How to Make a Stem Plot

When I’m done grading an exam, I like to see how well the class did as a whole. Most of the time there’s not a calculator handy, so I don’t start by calculating the mean or median
. Instead it’s helpful to see how the scores are distributed. Do they resemble a bell curve? Are the scores bimodal? One type of graph that displays these features of the data is called a stem and leaf plot, or stemplot. Despite the name, there is no flora or foliage involved. Instead the idea is that stems form one part of a number, and leaves are the rest of that number.
To make my stemplot, each score is broken into two pieces, the stem and leaf. In this particular case the tens digits are stems, and the ones digits form the leaves. The resulting stemplot produces a distribution of the data similar to a histogram, but all of the data values are retained in a compact form. Features of the students’ performance can be easily seen from the shape of the stem and leaf plot.
Suppose that my class had the following test scores: 84, 65, 78, 75, 89, 90, 88, 83, 72, 91, 90 and we wanted to see at a glance what features were present in the data. We rewrite the list of scores in order and then utilize a stem and leaf plot. The stems are 6, 7, 8, 9, corresponding to the tens place of the data. This is listed in a vertical column. The ones digit of each score is written in a horizontal row to the right of each stem:
9| 0 0 1
8| 3 4 8 9
7| 2 5 8
6| 2
The data can be easily read from the stemplot. For example, the top row contains the values 90, 90 and 91.

What’s the Stem and What’s the Leaf?

With test scores, as well as other data, that range between 0 and 100 points the above strategy works for choosing stems and leaves. But how do we know what to pick for a stem and a leaf with other sets of data? For data with more than two digits there are some options. It all comes down to how the data values are distributed.
If we wanted to make a stem and leaf plot for the data 100, 105, 110, 120, 124, 126, 130, 131, 132 what would we choose for a stem? If we say the highest place value, that is the hundreds digit is the stem, our resulting stemplot is not very helpful because none of the values are separated from any of the others:
1|00 05 10 20 24 26 30 31 32
Instead we could try to make the stem the first two digits of the data. The resulting stem and leaf plot does a better job at depicting the data:
13| 0 1 2
12| 0 4 6
11| 0
10| 0 5

Expanding and Condensing

The two stemplots in the previous section show how versatile stem and leaf plots are. They can be expanded or condensed by changing the form of the stem. One strategy for expanding a stemplot is to evenly split a stem into equally sized pieces. Consider the stemplot:
9| 0 0 1
8| 3 4 8 9
7| 2 5 8
6| 2
We expand this stem and leaf plot by splitting each stem into two. This results in two stems for each tens digit. The data with 0-4 in the ones place value are separated from those with digits 5-9 in the ones place:
9| 0 0 1
8| 8 9
8| 3 4
7| 5 8
7| 2
6|
6| 2
Here the 6 with no numbers to the right shows that there are no data values from 65 to 69.

Binomial Table For n = 7 to n = 9

The binomial distribution gives the probability of r successes in an experiment with a total of nindependent trials, each having probability of success p. The tables below are for n = 7, 8, and 9.
Each entry in the table is calculated by the formula C(n, r)pr(1 - p)n - r where C(n, r) is the formula for combinations.
Other binomial distribution tables: n = 2 to 6, n = 10 to 11

Tables for n = 7 to n = 9

n = 7
p.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
r0.932.698.478.321.210.133.082.049.028.015.008.004.002.001.000.000.000.000.000.000
1.066.257.372.396.367.311.247.185.131.087.055.032.017.008.004.001.000.000.000.000
2.002.041.124.210.275.311.318.299.261.214.164.117.077.047.025.012.004.001.000.000
3.000.004.023.062.115.173.227.268.290.292.273.239.194.144.097.058.029.011.003.000
4.000.000.003.011.029.058.097.144.194.239.273.292.290;268.227.173.115.062.023.004
5.000.000.000.001.004.012.025.047.077.117.164.214.261.299.318.311.275.210.124.041
6.000.000.000.000.000.001.004.008.017.032.055.087.131.185.247.311.367.396.372.257
7.000.000.000.000.000.000.000.001.002.004.008.015.028.049.082.133.210.321.478.698


n = 8
p.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
r0.923.663.430.272.168.100.058.032.017.008.004.002.001.000.000.000.000.000.000.000
1.075.279.383.385.336.267.198.137.090.055.031.016.008.003.001.000.000.000.000.000
2.003.051.149.238.294.311.296.259.209.157.109.070.041.022.010.004.001.000.000.000
3.000.005.033.084.147.208.254.279.279.257.219.172.124.081.047.023.009.003.000.000
4.000.000.005:018.046.087.136.188.232.263.273.263.232.188.136.087.046.018.005.000
5.000.000.000.003.009.023.047.081.124.172.219.257.279.279.254.208.147.084.033.005
6.000.000.000.000.001.004.010.022.041.070.109.157.209.259.296.311.294.238.149.051
7.000.000.000.000.000.000.001.003.008.016.031.055.090.137.198.267.336.385.383.279
8.000.000.000.000.000000.000.000.001.002.004.008.017.032.058.100.168.272.430.663


n = 9
rp.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
0.914.630.387.232.134.075.040.021.010.005.002.001.000.000.000.000.000.000.000.000
1.083.299.387.368.302.225.156.100.060.034.018.008.004.001.000.000.000.000.000.000
2.003.063.172.260.302.300.267.216.161.111.070.041.021.010.004.001.000.000.000.000
3.000.008.045.107.176.234.267.272.251.212.164.116.074.042.021.009.003.001.000.000
4.000.001.007.028.066.117.172.219.251.260.246.213.167.118.074.039.017.005.001.000
5.000.000.001.005.017.039.074.118.167.213.246.260.251.219.172.117.066.028.007.001
6.000.000.000.001.003.009.021.042.074.116.164.212.251.272.267.234.176.107.045.008
7.000.000.000.000.000.001.004.010.021.041.070.111.161.216.267.300.302.260.172.063
8.000.000.000.000.000.000.000.001.004.008.018.034.060.100.156.225.302.368.387.299
9.000.000.000.000.000.000.000.000.000.001.002.005.010.021.040.075.134.232.387.630

Combinations and Permutations How Can We Tell the Difference Between Them?

To figure out some probability problems, we must be able to count. Suppose we are given a total of n distinct objects and want to select r of them. This touches directly on an area of mathematics known as combinatorics, which is the study of counting. Two of the main ways to count are called permutations and combinations. These concepts are closely related to one another and easily confused.
How can we tell the difference between a combination and permutation? The key idea is that of order. A permutation pays attention to the order that we select our objects. The same set of objects, but taken in a different order will give us different permutations. With a combination we still select r objects from a total of n, but the order is no longer considered.

An Example of Permutations

To distinguish between these ideas, we will consider the following example: how many permutations are there of two letters from the set {a,b,c}?
Here we list all pairs of elements from the given set, all the while paying attention to the order. There are a total of six permutations. The list of all of these are: ab, ba, bc, cb, ac, ca. Note that as permutationsab and ba are different because in one case a was chosen first, and in the other a was chosen second.

An Example of Combinations

Now we will answer the following question: how many combinations are there of two letters from the set {a,b,c}?
Since we are dealing with combinations, we no longer care about the order. We can solve this problem by looking back at the permutations, and then eliminating those that include the same letters. As combinations, ab and ba are regarded as the same. Thus there are only three combinations: ab, ac, bc.

Formulas

For situations we encounter with larger sets it is too time consuming to list out all of the possible permutations or combinations and count the end result. Fortunately there are formulas that give us the number of permutations or combinations of n objects taken r at a time.
In these formulas we use the shorthand notation of n!, called n factorial. The factorial simply says to multiply all positive whole numbers less than or equal to n together. So, for instance, 4! = 4 x 3 x 2 x 1 = 24. By definition 0! = 1.
The number of permutations of n objects taken r at a time is given by the formula:
P(n,r) = n!/(n - r)!
The number of combinations of n objects taken r at a time is given by the formula:
C(n,r) = n!/[r!(n - r)!]

Formulas at Work

To see the formulas at work, let’s look at the initial example. The number of permutations of a set of three objects taken two at a time is given by P(3,2) = 3!/(3 - 2)! = 6/1 = 6. This matches exactly what we obtained by listing all of the permutations.
The number of combinations of a set of three objects taken two at a time is given by:
C(3,2) = 3!/[2!(3-2)!] = 6/2 = 3. Again, this lines up exactly with what we saw before.
The formulas definitely save time when we are asked to find the number of permutations of a larger set. For instance, how many permutations are there of a set of ten objects taken three at a time. It would take awhile to list all the permutations, but with the formulas we see that there would be:
P(10,3) = 10!/(10-3)! = 10!/7! = 10 x 9 x 8 = 720 permutations.

The Main Idea

What is the difference between permutations and combinations? The bottom line is that in counting situations that involve an order, permutations should be used. If the order is not important, then combinations should be utilized.

Binomial Probability Distribution Formula

The above formula indicates the probability of r successes out of n independent trials, where each success has probability p of occurring.
The formula can be thought of in the following way. The probability of r successes occuring, each with probability p is pr. This leaves n-r failures, and each failure has probability 1-p. We multiply all of these probabilities together because the trials are independent, and this gives us pr(1-p)n-r.
The factorial in the formula shows up because we need to find the combination of ways that we can have r successes out of n trials.
Most calculations involving the binomial distribution will be carried out with software or tables. However, it is important to realize that the values in the table do not fall out of the sky. Only a few mathematical ideas from probability and combinatorics are needed to derive tables of the binomial probability distribution.

Binomial Table For n = 2 to n = 6

The binomial distribution gives the probability of r successes in an experiment with a total of nindependent trials, each having probability of success p. The tables below are for n = 2, 3, 4, 5 and 6.
Each entry in the table is calculated by the formula C(nr)pr(1 - p)n - r where C(nr) is the formula for combinations.
Other binomial distribution tables: n = 7 to 9, n = 10 to 11.

Tables for n=2 to n=6

n = 2
p.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
r0.980.902.810.723.640.563.490.423.360.303.250.203.160.123.090.063.040.023.010.002
1.020.095.180.255.320.375.420.455.480.495.500.495.480.455.420.375.320.255.180.095
2.000.002.010.023.040.063.090.123.160.203.250.303.360.423.490.563.640.723.810.902


n = 3
p.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
r0.970.857.729.614.512.422.343.275.216.166.125.091.064.043.027.016.008.003.001.000
1.029.135.243.325.384.422.441.444.432.408.375.334.288.239.189.141.096.057.027.007
2.000.007.027.057.096.141.189.239.288.334.375.408.432.444.441.422.384.325.243.135
3.000.000.001.003.008.016.027.043.064.091.125.166.216.275.343.422.512.614.729.857


n = 4
p.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
r0.961.815.656.522.410.316.240.179.130.092.062.041.026.015.008.004.002.001.000.000
1.039.171.292.368.410.422.412.384.346.300.250.200.154.112.076.047.026.011.004.000
2.001.014.049.098.154.211.265.311.346.368.375.368.346.311.265.211.154.098.049.014
3.000.000.004.011.026.047.076.112.154.200.250.300.346.384.412.422.410.368.292.171
4.000.000.000.001.002.004.008.015.026.041.062.092.130.179.240.316.410.522.656.815


n = 5
p.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
r0.951.774.590.444.328.237.168.116.078.050.031.019.010.005.002.001.000.000.000.000
1.048.204.328.392.410.396.360.312.259.206.156.113.077.049.028.015.006.002.000.000
2.001.021.073.138.205.264.309.336.346.337.312.276.230.181.132.088.051.024.008.001
3.000.001.008.024.051.088.132.181.230.276.312.337.346.336.309.264.205.138.073.021
4.000.000.000.002.006.015.028.049.077.113.156.206.259.312.360.396.410.392.328.204
5.000.000.000.000.000.001.002.005.010.019.031.050.078.116.168.237.328.444.590.774


n = 6
p.01.05.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85.90.95
r0.941.735.531.377.262.178.118.075.047.028.016.008.004.002.001.000.000.000.000.000
1.057.232.354.399.393.356.303.244.187.136.094.061.037.020.010.004.002.000.000.000
2.001.031.098.176.246.297.324.328.311.278.234.186.138.095.060.033.015.006.001.000
3.000.002.015.042.082.132.185.236.276.303.312.303.276.236.185.132.082.042.015.002
4.000.000.001.006.015.033.060.095.138.186.234.278.311.328.324.297.246.176.098.031
5.000.000.000.000.002.004.010.020.037.061.094.136.187.244.303.356.393.399.354.232
6.000.000.000.000.000.000.001.002.004.008.016.028.047.075.118.178.262.377.531.735

What Is a Histogram?


A histogram is a type of graph that has wide applications in statistics. Histograms allow a visual interpretation of numerical databy indicating the number of data points that lie within a range of values, called a class or a bin. The frequency of the data that falls in each class is depicted by the use of a bar.

Histograms vs. Bar Graphs

At first glance, histograms look very similar to bar graphs. Both graphs employ vertical bars to represent data. The height of a bar corresponds to the relative frequency of the amount of data in the class. The higher the bar, the higher the frequency of the data. The lower the bar, the lower the frequency of data. But looks can be deceiving. It is here that the similarities end between the two kinds of graphs.
The reason that these kinds of graphs are different has to do with the level of measurement of the data. On one hand, bar graphs are used for data at the nominal level of measurement. Bar graphs measure the frequency of categorical data, and the classes for a bar graph are these categories. On the other hand, histograms are used for data that is at least at the ordinal level of measurement. The classes for a histogram are ranges of values.
Another key difference between bar graphs and histograms has to do with the ordering of the bars. In a bar graph it is common practice to rearrange the bars in order of decreasing height. However, the bars in a histogram cannot be rearranged. They must be displayed in the order that the classes occur.

Example of a Histogram

The diagram above shows us a histogram. Suppose that four coins are flipped and the results are recorded. The use of the appropriate binomial distribution table or straightforward calculations with the binomial formula shows the probability that no heads are showing is 1/16, the probability that one head is showing is 4/16. The probability of two heads is 6/16. The probability of three heads is 4/16. The probability of four heads is 1/16.
We construct a total of five classes, each of width one. These classes correspond to the number of heads possible: zero, one, two, three or four. Above each class we draw a vertical bar or rectangle. The heights of these bars correspond to the probabilities mentioned for our probability experiment of flipping four coins and counting the heads.

Histograms and Probabilities

The above example not only demonstrates the construction of a histogram, it also shows that discrete probability distributions can be represented with a histogram. Indeed, and discrete probability distribution can be represented by a histogram.
To construct a histogram that represents a probability distribution, we begin by selecting the classes. These should be the outcomes of a probability experiment. The width of each of these classes should be one unit. The heights of the bars of the histogram are the probabilities for each of the outcomes. With a histogram constructed in such a way, the areas of the bars are also probabilities.
Since this sort of histogram gives us probabilities, it is subject to a couple of conditions. One stipulations is that only nonnegative numbers can be used for the scale that gives us the height of a given bar of the histogram. A second condition is that since probability is equal to area, all of the areas of the bars must add up to a total of one, equivalent to 100%.

Histograms and Other Applications

The bars in a histogram need not be probabilities. Histograms are helpful in areas other than probability. Anytime that we wish to compare the frequency of occurrence of quantitative data a histogram can be used.

What Are Pie Charts?


One of the most common ways to represent data graphically is called a pie chart. It gets its name by how it looks, just like a circular pie that has been cut into several slices. This kind of graph is helpful when graphing qualitative data, where the information describes a trait or attribute and is not numerical. Each trait corresponds to a different slice of the pie. By looking at all of the pie pieces, you can compare how much of the data fits in each category. The larger a category, the bigger that its pie piece will be.

Big or Small Slices?

How do we know how large to make a pie piece? First we need to calculate a percentage. Ask what percent of the data is represented by a given category. Divide the number of elements in this category by the total number. We then convert this decimal into a percentage.
A pie is a circle. Our pie piece, representing a given category, is a portion of the circle. Because a circlehas 360 degrees all the way around, we need to multiply 360 by our percentage. This gives us the measure of the angle that our pie piece should have.

An Example

To illustrate the above, let’s think about the following example. In a cafeteria of 100 third graders, a teacher looks at the eye color of each student and records it. After all 100 students are examined, the results show that 60 students have brown eyes, 25 have blue eyes and 15 have hazel eyes.
The slice of pie for brown eyes needs to be the largest. And it needs to be over twice as large as the slice of pie for blue eyes. To say exactly how large it should be, first find out what percent of the students have brown eyes. This is found by dividing the number of brown eyed students by the total number of students, and converting to a percent. The calculation is 60/100 x 100% = 60%.
Now we find 60% of 360 degrees, or .60 x 360 = 216 degrees. This reflex angle is what we need for our brown pie piece.
Next look at the slice of pie for blue eyes. Since there are a total of 25 students with blue eyes out of a total of 100, this means that this trait accounts for 25/100x100% = 25% of the students. One quarter, or 25% of 360 degrees is 90 degrees, a right angle.
The angle for the pie piece representing the hazel eyed students can be found in two ways. The first is to follow the same procedure as the last two pieces. The easier way is to notice that there are only three categories of data, and we have accounted for two already. The remainder of the pie is corresponds to the students with hazel eyes.
The resulting pie chart is pictured above. Note that number of students in each category is written on each pie piece.

Limitations of Pie Charts

Pie charts are to be used with qualitative data, however there are some limitations in using them. If there are too many categories, then there will be a multitude of pie pieces. Some of these are likely to be very skinny, and can be difficult to compare to one another.
If we want to compare different categories that are close in size, a pie chart does not always help us to do this. If one slice has central angle of 30 degrees, and another has a central angle of 29 degrees, then it would be very hard to tell at a glance which pie piece is larger than the other.