5.5 Averages and Probability

Learning objectives

Learning objectives

• Calculate the mean of a set of numbers

• Find the median of a set of numbers

• Find the mode of a set of numbers

• Apply the basic definition of probability

One application of decimals that arises often is finding the average of a set of numbers. What do you think of when you hear the word average? Is it your grade point average, the average rent for an apartment in your city, the batting average of a player on your favorite baseball team? The average is a typical value in a set of numerical data. Calculating an average sometimes involves working with decimal numbers. In this section, we will look at three different ways to calculate an average.

Calculate the Mean of a Set of Numbers

The mean is often called the arithmetic average. It is computed by dividing the sum of the values by the number of values. Students want to know the mean of their test scores. Climatologists report that the mean temperature has, or has not, changed. City planners are interested in the mean household size.

Suppose Ethan’s first three test scores were $85,88,\mathrm{and}\,94.$ To find the mean score, he would add them and divide by $3.$

His mean test score is $89$ points.

Example 5.49

Solution

1. Write the mean formula. The mean of a data set equals the sum of all its values divided by the number of values, so $\mathrm{mean}=\frac{\mathrm{sum}}{n}$.
2. Count the values. The set $8,12,15,9,6$ contains $5$ numbers, so $n=5$.
3. Add the values. The numerator is $8+12+15+9+6=50$, giving $\mathrm{mean}=\frac{50}{5}$.
4. Divide the sum by the count. Dividing $50$ by $5$ gives $\mathrm{mean}=10$.
5. Check that the mean is typical. The value $10$ is greater than the minimum $6$ and less than the maximum $15$, as expected for the center.

Takeaway: The mean of $8,12,15,9,$ and $6$ is $10$ because their sum $50$ is divided by the $5$ values in the set.

Example 5.50

Solution

1. Count the ages. The data set $16,26,53,56,65,70,93,97$ contains $8$ ages, so the denominator of the mean is $8$.
2. Find the total age. Adding the values gives $16+26+53+56+65+70+93+97=476$.
3. Divide by the number of ages. The mean age is $\frac{476}{8}=59.5$ years.
4. Check the reasonableness of the result. The mean $59.5$ is between the smallest age $16$ and the largest age $97$, so it represents the center of the data.

Takeaway: The eight family ages total $476$, so their mean age is $\frac{476}{8}=59.5$ years, which lies within the range of the ages.

Did you notice that in the last example, while all the numbers were whole numbers, the mean was $59.5,$ a number with one decimal place? It is customary to report the mean to one more decimal place than the original numbers. In the next example, all the numbers represent money, and it will make sense to report the mean in dollars and cents.

Example 5.51

Solution

1. Count the bills. Daisy has $4$ monthly cell phone bills, so the mean is the sum of the four amounts divided by $4$.
2. Add the bill amounts. The total is $\text{\textdollar 42.75}+\text{\textdollar 50.12}+\text{\textdollar 41.54}+\text{\textdollar 48.15}=\text{\textdollar 182.56}$.
3. Divide the total by the count. Computing $\frac{182.56}{4}$ gives a mean bill of $\text{\textdollar 45.64}$.
4. Check that the mean is typical. The value $45.64$ is greater than the smallest bill $41.54$ and less than the largest bill $50.12$.

Takeaway: Daisy's four bills total $\text{\textdollar 182.56}$, so the mean monthly bill is $\frac{182.56}{4}=\text{\textdollar 45.64}$.

Find the Median of a Set of Numbers

When Ann, Bianca, Dora, Eve, and Francine sing together on stage, they line up in order of their heights. Their heights, in inches, are shown in Table 5.6.

Table

5.6

Dora is in the middle of the group. Her height, $65'',$ is the median of the girls’ heights. Half of the heights are less than or equal to Dora’s height, and half are greater than or equal. The median is the middle value.

What if Carmen, the pianist, joins the singing group on stage? Carmen is $62$ inches tall, so she fits in the height order between Bianca and Dora. Now the data set looks like this:

There is no single middle value. The heights of the six girls can be divided into two equal parts.

Statisticians have agreed that in cases like this the median is the mean of the two values closest to the middle. So the median is the mean of $62\,\mathrm{and}\,65,\,\frac{62+65}{2}.$ The median height is $63.5$ inches.

Notice that when the number of girls was $5,$ the median was the third height, but when the number of girls was $6,$ the median was the mean of the third and fourth heights. In general, when the number of values is odd, the median will be the one value in the middle, but when the number is even, the median is the mean of the two middle values.

Example 5.52

Solution

1. Arrange the seven values in increasing order. The original list is $12, 13, 19, 9, 11, 15,$ and $18$. Reordering it from least to greatest gives $9, 11, 12, 13, 15, 18, 19$.
2. Locate the middle value. There are $7$ values, so the middle is the fourth position. The figure shows the same ordered list with $13$ in the center, with three values $9, 11, 12$ below it and three values $15, 18, 19$ above it.
3. State the median. Because the center value is $13$ and the numbers on either side are balanced, the median of the data set is $13$.

Takeaway: For an odd number of ordered values, the median is the middle value, with equal numbers of data points on each side.

Example 5.53

Solution

1. Order the ten quiz scores from least to greatest. The scores become $64, 76, 79, 83, 85, 86, 88, 90, 92, 100$. The figure displays these ten values in this same increasing order.
2. Find the two middle values. With $10$ values, the middle positions are the fifth and sixth. They are $85$ and $86$, and the figure divides the ordered scores into two groups of five with these two center values highlighted.
3. Average the middle values. The median is $\frac{85+86}{2}=\frac{171}{2}=85.5$. Thus Kristen’s median quiz score is $85.5$ points.

Takeaway: For an even number of ordered values, average the two middle values to obtain the median.

Identify the Mode of a Set of Numbers

The average is one number in a set of numbers that is somehow typical of the whole set of numbers. The mean and median are both often called the average. Yes, it can be confusing when the word average refers to two different numbers, the mean and the median! In fact, there is a third number that is also an average. This average is the mode. The mode of a set of numbers is the number that occurs the most. The frequency, is the number of times a number occurs. So the mode of a set of numbers is the number with the highest frequency.

Suppose Jolene kept track of the number of miles she ran since the start of the month, as shown in Figure 5.7.

If we list the numbers in order it is easier to identify the one with the highest frequency.

Jolene ran $8$ miles three times, and every other distance is listed only once. So the mode of the data is $8$ miles.

Example 5.54

Solution

1. Count each repeated age. The frequencies include $4$ students aged $18$, $3$ aged $19$, $7$ aged $20$, $2$ aged $21$, and $5$ aged $22$.
2. Check the remaining frequencies. The ages $23,24,25,29,30,40,$ and $44$ each occur at most $2$ times, so none has frequency $7$.
3. Identify the greatest frequency. The largest frequency in the data set is $7$, which corresponds to the age $20$.
4. State the mode. Because the mode is the value occurring most often, the mode age is $20$ years.

Takeaway: The age $20$ occurs $7$ times, more often than every other age, so it is the mode of the college math class.

Example 5.55

Solution

1. Count the occurrences of each height. The table lists the heights $56, 60, 61, 62, 63, 64, 65, 66, 67, 70,$ and $74$. The frequency table in the figure shows that $56$ occurs $1$ time, $60$ occurs $2$ times, $61$ occurs $2$ times, $62$ occurs $1$ time, $63$ occurs $5$ times, $64$ occurs $4$ times, $65$ occurs $3$ times, $66$ occurs $5$ times, $67$ occurs $6$ times, $70$ occurs $1$ time, and $74$ occurs $1$ time.
2. Compare the frequencies. The greatest frequency is $6$, because the height $67$ appears six times in the data. Every other height appears at most five times, so no other value competes with $67$ for the highest frequency.
3. Identify the mode. The figure agrees with the data by pairing each height with its frequency and highlighting the row in which $67$ has the greatest count. The mode is therefore $67$ inches.

Takeaway: The mode is the value with the greatest frequency, not merely the largest or most central value.

Some data sets do not have a mode because no value appears more than any other. And some data sets have more than one mode. In a given set, if two or more data values have the same highest frequency, we say they are all modes.

Use the Basic Definition of Probability

The probability of an event tells us how likely that event is to occur. We usually write probabilities as fractions or decimals.

For example, picture a fruit bowl that contains five pieces of fruit - three bananas and two apples.

If you want to choose one piece of fruit to eat for a snack and don’t care what it is, there is a $\frac{3}{5}$ probability you will choose a banana, because there are three bananas out of the total of five pieces of fruit. The probability of an event is the number of favorable outcomes divided by the total number of outcomes.

Converting the fraction $\frac{3}{5}$ to a decimal, we would say there is a $0.6$ probability of choosing a banana.

This basic definition of probability assumes that all the outcomes are equally likely to occur. If you study probabilities in a later math class, you’ll learn about several other ways to calculate probabilities.

Example 5.56

Solution

1. Identify the favorable outcome. Cherie owns exactly $1$ ticket, and that ticket is the only favorable outcome for her winning the prize.
2. Count all possible outcomes. The club sold $100$ tickets, and each equally likely ticket could be the one selected from the jar.
3. Apply the probability definition. Dividing favorable outcomes by total outcomes gives $P(\text{Cherie wins})=\frac{1}{100}$.
4. Convert the probability to a decimal. Since $\frac{1}{100}=0.01$, Cherie's chance of winning the raffle prize is $0.01$.
5. Check the result. The probability is positive and no greater than $1$, as required for an event with one favorable outcome among $100$ possibilities.

Takeaway: Cherie has $1$ favorable ticket among $100$ equally likely tickets, so her probability of winning is $\frac{1}{100}=0.01$.

Example 5.57

Solution

1. Count the favorable candidates. There are $3$ women, and any of these $3$ people would make the favorable outcome for the requested event.
2. Count every possible candidate. The interview pool contains $3$ women and $5$ men, for a total of $3+5=8$ candidates.
3. Apply the probability definition. Dividing favorable outcomes by total outcomes gives $P(\text{a woman is offered the job})=\frac{3}{8}$.
4. Convert the probability to a decimal. Dividing $3$ by $8$ gives $0.375$, so the probability that the job is offered to a woman is $0.375$.
5. Check the result. Since $3$ favorable outcomes are fewer than all $8$ outcomes, the probability is correctly between $0$ and $1$.

Takeaway: There are $3$ women among $8$ candidates, so the probability that a woman receives the job is $\frac{3}{8}=0.375$.

Section 5.5 Exercises

Section 5.5 Exercises

Calculate the Mean of a Set of Numbers

In the following exercises, find the mean.

Find the Median of a Set of Numbers

In the following exercises, find the median.

Identify the Mode of a Set of Numbers

In the following exercises, identify the mode.

Use the Basic Definition of Probability

In the following exercises, express the probability as both a fraction and a decimal. (Round to three decimal places, if necessary.)

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well prepared for the next section? Why or why not?

1. The ages of the four students in Ben’s carpool are $25,18,21,\mathrm{and}\,22.$ Find the mean age of the students.

2. Last week Ray recorded how much he spent for lunch each workday. He spent $\text{\textdollar 6.50},\text{\textdollar 7.25},\text{\textdollar 4.90},\text{\textdollar 5.30},\mathrm{and}\,\text{\textdollar 12.00}.$ Find the mean of how much he spent each day.

1. Lisa has kept the receipts from the past four trips to the gas station. The receipts show the following amounts: $\text{\textdollar 34.87},\text{\textdollar 42.31},\text{\textdollar 38.04},\mathrm{and}\,\text{\textdollar 43.26}.$ Find the mean.

2. Ignaly is attending a fashion show where the guests are seated at tables of ten. One guest from each table will be selected at random to receive a door prize. ⓐ Find the probability Ignaly will win the door prize for her table. ⓑ Convert the fraction to a decimal.

1. Hoang is among $20$ people available to sit on a jury. One person will be chosen at random from the $20.$ ⓐ Find the probability Hoang will be chosen. ⓑ Convert the fraction to a decimal.

2. A bowl of Halloween candy contains $5$ chocolate candies and $3$ lemon candies. Tanya will choose one piece of candy at random. ⓐ Find the probability Tanya will choose a chocolate candy. ⓑ Convert the fraction to a decimal.

1. Dan has $2$ pairs of black socks and $6$ pairs of blue socks. He will choose one pair at random to wear tomorrow. ⓐ Find the probability Dan will choose a pair of black socks ⓑ Convert the fraction to a decimal.

2. Four girls leaving a mall were asked how much money they had just spent. The amounts were $\text{\textdollar 0}$, $\text{\textdollar 14.95}$, $\text{\textdollar 35.25}$, and $\,\text{\textdollar 25.16}$. Find the mean amount of money spent.

1. The number of minutes it took Jim to ride his bike to school for each of the past six days was $21$, $18$, $16$, $19$, $24$, and $19$. Find the mean number of minutes.

2. Last week Ray recorded how much he spent for lunch each workday. He spent $\text{\textdollar 6.50}$, $\text{\textdollar 7.25}$, $\text{\textdollar 4.90}$, $\text{\textdollar 5.30}$, and $\text{\textdollar 12.00}$. Find the median.

1. Brian is teaching a swim class for $6$ three-year olds. Their ages, in months, are $38,41,45,36,40,\mathrm{and}\,42.$ Find the median age.

2. The number of children per house on one block: $1$, $4$, $2$, $3$, $3$, $2$, $6$, $2$, $4$, $2$, $0$, $3$, $0.$

1. The number of units being taken by students in one class: $12$, $5$, $11$, $10$, $10$, $11$, $5$, $11$, $11$, $11$, $10$, $12$.

2. Josue is in a book club with $20$ members. One member is chosen at random each month to select the next month’s book. Find the probability that Josue will be chosen next month.

1. There are $24$ people who work in Dane’s department. Next week, one person will be selected at random to bring in doughnuts. Find the probability that Dane will be selected. Round your answer to the nearest thousandth.

2. Michel has four rock CDs and six country CDs in his car. He will pick one CD to play on his way to work. Find the probability Michel will pick a rock CD.

1. Donovan is considering transferring to a $\mathrm{4-year college}.$ He is considering $10$ out-of state colleges and $4$ colleges in his state. He will choose one college at random to visit during spring break. Find the probability that Donovan will choose an out-of-state college.

2. Joaquin gets paid every Friday. His paychecks for the past $8$ Fridays were $\text{\textdollar 315},$$\text{\textdollar 236.25},$$\text{\textdollar 236.25},$$\text{\textdollar 236.25},\text{\textdollar 315},$$\text{\textdollar 315},$$\text{\textdollar 236.25},$$\text{\textdollar 393.75}.$ Find the ⓐ mean, ⓑ median, and ⓒ mode.

1. Explain in your own words the difference between the mean, median, and mode of a set of numbers.