;\text{ }3\\&4;\text{ }4;\text{ }4;\text{ }4;\text{ }4;\text{ }4\\&5;\text{ }5;\text{ }5;\text{ }5;\text{ }5\\&6;\text{ }6\end{align*}[/latex]

Solution

Note that ‘number of books’ is discrete data, since books are counted.

To draw the histogram you must decide how many bars or intervals, also called classes, represent the data.

Calculate the number of bars as follows: [latex]\scriptsize \displaystyle \displaystyle \frac{{\text{largest value}-\text{smallest value}}}{{\text{number of bars}}}=\text{width of bar/interval}[/latex].

Before you can calculate the number of bars you must choose a starting point for the first interval to be less than the smallest data value.

With discrete data, all the data values happen to be integers. In this example the smallest value is [latex]\scriptsize 1[/latex], so a convenient starting point will be [latex]\scriptsize 1-0.5=0.5[/latex]. The largest value is [latex]\scriptsize 6[/latex] so we add [latex]\scriptsize 0.5[/latex] to [latex]\scriptsize 6[/latex] to get the end point of [latex]\scriptsize 6.5[/latex].

Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar is the most convenient.

Since the data consist of the numbers [latex]\scriptsize \displaystyle 1;\text{ }2;\text{ }3;\text{ }4;\text{ }5;\text{ }6[/latex] and the starting point is [latex]\scriptsize 0.5[/latex], a width of one places the [latex]\scriptsize 1[/latex] in the middle of the interval from [latex]\scriptsize 0.5[/latex]to [latex]\scriptsize 1.5[/latex], the [latex]\scriptsize 2[/latex] in the middle of the interval from [latex]\scriptsize 1.5[/latex] to [latex]\scriptsize 2.5[/latex], and so on.

Therefore, [latex]\scriptsize \displaystyle \frac{{6.5-0.5}}{\text{number of bars}}=1[/latex] so the number of bars is [latex]\scriptsize 6[/latex].

The histogram displays the number of books on the x-axis and the frequency on the y-axis.

The height of the bars gives the frequency of each interval and the intervals are listed on the x-axis. Now that you’ve drawn the histogram, you get a better overall picture of the data because you can visualise it. We can also see at a glance some important things that the data reveals.

The tallest bar shows that [latex]\scriptsize 3[/latex] books was the most popular observation, with a frequency of [latex]\scriptsize 16[/latex].

The following data represent the heights of [latex]\scriptsize \displaystyle 16[/latex] people, in centimetres.

[latex]\scriptsize \displaystyle 162\text{ };\text{ }168\text{ };\text{ }177\text{ };\text{ }147\text{ };\text{ }189\text{ };\text{ }171\text{ };\text{ }173\text{ };\text{ }168;\text{ }178\text{ };\text{ }184\text{ };\text{ }165\text{ };\text{ }173\text{ };\text{ }179\text{ };\text{ }166\text{ };\text{ }168\text{ };\text{ }165[/latex]

Use class intervals that start at [latex]\scriptsize \displaystyle 140\text{ cm}[/latex] and end at [latex]\scriptsize \displaystyle 190\text{ cm}[/latex] to draw a histogram.

Solution

Step 1: Determine intervals

We need intervals of the same length between [latex]\scriptsize \displaystyle 140\text{ cm}[/latex] and [latex]\scriptsize \displaystyle 190\text{ cm}[/latex] so we could go up by [latex]\scriptsize \displaystyle 10[/latex] centimetres each time. Calculate the number of intervals, which corresponds to the number of bars, using:

[latex]\scriptsize \displaystyle \displaystyle \frac{{\text{largest value}-\text{smallest value}}}{{\text{number of bars}}}=\text{width of bar/interval}[/latex]

[latex]\scriptsize \displaystyle \begin{align*}\displaystyle \frac{{190-140}}{{\text{number of bars}}}&=10\\10\times \text{number of bars}&=50\\\therefore \text{number of bars}&=5\end{align*}[/latex]

Note: If you know how many bars you’d like to use you can use the method above to calculate the bar width.

We can use round and square brackets to show which values are excluded and included in the intervals. Round brackets are used to show values that are not included and square brackets are used to show included values. Interval [latex]\scriptsize (140;150][/latex] means all values greater than [latex]\scriptsize 140\text{ cm}[/latex]and less than or equal to [latex]\scriptsize 150\text{ cm}[/latex].

Step 2: Include the frequency in each interval

The following frequency table summarises the number of data values in each of the intervals.

Step 3: Draw the histogram

The interval [latex]\scriptsize (150;160][/latex] has no entries so we do not draw a bar in that interval.

Which box-and-whisker plot represents the shape of the histogram below?

Solution

The histogram has a longer tail on the right so it is positively skewed, which shows that the mean is greater than the median.

Box plot A: The median is in the middle of the box, and the whiskers are about the same length on either side of the box so the distribution is symmetric or normal.

Box plot B: The median is pulled towards the upper quartile, and the whisker is shorter on the upper end of the box so the distribution is negatively skewed (skewed left).

Box plot C: The median is pulled towards the lower quartile, and the whisker is shorter on the lower end of the box so the distribution is positively skewed (skewed right).

Therefore, box plot C is a possible representation of the histogram in this example.