# Word Problems Involving the Mean of a Data Set

As we delve deeper into the world of statistics and probability, one specific term becomes very important to us: The mean. You've probably encountered this concept before -- even if you weren't aware of it. "Mean" essentially means the same thing as "average," and this number becomes very useful in a number of different fields. A baseball player's "batting average" tells you how skilled he or she is. The "average" return of an investment helps us understand how profitable it really is. As we can see, the mean is very useful in real-world applications. Let's tackle some word problems that show us how we can apply means to these real-world situations.

## What is the mean?

We can find the mean of any data set with two easy steps:

- Add all the data points together
- Divide by the number of data points

## Solving word problems involving means

Now let's put our knowledge to good use and solve a few word problems that involve means:

Consider the following data set: $4,8,20,25,32$

If we add one more number $\left(x\right)$ to this data set, we are left with a mean of 15. What is the value of this additional number?

First, let's try to add all of our numbers together: $4+8+20+25+32+x$

We don't know the value of x, so we can't complete this addition. But let's continue with the operation anyway:

If we add one more value (x) to our current data set, we get 6 total data points. This means that we need to divide the above sum by six:

$4+8+20+25+32+x+6$

We know that the *end* result is a mean of 15. So how can we solve this equation? Easy: We simplify it.

$4+8+20+25+32+x+6=15$

$89+x/6=15$

Now, all we need to do is multiply each side of the equation by 6:

$6\times (89+x)/\left(6\right)=15\times 6$

$89+x=90$

The answer is 1.

We can even use our knowledge of means to help us set higher goals!

What happens if we know that we need an average of 90% or higher to get an "A" in math class?

What if our past test scores were 80%, 85%, 88%, and 95%?

We have one more test for the term. What score do we need to get an "A" average?

Right away, we know that our end result needs to be at least 90.

We also know that the total number of tests is 5.

With these values, we can find our minimum test score for an A average (x).

$80+85+88+95+x/5$

We can simplify this as $348+x/5$

Now, all we need to do is multiply each side of the equation by 5:

$5\times (348+x)/\left(5\right)=90\times 5$

Or: $348+x=450$

Uh oh, looks like that value is 102. As we know, the highest possible score on a test is 100%. This means that it's theoretically impossible for us to get an A average this term -- unless, of course, we speak to our teacher about the possibility of doing additional work for extra credit. This highlights the importance of calculating means in order to plan effectively for the future.

Exercises such as this also show us how even a few low scores can seriously affect the overall average or mean.

The especially low score of 80% is 10 percentage points below our target grade, and this skews the data toward a lower mean. We can consider this value an outlier.

## Topics related to the Word Problems Involving the Mean of a Data Set

## Flashcards covering the Word Problems Involving the Mean of a Data Set

Common Core: High School - Statistics and Probability Flashcards

## Practice tests covering the Word Problems Involving the Mean of a Data Set

Probability Theory Practice Tests

Common Core: High School - Statistics and Probability Diagnostic Tests

## Tutoring helps your student gain confidence when solving word problems involving means

Tutoring gives your student the opportunity to go over a few more examples of word problems involving means. Sometimes, all it takes to reach for confidence in these concepts is a little extra time outside of class. Tutors can even cater to your student's learning style during these sessions, whether they prefer verbal or visual cues. Speak with Varsity Tutors' Educational Directors today to learn more about the possibilities of tutoring. We'll match your student with a suitable tutor.