Introduction to Using Algebra in Statistics and Probability
42.7% of all statistics are made up on the spot.
—Steven Wright (1955– ) American Comedian
In this lesson, you'll learn how to use algebra to solve statistics and probability problems, such as finding the missing value in a data set or how a probability changes with the addition of more data.
Algebra can be used in various fields of mathematics. Anytime you have an unknown value, you can represent it with x (or any other letter).
In statistics, data can be analyzed by finding the mean, median, mode, and range of a set. Algebra can help us find the mean and the range of a set as new values are added to the set, and it can help us find a missing value in a set.
The mean of a set can be found by dividing the sum of all the values in the set by the number of values in the set. For example, the mean of the set {1, 1, 2, 5, 7, 8} is 4, because the sum of the values in the set is 24, the number of values in the set is 6, and 24 ÷ 6 = 4.
We did not need algebra to tell us that, but what if we were told that a seventh number was added to the set that changed the mean to 5? How could we find that added number? Because we don't know what the number is, we represent it with x, and add x to the set: {1, 1, 2, 5, 7, 8, x}. We find the mean by adding the values in the set and dividing by the number of values, and even though our set now contains a variable, we can still do that: 1 + 1 + 2 + 5 + 7 + 8 + x = 24 + x. We are told that the new mean is 5. We now have seven values, so we know that the sum, 24 + x, divided by 7 is 5. We can write an equation and solve for x:
= 5
The number that was added to the set was 11. The set is now {1, 1, 2, 5, 7, 8, 11}.
Finding a Median
The median of a data set is the value that is in the middle of the set after the set is put in order from least to greatest. If the set has an even number of values, then the median is the average of the two middle values. In the set {1, 3, 5, 5, 7}, 5 is the median, because it is the middle value.
We are told that a new value is added to the set, and the median is now 4. What could have been that new value? How can we represent the set of numbers now? If 4 had been added to the set, we would have {1, 3, 4, 5, 5, 7}, and the median would have been the average of 4 and 5, which is 4.5. Because the new median is 4, not 4.5, the new value could not have been 4. Therefore, the median must be the average of two numbers, neither of which is 4. Look at the set again: {1, 3, 5, 5, 7}. If the new value is 5 or greater, the median will still be 5. However, if the new value is 3, the set becomes {1, 3, 3, 5, 5, 7}, and the median is the average of 3 and 5, which is 4. The new value could be 3, but hold on. If the new value is less than 3, the median will still be 4, because the two middle values in the set will still be 3 and 5. Let's represent the new value with x. The new value can be any number less than or equal to 3: x ≤ 3.
Finding a Range
The range of a data set is the difference between its greatest value and its least value. In the set {1, 3, 5, 5, 7}, the range is 6, because the greatest value is 7, the least value is 1, and 7 – 1 = 6.
Now we are told that a new value has been added to the set, and the range is now 10. Use x to represent the new value. What could that value be? Because the range has changed, the new value is now either the smallest value in the set or the largest value in the set. If the new value is the smallest value, then the range, 10, is equal to 7 – x. If the new value is the largest value, then the range is equal to x – 1:
7 – x = 10 
x – 1 = 10 
–x = 3 
x = 11 
x = –3 
The new value is either –3 or 11.
Probability
Probability is the likelihood that an event or events will occur. We usually write probability as a fraction. The denominator is the total number of possibilities, and the numerator is the number of possibilities that make an event true. For example, a coin has two sides, heads and tails, so the probability of a coin landing on heads is , because there are two possibilities (heads or tails) and only one possibility that makes the event "landing on heads" true.
Example
A gumball machine contains 12 red gumballs, 8 green gumballs, 5 yellow gumballs, and an unknown number of orange gumballs. If the probability of selecting a green gumball is , how many orange gumballs are in the machine?
The probability of selecting a green gumball is . We know that this probability is equal to the number of green gumballs, 8, divided by the total number of gumballs. We can let x represent the number of orange gumballs. The total number of gumballs is equal to 12 + 8 + 5 + x = 25 + x. The fraction is equal to , because both fractions represent the probability of selecting a green gumball. When two fractions equal each other, we can cross multiply: Multiply the numerator of the first fraction by the denominator of the second fraction, and multiply the numerator of the second fraction by the denominator of the first fraction. Then, set those products equal to each other:
=
Now we have an equation we can solve. By subtracting 25 from both sides, we find that x = 7. There are 7 orange gumballs in the machine.
Tip:
When a new value is added to a data set, that new value will likely change the denominator of probabilities for the data set, and you will have to add x to the denominator of these probabilities. If you are looking for the probability of selecting that new value, you may have to add x to the numerator of these probabilities, too.

Example
The gumball machine contains 12 red gumballs, 8 green gumballs, 5 yellow gumballs, and 7 orange gumballs. How many green gumballs must be added to make the probability of selecting a green gumball ?
The probability of selecting a green gumball is equal to the number of green gumballs divided by the total number of gumballs. Let x represent the number of green gumballs added to the machine. That means there are now 8 + x green gumballs, and 12 + 8 + 5 + 7 + x = 32 + x total gumballs. The probability of selecting a green gumball is equal to . The value of x will make this probability equal to , so set the two fractions equal to each other and solve for x:
=
Cross multiply:
For the probability of selecting a green gumball to become , 16 green gumballs must be added to the gumball machine.
Find practice problems and solutions for these concepts at Using Algebra in Statistics and Probability Practice Questions.
View Full Article
From Algebra in 15 Minutues A Day. Copyright © 2009 by LearningExpress, LLC. All Rights Reserved.