QuanticaPrealgebraOpen in the courseProbability gave one number for how likely something is. A whole list of numbers deserves the same treatment, one number that stands in for the crowd. There are three natural ways to name the center of a data set, and each answers a slightly different question. The mean shares everything out fairly, the median finds the value sitting in the middle, and the mode picks the one that shows up most.
Problem
Five aquarium tanks along a wall hold \(12\), \(7\), \(15\), \(9\), and \(12\) guppies. A keeper scoops guppies from tank to tank until every tank holds the same number, with none added and none removed overall. How many guppies end up in each tank? Give the number.
Show a hint
- Since no guppies are added or removed, the total across all five tanks stays fixed no matter how they get shuffled around. Start by adding up how many guppies there are altogether.
- Every tank must end with the same count, so take that total and split it evenly among the \(5\) tanks. That single equal share is the number each tank ends up holding.
Show the full solution
Moving guppies around does not change the total, so add the five tanks, \(12 + 7 + 15 + 9 + 12 = 55\), then split that evenly among the \(5\) tanks, $$\frac{55}{5} = \boxed{11}.$$ That equal share is the mean, the number you get when you level every value out.
Problem
A kite flyer records how many minutes her kite stayed up on four windy afternoons: \(18\), \(25\), \(22\), and \(30\) minutes. What is the mean number of minutes the kite stayed up per afternoon? Give your answer as a decimal.
Show a hint
- The mean is the total of all the values divided by how many values there are, so first add the four times \(18 + 25 + 22 + 30\).
- The sum is \(95\), and there are \(4\) afternoons, so divide \(95\) by \(4\) and notice the answer does not have to match any single afternoon.
Show the full solution
Add the four recorded times, \(18 + 25 + 22 + 30 = 95\), then divide by the \(4\) afternoons,
$$\text{mean} = \frac{95}{4} = \boxed{23.75}.$$
The mean is a balancing point for the whole set, so it can land between the values instead of on one of them.
Problem
A bakery records how many loaves it sells each day from Monday through Friday. On four of the days it sold \(40\), \(52\), \(38\), and \(61\) loaves, and the five-day average came out to exactly \(49\) loaves. How many loaves were sold on the fifth day? Give the number.
Show a hint
- The mean is the total divided by the number of days, so if the average of five days is \(49\), the five daily counts must add up to \(49 \times 5\). Work out that total first.
- You know four of the five values, so subtract their sum from the full total. Whatever is left over is the number of loaves sold on the missing day.
Show the full solution
The average of the five days is the total divided by \(5\), so the five days add to \(49 \times 5 = 245\). The four known days give \(40 + 52 + 38 + 61 = 191\), so the missing day is $$245 - 191 = \boxed{54}.$$ A mean and a count together always hand you the total, which is what makes a missing value findable.
Problem
A coding club has \(12\) beginners who averaged \(30\) points on a puzzle, and \(8\) advanced members who averaged \(55\) points. What is the average score of all \(20\) members together? It is tempting to just average \(30\) and \(55\), but that ignores how many people are in each group, so think about the total points first. Give the number.
Show a hint
- The mean of a group is total points divided by number of people, so you can run that backward. A group's total is its mean times its size, which gives \(12 \times 30\) points for the beginners and \(8 \times 55\) points for the advanced members.
- Add both totals to get the points scored by all \(20\) members, then divide that sum by \(20\). Averaging \(30\) and \(55\) to get \(42.5\) would be wrong here because the two groups are not the same size, and the larger beginner group pulls the true average down.
Show the full solution
Rebuild the totals. The beginners scored \(12 \times 30 = 360\) points and the advanced members scored \(8 \times 55 = 440\) points, so all \(20\) together scored \(800\) points and the average is $$\frac{12 \times 30 + 8 \times 55}{20} = \frac{800}{20} = \boxed{40}.$$ Averaging \(30\) and \(55\) would give \(42.5\), but the groups are different sizes, and the larger beginner group pulls the combined average down toward its own score.
Problem
Seven kids skip stones across a pond and count the skips: \(4, 9, 6, 3, 8, 7, 11\). The median is the middle score once the list is sorted from smallest to largest. What is the median number of skips? Give the number.
Show a hint
- The list is scrambled, so sort it from smallest to largest before you look for the middle. Sorting gives \(3, 4, 6, 7, 8, 9, 11\).
- With \(7\) values in order, the middle one is the \(4\)th, since three values sit below it and three sit above it.
Show the full solution
Sort the counts from smallest to largest, $$3,\ 4,\ 6,\ \underline{7},\ 8,\ 9,\ 11.$$ With \(7\) values the middle one is the \(4\)th, with three below it and three above it, so the median is \(\boxed{7}\). Sorting comes first every time, since middle only means something once the list is in order.
Problem
Six friends compare how many books they read over the summer: \(5, 12, 8, 3, 9, 14\). What is the median number of books? Give your answer as a decimal.
Show a hint
- The median is the middle value once the data is in order, so start by sorting the six counts from smallest to largest.
- With six values there is no single middle number, so look at the two values sitting in the middle and take their mean.
Show the full solution
Sort the counts, giving \(3, 5, 8, 9, 12, 14\). Six values means no single number sits in the middle, so take the two middle ones, \(8\) and \(9\), and average them, $$\frac{8 + 9}{2} = \frac{17}{2} = \boxed{8.5}.$$ With an even count the median is often not one of the numbers in the list.
Problem
A jacket shop records the sizes it sold today: \(6, 8, 8, 10, 8, 12, 8, 10, 6, 8\). They want to reorder the most of whichever size sells best, so they need the size that appears most often in the list. Which size should they reorder the most of? Give the size.
Show a hint
- The size they should reorder most is the mode, the value that appears most often, so tally how many times each size shows up in the list.
- Count carefully. Size \(6\) appears twice, size \(10\) appears twice, and size \(12\) appears once, but one size beats them all, and that highest count is the mode.
Show the full solution
Tally the sizes in \(6, 8, 8, 10, 8, 12, 8, 10, 6, 8\). Size \(6\) shows up twice, size \(10\) twice, size \(12\) once, and size \(8\) five times, so the most frequent size is $$\boxed{8}.$$ The mode is the right summary here, since the shop cares about which size sells most, not about an average size.
Problem
A list can have more than one mode. A survey records these numbers: \(2, 5, 2, 7, 5, 9, 5, 2\). Both \(2\) and \(5\) appear three times, which is more often than any other value. A value counts as a mode when it ties for the most frequent. How many values are modes?
Show a hint
- Tally how many times each different value shows up. Count the \(2\)s, the \(5\)s, the \(7\), and the \(9\) separately.
- The mode is the value with the highest count, but if two values share that same top count, they are both modes. Look for which values reach the largest tally and count how many there are.
Show the full solution
Count how often each value appears. The value \(2\) shows up three times, \(5\) shows up three times, and \(7\) and \(9\) show up once each. Two values tie for the top count of three, so both are modes and the number of modes is $$\boxed{2}.$$ A tie for most frequent does not cancel out, it just gives the list two modes, which is what "multimodal" means.
Problem
A five-person startup pays its four employees \(42\), \(48\), \(45\), and \(51\) thousand dollars a year, while the founder earns \(314\) thousand. By how many thousand dollars does the mean salary exceed the median salary? Give the number (of thousand dollars).
Show a hint
- The mean is the total of all five salaries divided by \(5\), so add \(42 + 48 + 45 + 51 + 314\) and split the sum evenly across the five people.
- The median is the middle value once the salaries are lined up in order, so sort them as \(42, 45, 48, 51, 314\) and read off the one in the center, then subtract it from the mean.
Show the full solution
The mean adds all five salaries and divides by \(5\),
$$\text{mean} = \frac{42 + 48 + 45 + 51 + 314}{5} = \frac{500}{5} = 100.$$
Ordering the salaries as \(42, 45, 48, 51, 314\) puts \(48\) in the middle, so the median is \(48\) and the gap is
$$100 - 48 = \boxed{52}.$$
The four ordinary salaries hold the median in place, while the founder's \(314\) thousand pulls the mean far above it. That is what a single large value does to an average.
Problem
A gymnast's five routine scores average \(9.2\). She has one routine left, and she wants her average across all six scores to stay exactly \(9.2\). What score must she earn on the sixth routine? Give your answer as a decimal.
Show a hint
- The mean is the total of all scores divided by how many there are. With five scores averaging \(9.2\), the current total is \(5 \times 9.2 = 46\).
- After the sixth routine you will divide a new total by \(6\), and you want that to equal \(9.2\). Ask what sixth score keeps the total-over-count balanced, and notice that adding one more copy of the mean itself leaves the mean untouched.
Show the full solution
Five scores averaging \(9.2\) total \(5 \times 9.2 = 46\). Call the sixth score \(x\), so the six-score average is
$$\frac{46 + x}{6} = 9.2.$$
Multiply both sides by \(6\) to get \(46 + x = 55.2\), then subtract \(46\), so she must score \(\boxed{9.2}\). Adding one more value equal to the current mean leaves the mean exactly where it was, since you are adding an average-sized share to an average-sized pile.
Problem
A staircase has an LED on every step. The steps show \(7, 11, 15, 19,\) and so on, going up by \(4\) each step, with the top step showing \(43\). What is the mean of all the numbers shown? Give the number.
Show a hint
- The numbers are evenly spaced, each one \(4\) more than the last, so pair them from the outside in. The smallest \(7\) pairs with the largest \(43\), the next \(11\) pairs with \(39\), and so on, just like pairing the ends when you add \(1\) up to \(n\).
- Every pair adds to the same total, \(7+43 = 50\), so every number sits the same distance above or below the center. That means you never have to add the whole list, since the mean is just the value halfway between the two ends.
Show the full solution
The numbers are evenly spaced, each one \(4\) above the last, so pair them from the ends inward. \(7\) pairs with \(43\), \(11\) with \(39\), \(15\) with \(35\), and every pair sums to \(50\). The values balance around a single center, so the mean is the average of the two ends,
$$\frac{7 + 43}{2} = \frac{50}{2} = \boxed{25}.$$
For any evenly spaced list the mean equals the median and equals the average of the first and last term, so you never have to add the whole thing.
Problem
A gym sets your membership tier by the median of your nine monthly visit counts. Seven months in, your counts are \(20, 8, 16, 5, 22, 12, 16\). You have two months left, and you can visit at most \(30\) times in a month. What is the highest median you can reach across all nine months? Give the number.
Show a hint
- With nine values, the median is the \(5\)th number once they are sorted, so the whole question is where that middle slot lands.
- Each new month can be at most \(30\), which is larger than everything you already have, so both new values drop in above your current numbers and none of them can slide into the \(5\)th slot from below.
Show the full solution
Sort the seven known counts, $$5,\ 8,\ 12,\ 16,\ 16,\ 20,\ 22.$$ With nine values the median is the \(5\)th, and each new month is at most \(30\), which is bigger than everything already on the list, so both new counts land at the top. Adding only above the middle leaves the \(5\)th value right where it was, at $$\boxed{16}.$$ Pushing the new months higher stretches the top of the list, never the middle, which is exactly why the median resists big values.
Problem
A set of five positive whole numbers has a mean of \(16\), a median of \(15\), and \(15\) is its only mode. What is the greatest value the largest number in the set could be? Give the number.
Show a hint
- A mean of \(16\) across five numbers fixes the total. Multiply to get the sum, \(5 \times 16 = 80\), so however you fill the five slots they must add to \(80\).
- Sort the five numbers. The median is the 3rd value, so the middle slot is \(15\). For \(15\) to be the only mode it has to appear at least twice, so make the 4th slot \(15\) too, and to leave the largest as big as possible make the two smallest slots the tiniest distinct positive integers that do not tie \(15\), namely \(1\) and \(2\).
Show the full solution
Five numbers averaging \(16\) add to \(5 \times 16 = 80\). Sort the set, so the 3rd slot is the median and holds \(15\), and \(15\) needs a second copy to be a mode, so slot 4 is \(15\) too. To leave the top number as large as possible, make the first two slots the smallest positive whole numbers that do not tie \(15\) for most frequent, namely \(1\) and \(2\). The largest number takes the rest of the total,
$$80 - (1 + 2 + 15 + 15) = \boxed{47}.$$
Checking, \(\{1, 2, 15, 15, 47\}\) sums to \(80\) for a mean of \(16\), has middle value \(15\), and repeats only \(15\).
Practice these ideas
Practice
Find the mean of the four numbers \(14\), \(9\), \(22\), and \(15\). Give the number.
Show the solution
Add the four numbers, \(14 + 9 + 22 + 15 = 60\), then divide by the \(4\) values, $$\text{mean} = \frac{60}{4} = \boxed{15}.$$
Practice
Find the median of the data set \(7, 2, 9, 4, 6\). Give the number.
Show the solution
Sort the five values, giving \(2, 4, 6, 7, 9\). With five numbers there is a single middle position, the third, with two numbers below it and two above it, so the median is \(\boxed{6}\).
Practice
Find the mode of this data set: \(3, 7, 3, 9, 3, 7, 5\). Give the number.
Show the solution
Tally how often each number appears. The value \(3\) shows up three times, \(7\) twice, and \(9\) and \(5\) once each, so the value appearing most often is \(\boxed{3}\).
Practice
Find the mean of the four numbers \(10\), \(13\), \(14\), and \(10\). Give your answer as a decimal.
Show the solution
Add the four numbers, \(10 + 13 + 14 + 10 = 47\), then divide by \(4\), $$\text{mean} = \frac{47}{4} = \boxed{11.75}.$$ A mean does not have to be a whole number even when every value in the list is.
Practice
A weather station takes five daily temperature readings, and their mean is \(20\) degrees. Four of the readings are \(15\), \(22\), \(18\), and \(25\). What is the fifth reading? Give the number.
Show the solution
Five readings with a mean of \(20\) must total \(5 \times 20 = 100\). The four known readings add to \(15 + 22 + 18 + 25 = 80\), so the fifth is
$$100 - 80 = \boxed{20}.$$
It lands right on the mean, which is what happens when the values you already have average out to \(20\) on their own.
Practice
Find the median of these six numbers: \(12, 4, 8, 20, 6, 10\). Give the number.
Show the solution
Sort the six numbers, giving \(4, 6, 8, 10, 12, 20\). With an even count no single value sits in the middle, so average the \(3\)rd and \(4\)th, $$\frac{8 + 10}{2} = \boxed{9}.$$
Practice
On a test, a group of \(10\) students averaged \(80\) points and a separate group of \(15\) students averaged \(90\) points. What is the average score of all \(25\) students combined? Give the number.
Show the solution
Rebuild each group's total from its average. The \(10\) students averaging \(80\) scored \(10 \times 80 = 800\) points, and the \(15\) students averaging \(90\) scored \(15 \times 90 = 1350\) points. Divide the combined total by all \(25\) students, giving
$$\frac{800 + 1350}{25} = \frac{2150}{25} = \boxed{86}.$$
The result sits closer to \(90\) than to \(80\) because the larger group pulls the mean toward its own score.
Practice
A data set has the values \(8\), \(10\), \(9\), \(11\), and \(62\). The mean of these five numbers is \(20\), and the median is \(10\). By how much does the mean exceed the median? Give the number.
Show the solution
The mean is \(\frac{8 + 10 + 9 + 11 + 62}{5} = \frac{100}{5} = 20\), and sorting the list to \(8, 9, 10, 11, 62\) puts \(10\) in the middle, so the difference is $$20 - 10 = \boxed{10}.$$ The outlier \(62\) drags the mean well above the typical value while leaving the median untouched, which is why the two centers disagree.
Practice
The ten multiples of \(3\) from \(3\) up to \(30\) are \(3, 6, 9, \dots, 30\), and they add up to \(165\). Remove one number so that the nine numbers left have a mean of \(16\). Which number is removed? Give the number.
Show the solution
Nine numbers with a mean of \(16\) must add to \(9 \times 16 = 144\). All ten add to \(165\), so the removed number is the gap between those totals, $$165 - 144 = \boxed{21}.$$ And \(21\) really is on the list, since it is a multiple of \(3\) between \(3\) and \(30\).
Practice
Six different positive whole numbers have a mean of \(10\), so they add up to \(60\). What is the largest that any single one of them can be? Give the number.
Show the solution
A mean of \(10\) across six numbers fixes the total at \(6 \times 10 = 60\). To push one value as high as possible, make the other five as small as they can be, and since they must be different positive whole numbers that is \(1, 2, 3, 4, 5\), which adds to \(15\). The rest goes to the sixth number,
$$60 - 15 = \boxed{45}.$$
Practice
The mean of five numbers is \(12\). A sixth number is added, and the mean of all six numbers rises to \(14\). What is the sixth number? Give the number.
Show the solution
The first five numbers total \(5 \times 12 = 60\), and all six total \(6 \times 14 = 84\), so the sixth number is $$84 - 60 = \boxed{24}.$$ Turning each mean back into a total is what makes the added value easy to read off.
Practice
A shoe shop sold these sizes over one afternoon: \(7, 8, 8, 9, 8, 10, 7, 8\). The owner wants to restock the most common size, since that is what customers buy the most. Which size should the shop restock the most of? Give the size.
Show the solution
Tally the sizes sold. Size \(7\) appears twice, \(9\) once, \(10\) once, and \(8\) appears in four of the eight sales, so the mode is $$\boxed{8}.$$ The shop should restock size \(8\), the size customers bought most often.
Practice
Find the average of all the whole numbers from \(41\) to \(59\), including both ends. Give the number.
Show the solution
The numbers \(41, 42, 43, \dots, 59\) are evenly spaced, one apart, so the mean is the average of the two ends, $$\frac{41+59}{2}=\frac{100}{2}=\boxed{50}.$$ That saves adding all nineteen numbers, and it works for any evenly spaced list because the values balance around their center.
Practice
A set of five positive whole numbers has a mean of \(6\), a median of \(6\), and \(8\) as its only mode. What is the difference between the largest and smallest values in the set? Give the number.
Show the solution
A mean of \(6\) across five numbers fixes the total at \(5 \times 6 = 30\), and a median of \(6\) puts \(6\) in the middle slot once they are sorted. For \(8\) to be the only mode it must appear more than any other value, so the two largest slots are both \(8\), giving \(6, 8, 8\) and a running total of \(22\). The two smallest must add to \(30 - 22 = 8\) while staying different from each other and from \(8\), and the pair \(3\) and \(5\) works, so the set is \(3, 5, 6, 8, 8\) and
$$8 - 3 = \boxed{5}.$$