From a Distribution Table to a Picture
In Building a Distribution from a Sample Space, you learned to list each possible value of a random variable and its probability. A probability histogram displays that same information visually: the horizontal position identifies a value, and the height of its bar shows how likely that value is. The table and histogram are two ways to represent one distribution; the graph does not change the probabilities.
Suppose \(X\) is a discrete random variable with possible values \(0,1,2,\ldots\). To draw its probability histogram, put the possible values on a numerical horizontal axis and \(P(X=x)\) on a vertical probability axis. Draw one bar for each possible value, centered at that value. For integer-valued variables with unit-width bars, the height of the bar at \(x\) is \(P(X=x)\).
The vertical scale is a probability scale, not a count scale. Label it with values such as 0, 0.1, 0.2, and so on, choosing intervals that make the heights easy to read. The scale should begin at 0 so that bar heights can be compared fairly. A probability such as 0.35 means a 35% chance of that exact value; it does not mean 0.35 observations.
For consecutive integer values, unit-width bars meet at their edges. If the possible values are separated, keep the numerical spacing on the horizontal axis: bars centered at 0 and 2 should not be drawn as if the values were neighboring categories. With unit-width bars, there will be a gap between them. A possible value with probability 0 has zero height and is not visible, but its location still matters on the numerical axis.
How to Draw and Check the Histogram
First identify the random variable and read the table’s possible values and probabilities. Mark the horizontal axis with a consistent numerical scale, including any gaps between possible values. Then label the vertical axis “Probability” and choose a scale that reaches at least the largest probability. Draw each bar centered at its value, using the table probability as its height when the width is 1.
A sketch does not need to make a bar’s height look exact by eye if the axis is carefully labeled and the table gives the exact value. Still, the plotted heights should be consistent with the scale. For example, a probability of 0.4 should be twice as high as a probability of 0.2. The height is not the probability of being at or below that value; it is the probability of that exact value.
The area property gives a useful visual check. Each unit-width bar has area equal to its probability. Because the probabilities in a valid distribution add to 1, the combined area of all bars is 1. As in Checking Whether a Probability Distribution Is Valid, the table should already pass the probability checks before you use it; a graph cannot repair an invalid table.
Worked Example: Draw a Histogram for Bus Delays
Worked Example: Draw a Histogram for Bus Delays
For a simple chance model, let \(X\) be the number of buses in a randomly selected group of four that arrive more than five minutes late. The model gives this distribution:
| Number late, \(x\) | \(P(X=x)\) |
|---|---|
| 0 | 0.15 |
| 1 | 0.35 |
| 2 | 0.30 |
| 3 | 0.20 |
State. Draw a probability histogram for the number of late buses, \(X\).
Plan. Put the possible values 0 through 3 on the horizontal axis. Use probability on the vertical axis, with a scale from 0 to at least 0.35. Since the values are consecutive integers, draw unit-width bars centered at each value and use the table entries as their heights.
Do. Mark 0, 1, 2, and 3 at equal intervals on the horizontal axis. Label the vertical axis “Probability” and mark, for example, 0, 0.1, 0.2, 0.3, and 0.4. Draw the bar centered at 0 to height 0.15, the bar at 1 to height 0.35, the bar at 2 to height 0.30, and the bar at 3 to height 0.20. Adjacent bars meet because the values are one unit apart and each bar has width 1.
The heights can be checked against one another: the bar at 1 is the tallest, at 0.35; the bar at 2 is slightly shorter, at 0.30; and the bar at 0 is the shortest, at 0.15. The total area is \(1(0.15)+1(0.35)+1(0.30)+1(0.20)=1.00\).
Conclude. The histogram has unit-width bars at 0, 1, 2, and 3 with heights 0.15, 0.35, 0.30, and 0.20, respectively. It visually shows that exactly one late bus has the greatest probability in this model.
Respect the Numerical Scale and Gaps
A histogram’s horizontal axis is quantitative. The distance between values on that axis should match the numerical distance between them. If a distribution has possible values 0, 2, and 4, those locations are two units apart. Drawing their bars at three equally spaced category positions would change the visual impression of the distribution.
For unit-width bars centered at 0, 2, and 4, the bars occupy intervals from \(-0.5\) to \(0.5\), from \(1.5\) to \(2.5\), and from \(3.5\) to \(4.5\). The spaces between bars indicate that the intervening integer values are not possible in this model. Do not add bars at those values unless the distribution assigns them positive probability.
Worked Example: Plot Values Separated by Gaps
Let \(Y\) be the number of replacement parts needed for a particular repair under a simplified model. The only possible values and their probabilities are:
| Parts needed, \(y\) | \(P(Y=y)\) |
|---|---|
| 0 | 0.50 |
| 2 | 0.30 |
| 4 | 0.20 |
State. Describe how to draw a probability histogram for \(Y\), including the locations of its bars.
Plan. Use the actual numerical values 0, 2, and 4, rather than treating them as three adjacent categories. Draw a unit-width bar at each possible value with height equal to its probability.
Do. Label the horizontal axis with a numerical scale from at least 0 to 4, marking each integer so the two-unit distances are clear. Label the vertical axis “Probability,” with a scale reaching at least 0.50. Draw a bar centered at 0 with height 0.50, a bar centered at 2 with height 0.30, and a bar centered at 4 with height 0.20. Each bar has width 1. The locations 1 and 3 have no bars because they are not possible values in the table.
Conclude. The bars should be centered at 0, 2, and 4, with visible gaps between them. The bar at 0 is tallest, and the total area is 1, as expected for a probability histogram with unit-width bars.
Read Probabilities from the Bars
A bar’s height gives the probability of the exact value at its center. To find the probability of an event containing several possible values, add the heights of the bars for those values. This is the same addition of probabilities you used when working from a distribution table; the histogram simply makes the included values easier to see.
For example, if an event is \(X\geq2\), include the bars centered at 2 and at every larger possible value. Do not use the height of one bar to represent the combined event, and do not add the heights of bars outside the event. A useful practice is to mark which values satisfy the event before adding their probabilities.
Worked Example: Use a Histogram to Find an Event Probability
A small device sends a number \(X\) of status alerts in an hour according to this model:
| Alerts, \(x\) | \(P(X=x)\) |
|---|---|
| 0 | 0.10 |
| 1 | 0.20 |
| 2 | 0.40 |
| 3 | 0.30 |
State. Use the probability histogram to find the probability that the device sends at least two alerts in an hour.
Plan. “At least two” means \(X=2\) or \(X=3\). On the histogram, identify the bars centered at 2 and 3 and add their heights.
Do. The bar at 2 has height 0.40, and the bar at 3 has height 0.30. Therefore, \(P(X\geq2)=P(X=2)+P(X=3)=0.40+0.30=0.70\). As a check, the other bars, at 0 and 1, have total height 0.10+0.20=0.30, so the probability of their complement is \(1-0.30=0.70\).
Conclude. The model assigns probability 0.70 to the event that the device sends at least two alerts in an hour. The relevant histogram bars are the ones centered at 2 and 3.
Common Mistakes and AP Exam Tips
- Using counts or percentages as the vertical scale. The vertical axis of a probability histogram is probability. If the table lists 0.30, the bar height is 0.30, not 30 or the number of trials.
- Confusing height with area. With unit-width bars, height and area have the same numerical value. Keep the width in mind: the area is width times height, and it represents probability.
- Putting bars at equally spaced category positions. Values such as 0, 2, and 4 are separated by two units. Preserve that distance on a numerical axis instead of compressing the gaps.
- Drawing a bar for an impossible value. Include only values in the distribution table. If an integer between listed values is impossible, it gets no positive-height bar.
- Adding bar heights for the wrong event. Translate the event into values first. For “at least two,” include 2 and larger values, not values below 2.
- Leaving the axes unlabeled or the scale unclear. Label the horizontal axis with the random variable’s values and the vertical axis “Probability.” Mark a consistent scale that includes the tallest bar.
For a full-credit response, identify the variable, show a numerical horizontal axis and a probability vertical axis, and give each possible value a bar with the correct location and height. When asked to interpret a probability, name the event and state its probability in context. If the task asks for a combined event, identify and add exactly the probabilities for values in that event.
Key Takeaway
A discrete distribution table becomes a probability histogram when each possible value is placed on a numerical horizontal axis and its probability is represented by a bar. For unit-width bars, the height equals \(P(X=x)\), and the total bar area equals 1.
Check Your Understanding
Use the distribution below for \(W\), the number of items selected for a quality check that need adjustment.
| Items needing adjustment, \(w\) | \(P(W=w)\) |
|---|---|
| 0 | 0.25 |
| 1 | 0.45 |
| 3 | 0.30 |
- What should the horizontal and vertical axes be labeled?
- Describe the location and height of each bar in a unit-width probability histogram. Where should the gaps appear?
- What is the area of the bar centered at 1, and what probability does it represent?
- Find \(P(W\geq1)\) by identifying and adding the appropriate bar heights.
- Explain why the total area of the three bars is 1.