QuanticaPrealgebraOpen in the courseYou can now find the mean, the median, and the mode of any list of numbers. This lesson asks the harder question, which is what those single numbers leave out. Every statistic is a summary, and a summary always drops information. Sometimes what it drops is minor. Sometimes it is the whole story.
Problem
Two basketball teams both average 20 points a game. The Hawks scored 19, 20, and 21 points in their three games. The Falcons scored 5, 20, and 35. Find the range of the Falcons' scores, meaning their highest score minus their lowest score. Give the number.
Show a hint
- Both teams share the same mean, so the average alone tells you nothing about how the three games differed. Look instead at how far apart the Falcons' scores stretch.
- The range measures spread, not center. Subtract the smallest Falcons score from the largest and see how much wider it is than anything the Hawks did.
Show the full solution
The Falcons scored 5, 20, and 35, so their range is $$35 - 5 = \boxed{30}.$$ The Hawks' range is only 2. Two teams with the same average can be completely different in how steady they are, and the mean says nothing about that.
Problem
A company advertises that its "average salary" is 80 thousand dollars, and technically that is true. But that mean is pulled up by a handful of executives who earn far more than everyone else. The median salary is 45 thousand, and nearly every worker earns close to that. Which number honestly describes what a typical worker at this company earns, in thousands? Give the number (in thousands).
Show a hint
- A statistic can be true and still misleading. Ask what the number is standing in for here, which is the pay of a normal, in-the-middle worker, and then ask which measure actually sits in the middle.
- A few executive salaries are large enough to drag the mean upward, away from where most people actually are. The measure that ignores how extreme those top values get, and just marks the middle of the line, is the one that stays honest.
Show the full solution
Nearly every worker earns close to the median, so the honest number is \(\boxed{45}\) thousand. The mean of 80 is real but misleading, since a handful of executive salaries get spread across the total and lift it. The median just marks the middle worker, so huge values at the top leave it alone.
Problem
Nine students turned in a project, each scored out of 10. Their scores were 3, 3, 3, 3, 7, 8, 9, 10, and 10. The mode is 3, which makes it sound like the class struggled. Find the median instead, since it better reflects how the class actually did. Give the number.
Show a hint
- The mode is just the score that shows up most often. Four students landed on a 3, so it wins the count, but ask yourself whether most of the class was really down there.
- The median is the middle value once the scores are lined up in order. With nine scores, count in to the 5th one from either end.
Show the full solution
The nine scores are already in order, so the median is the 5th value.
$$3,\ 3,\ 3,\ 3,\ \boxed{7},\ 8,\ 9,\ 10,\ 10$$
The top half is 7, 8, 9, 10, 10, so most of the class did well. The mode only reports which score appears most often, and here that one clump of 3's is not typical of the class.
Problem
Anna took seven quizzes and scored 94, 93, 90, 93, 92, and 91. She missed the last quiz and scored a 0. Because of that single 0, her mean sits at about 79, which makes her look like a middling student. Find the median of her seven scores, the summary number that shrugs off the lone 0. Give the number.
Show a hint
- The mean got dragged down because it adds in the 0 and shares the damage across every score. The median works differently, so ask yourself what the median actually looks at.
- Put all seven scores in order from lowest to highest. The median is the single value sitting exactly in the middle, so the one stray 0 at the bottom of the list barely gets a vote.
Show the full solution
Order the seven scores, giving $$0,\ 90,\ 91,\ 92,\ 93,\ 93,\ 94.$$ With seven values the median is the 4th, the one with three scores on each side, so it is \(\boxed{92}\).
The single 0 is an outlier, and it drags the mean down to about 79. The median depends only on position, so it steps right past that missed quiz.
Problem
A basketball player's five games scored 40, 45, 50, 55, and 110 points. Find the mean and the median of these five scores, then give the mean minus the median. Give the number.
Show a hint
- Line the five scores up in order and pick the middle one for the median, then add all five and divide by five for the mean. Notice they will not come out equal.
- One score, the 110, sits far above the rest. Think about which statistic that giant number drags upward and which one it barely touches.
Show the full solution
Sorted, the scores are 40, 45, 50, 55, 110, so the median is 50. The mean is $$\frac{40+45+50+55+110}{5}=\frac{300}{5}=60,$$ so the mean minus the median is $$60-50=\boxed{10}.$$ The one 110-point game left the middle spot alone but pulled the mean up. A gap like this is a sign that a few large values are stretching the average.
Problem
Richville has an average wealth of 150 thousand dollars per person, and Poorville has an average of 20 thousand dollars per person. The two towns merge into one. If Richville and Poorville have exactly the same number of people, what is the merged town's average wealth, in thousands? Give the number.
Show a hint
- When two groups have the same number of people, each group's average counts for exactly the same share of the combined total, so the merged average sits right in the middle of the two.
- Watch what makes this work. It is the equal sizes. Add the two averages and split the result in half, and ask yourself whether you could still do that if one town were much larger than the other.
Show the full solution
The two towns are the same size, so each contributes an equal share of the merged total and you can average the two averages.
$$\frac{150 + 20}{2} = \boxed{85}$$
The merged average is 85 thousand per person. That shortcut works only because the sizes match. If Poorville had ten times as many people, the merged average would sit far closer to 20.
Problem
Two towns are merging into one. Richville has 3 residents with an average wealth of 150 thousand dollars each. Poorville has 7 residents with an average wealth of 20 thousand dollars each. After the merge the new town has all 10 people. Find the average wealth of the merged town, in thousands. Give the number.
Show a hint
- You cannot just average the two town averages, because the towns are not the same size. To find the total wealth of a group, multiply its average by how many people it holds.
- Poorville has more than twice as many residents as Richville, so its lower average gets far more say. Add up all the wealth, then divide by all 10 people.
Show the full solution
Richville holds \(3 \times 150 = 450\) thousand in total wealth and Poorville holds \(7 \times 20 = 140\) thousand. Divide the combined wealth by all 10 residents.
$$\frac{450 + 140}{10} = \boxed{59}$$
That sits well below the midpoint of 85, because Poorville has more than twice as many people and pulls harder. This is a weighted mean, and group size is what decides where it lands.
Problem
Five houses on a block are worth 10, 14, 20, 26, and 30 in ten-thousands of dollars, so their median value is 20. A developer builds a sixth house on the block, a mansion worth 5000 in the same units (fifty million dollars). Find the new median of all six houses, in ten-thousands of dollars. Give the number.
Show a hint
- The median depends only on the position of the middle value, not how far the largest number sits from the others. With six houses now, there is no single middle, so the median is the average of the two central values once everything is in order.
- Sort all six values and find the two that land in the middle. The giant mansion sits at the far end, so it never becomes a middle value, and it can only push the median a little.
Show the full solution
Sorted, the six values are 10, 14, 20, 26, 30, 5000. With an even count the median is the average of the two middle numbers, 20 and 26.
$$\frac{20 + 26}{2} = \boxed{23}$$
A fifty million dollar mansion moved the median from 20 to 23. The mean would rocket past 800, which is exactly why the median resists an extreme value.
Problem
A basketball team of 11 players has a median height of 76 inches. Someone claims the tallest player must therefore be well above 76 inches. What is the shortest the tallest player could possibly be? Give the number in inches.
Show a hint
- The median is the height of the 6th player once you line all 11 up in order. The five players above that spot are each at least as tall as the median, and the tallest is the top of that group. So all you know for sure is that the tallest is at least 76 inches.
- Ask yourself whether anything forces the top players to be strictly taller. Ties are allowed, so imagine the whole top half measuring exactly 76 inches. That is a legal team with a median of 76, and its tallest player is 76 too.
Show the full solution
With 11 heights in order the median is the 6th, so it is 76 inches and the five players above it are each at least 76. Ties are allowed, so every one of them can sit at exactly 76.
$$76,\ 76,\ 76,\ 76,\ 76,\ \underbrace{76}_{\text{median}},\ 76,\ 76,\ 76,\ 76,\ 76$$
The median is still 76, so the tallest player can be as short as \(\boxed{76}\) inches.
A median fixes the middle and leaves the extremes free, so a median of 76 fits a tallest player of 76 or of 90 equally well.
Problem
Five families have a median income of 40 thousand dollars. The two lowest earners make 10 thousand and 25 thousand. What is the smallest the total income of all five families could be? Give the number (in thousands).
Show a hint
- The median only pins down the middle family, so the third income when the five are sorted is fixed at 40. The two lowest are already given as 10 and 25, both below it.
- The two families above the median can earn anything at or above 40. To make the total as small as possible, push both of them down as low as they are allowed to go, which is exactly 40.
Show the full solution
Sort the five incomes. The median of 40 is the third value, and the two given low earners, 10 and 25, sit below it. The top two must each be at least 40, so make both exactly 40.
$$10 + 25 + 40 + 40 + 40 = \boxed{155}$$
The smallest possible total is 155 thousand. A median pins the middle and nothing above it, so those two families could be at 40 or at 4000 and the median would not change.
Problem
A class of 20 students took a test scored out of 100, and their mean was exactly 75. Nothing else about the individual scores is known. Consider these four statements. (1) At least one student scored 75 or higher. (2) The twenty scores add up to 1500. (3) At least one student scored below 75. (4) No one scored a perfect 100. How many of the four statements must be true? Give the number.
Show a hint
- A mean of 75 pins down the total, since the total is always the mean times the count. But knowing the total tells you nothing about how any single score is spread around 75.
- For each statement, try to build a class that breaks it. If you cannot break it, it must be true. Ask yourself whether every score could sit exactly at 75, or whether one score could be jammed all the way up to 100.
Show the full solution
Total equals mean times count, so the scores add to \(75 \times 20 = 1500\) and (2) is forced. If every student scored below 75 the mean would land below 75, so (1) is forced too. (3) fails when all twenty score exactly 75, and (4) fails when one student scores 100 and the other nineteen keep the total at 1500. That leaves $$\boxed{2}$$ statements that must be true.
A mean fixes the total and forces at least one value at or above it. Everything about spread and extremes stays open.
Problem
A company has 10 people. Their mean salary is 100 thousand dollars and their median salary is 40 thousand dollars. The nine non-executive workers all earn exactly the same amount, and the tenth person is the CEO. How much does the CEO earn? Give the number (in thousands).
Show a hint
- The mean is the total shared out evenly, so run it backward. If 10 people average 100, the whole payroll must add up to \(100 \times 10 = 1000\) thousand.
- With 10 people the median sits between the 5th and 6th salaries. Both of those are ordinary workers, and every worker earns the same, so each worker's salary is just 40. That pins down nine of the ten numbers.
Show the full solution
Ten people averaging 100 means the payroll adds to \(100 \times 10 = 1000\) thousand. With ten salaries the median is the average of the 5th and 6th, both ordinary workers, so all nine workers earn 40 each, or \(9 \times 40 = 360\) thousand. The CEO takes what is left.
$$1000 - 360 = \boxed{640}$$
A mean of 100 against a median of 40 is the fingerprint of one enormous outlier, and here the two summaries together pin down its exact size.
Practice these ideas
Practice
Two chess clubs each average 50 points per match. Club A scored 50, 50, and 50. Club B scored 10, 50, and 90. Find the range of Club B's scores. Give the number.
Show the solution
Club B's high is 90 and its low is 10, so the range is $$90 - 10 = \boxed{80}.$$ Club A's range is 0. Both clubs average 50, yet one never moved and the other swung wildly. Spread is what the average leaves out.
Practice
Ben played five basketball games and scored 88, 90, 85, 92, and 0 points. He fouled out early in one game, which is why one of his scores is 0. Find the median of his five scores. Give the number.
Show the solution
Sort the five scores from smallest to largest and take the third one.
$$0,\ 85,\ \boxed{88},\ 90,\ 92$$
Ben's median is 88 points. The 0 from fouling out drags the mean down to \(\frac{0+85+88+90+92}{5}=71\), but the median barely notices it and stays right at his typical game.
Practice
Five houses on a street sold for 20, 22, 24, 26, and 58 ten-thousands of dollars. Find the mean and the median of these five prices, then give the mean minus the median. Give the number.
Show the solution
The mean is $$\frac{20+22+24+26+58}{5}=\frac{150}{5}=30,$$ and the median is the third price in order, which is 24. The difference is $$30-24=\boxed{6}.$$ Four of the five houses sold between 20 and 26, yet the mean says 30. When mean and median disagree this much, one large value is hiding inside the average.
Practice
A team of 9 players has a median height of 70 inches. Someone reads that and pictures nine tall people. But the median only pins down the middle player once the heights are lined up in order. Four players are shorter than the middle one and four are taller, and nothing says how tall those four at the top can climb. What is the smallest possible height, in inches, of the tallest player on the team? Give the number in inches.
Show the solution
With 9 heights in order the median is the 5th, so that player is exactly 70 inches and the four above are each at least 70. To make the tallest as short as possible, let all four tie the median, so the top of the list reads $$70,\ 70,\ 70,\ 70.$$ The tallest can be no shorter than \(\boxed{70}\) inches.
The median fixes the middle and says nothing about the tails, so it cannot tell you whether the team is all one height or stretches far above it.
Practice
A server's five weekly tips, in dollars, are 4, 8, 10, 14, and 20, so the median is 10. One night a grateful customer leaves a huge 9000 dollar tip, added as a sixth value. Find the median of all six tips. Give the number.
Show the solution
Sorted, the six tips are 4, 8, 10, 14, 20, 9000. With an even count the median is the average of the two middle ones, 10 and 14, so $$\frac{10 + 14}{2} = \boxed{12}.$$ A tip 9000 dollars above everything else moved the median only from 10 to 12. The mean would swing enormously, while the median just slides to the next spot in the sorted list.
Practice
Five numbers have a mean of 12. Consider these four statements about them. (1) The five numbers add to 60. (2) At least one of them is 12 or more. (3) They are not all equal. (4) At least one of them is 12 or less. Some of these are forced by the mean and some are not. How many of the four statements must be true?
Show the solution
The five numbers add to \(5 \times 12 = 60\), so (1) is forced. If all five were below 12 the total would fall short of 60, and if all were above 12 it would overshoot, so (2) and (4) are forced too. (3) fails, since five 12's average to 12 and are all equal. That gives $$\boxed{3}$$ statements that must be true.
The mean hands you the total and the center, but it hides the spread. Identical values and wildly scattered ones can share a mean.
Practice
Team A has 20 players who average 90 points each. Team B has 20 players who average 30 points each. The two teams merge into one group of 40 players. Find the average of the merged group. Give the number.
Show the solution
Both teams have 20 players, so they carry equal weight and you can average the averages. Halfway between 90 and 30 is $$\frac{90+30}{2}=\boxed{60}.$$ No player on either team actually scored 60. The average lands in the empty gap between the two clusters, which is what a single number hides.
Practice
A gym advertises that its members' average number of monthly visits is 70. But a few members come almost every day, and their high counts pull that average up. The median number of visits is 40, which sits much closer to what most members actually do. Which number honestly describes a typical member, the average of 70 or the median of 40? Give the number.
Show the solution
A few near-daily regulars log huge visit counts, and the mean gets pulled toward them. The median just marks the middle of the group, so the honest number for a typical member is \(\boxed{40}\) visits.
When a handful of values sit far above the rest, the mean drifts toward them and stops describing anyone real.
Practice
A tutoring center has two study groups. Group A has 4 students who averaged 100 on a test, and Group B has 6 students who averaged 55 on the same test. All 10 students are combined into one list. Find the average score of the combined group. Give the number.
Show the solution
An average times the count gives back the total. Group A scored \(4 \times 100 = 400\) points and Group B scored \(6 \times 55 = 330\). Divide the combined total by all 10 students.
$$\frac{400 + 330}{10} = \frac{730}{10} = \boxed{73}$$
It lands closer to 55 than to 100 because the larger group pulls harder. A plain midpoint of 77.5 would hide that entirely.
Practice
Five stores report their daily sales in hundreds of dollars. The median of the five values is 30, and the two lowest stores sold 5 and 12. What is the smallest possible total for all five stores? Give the number.
Show the solution
The median of 30 is the third value, so the top two stores can be no lower than 30. Push both down to exactly 30, giving the values 5, 12, 30, 30, 30.
$$5 + 12 + 30 + 30 + 30 = \boxed{107}$$
The smallest possible total is 107 hundreds of dollars. The median puts a floor under the total but no ceiling, since those top two stores could have sold thousands and the median would not budge.
Practice
Nine people rated a movie on a scale from 1 to 13. Their ratings were 2, 2, 2, 2, 9, 10, 11, 12, and 13. Someone reports the mode and says the typical rating was low. Find the median instead. Give the number.
Show the solution
The nine ratings are already in order, so the median is the 5th value.
$$2,\ 2,\ 2,\ 2,\ \boxed{9},\ 10,\ 11,\ 12,\ 13$$
The median is 9, far above the mode of 2. The mode only counts which value appears most, and here that clump of low ratings hides the fact that most people liked the movie.
Practice
Ten test scores have a mean of 60. Nine of the ten scores are each 50. Find the tenth score. Give the number.
Show the solution
Ten scores averaging 60 must sum to \(10 \times 60 = 600\). Nine scores of 50 contribute \(9 \times 50 = 450\), so the tenth score is $$600 - 450 = \boxed{150}.$$ A mean of 60 sounds like a room full of ordinary scores near 60, but here nine sit far below it and one value does all the heavy lifting.