Unit 17 · Statistics and Probability

Ask one friend, "How many pets do you have?" You get one answer, maybe 2. Ask the same question to everyone in your class and you get a whole set of answers: 0, 0, 1, 2, 3, 0, 1, and so on. That set of answers is data. A statistical question is a question you expect to have many different answers, so you need data to answer it. "How many pets does Ana have?" is not statistical. "How many pets do students in our class have?" is.

The word for the differences among the answers is variability. Heights vary, bus wait times vary, and the number of pets varies. If there is no variability, you do not need statistics. "How many days are in a week?" has one answer, so it is not a statistical question. "How many minutes did students in Illinois spend on homework last night?" has thousands of answers, so it is.

To answer a statistical question, follow four steps. First, ask the question clearly. Second, collect the data from a group. Third, organize and picture the data with a plot. Fourth, describe the data with numbers such as the center and the spread. A common mistake is to stop after one answer, like "My cousin has 3 pets, so seventh graders have 3 pets." One answer is a fact about one person, not a description of a group.

Ten students report their number of pets: 1, 0, 2, 1, 3, 0, 1, 2, 0, 1. A dot plot shows this on a number line from 0 to 3. Put one dot above the number for each answer and stack repeats. Above 0 you stack 3 dots, above 1 you stack 4 dots, above 2 you stack 2 dots, and above 3 you place 1 dot. Count the dots: 3 + 4 + 2 + 1 = 10, which matches the ten students. Always check that total.

The dot plot answers questions at a glance. The tallest stack shows the most common value, called the mode; here it is 1 pet. The dots spread from 0 to 3, so the smallest and largest values are easy to spot. You can also count how many students have at least 2 pets: 2 + 1 = 3 students, which is 3 out of 10. The height of each stack is that value’s frequency.

A common mistake is to put the dot in the wrong place, for example stacking a dot at 2 when the student said 3 pets. Another is skipping numbers on the line. The line must include every whole number from the smallest value to the largest, even if a number has no dots. A gap with no dots is information too. Dot plots work best for small sets of whole numbers, like pets or siblings. For hundreds of values, or measurements like 157.3 cm, a histogram works better.

Twenty students record last night’s homework time in minutes; the values range from 5 to 75. A dot plot with a dot at 5, 12, 18, 23, 25, 27 and so on would be hard to read. Instead, group the values into equal intervals: 0–19, 20–39, 40–59 and 60–79 minutes. Count how many values land in each: 3, 8, 6 and 3. Check: 3 + 8 + 6 + 3 = 20. A histogram draws a bar for each interval, with height equal to the count. The bars touch because the intervals sit next to each other on the number line.

Reading the histogram: the tallest bar is 20–39 minutes, so most students spent between 20 and 39 minutes. Nine students (6 + 3) spent 40 minutes or more. Nine out of twenty is 9/20 = 0.45, or 45% of the class. You cannot see exact values, only the interval each value belongs to. That is the trade-off: a histogram shows the shape clearly but hides the details.

Two common mistakes: using intervals of different widths, which makes bars unfair to compare, and putting a value on the wrong side of a boundary. Decide ahead of time that 20 minutes belongs to 20–39, not 0–19, and stick with it. Changing the interval width changes the picture. Intervals of 10 minutes would give eight bars and more detail; intervals of 40 would give two bars and almost no detail.

Five friends bring cookies to share: 3, 7, 4, 6 and 5 cookies. If they pool the cookies and share equally, how many does each get? Add: 3 + 7 + 4 + 6 + 5 = 25. Divide by the 5 friends: 25 ÷ 5 = 5. Each friend gets 5 cookies. That fair-share number is the mean, also called the average. The mean is the value every data point would have if the total were spread out evenly.

Try it with four test scores: 80, 90, 70 and 100. The sum is 80 + 90 + 70 + 100 = 340. Divide by 4 scores: 340 ÷ 4 = 85. The mean score is 85. Notice that the mean does not have to be one of the values, and it can be a decimal. The mean of 2, 3 and 5 is 10 ÷ 3, which is about 3.33.

You can also work backward. Suppose four numbers have a mean of 10, and three of them are 7, 9 and 12. The total must be 4 × 10 = 40. The three known numbers add to 28, so the missing number is 40 − 28 = 12. A common mistake is dividing by the wrong count, such as dividing 340 by 3 because you skipped a score. Another is forgetting a zero: if one friend brought 0 cookies, that friend still counts in the division.

Five students’ heights, in centimeters, are 155, 150, 160, 152 and 158. To find the median, first put the values in order: 150, 152, 155, 158, 160. The middle value is 155, with two values below and two above. The median height is 155 cm. The ordering step matters. If you take the middle of the unordered list you get 160, which is wrong.

When the count is even there are two middle values, and the median is their mean. Six values: 2, 3, 5, 8, 9, 12. The two middle values are 5 and 8. Their mean is (5 + 8) ÷ 2 = 13 ÷ 2 = 6.5. So the median is 6.5, even though 6.5 is not in the list. A quick way to find the middle: cross off the smallest and largest together, again and again, until one or two values remain.

The median is a position, not a total. Change the largest height from 160 to 200 cm. The median does not move; it is still 155. That makes the median a good choice when a few values are far from the rest. Watch out for repeated values: in 4, 4, 4, 9, 10, the median is 4, and each 4 counts as its own value in the ordered list.

Six workers at a small shop earn these yearly amounts, in thousands of dollars: 30, 32, 35, 38, 40 and 200. The 200 is the owner. The mean is (30 + 32 + 35 + 38 + 40 + 200) ÷ 6 = 375 ÷ 6 = 62.5, so "the average worker earns $62,500." But five of the six earn under $41,000. The median is the mean of the two middle values, (35 + 38) ÷ 2 = 36.5, or $36,500. The median describes a typical worker far better.

A value far from the rest is called an outlier. Outliers pull the mean toward them because the mean uses every value in its sum. The median ignores how far away an outlier is; it only counts it as one value. When the data have outliers or a long tail on one side, the median is usually the better measure of center. When the data are roughly symmetric, the mean and median are close, and either works.

Ask two questions before choosing. Are there outliers? Does the total matter? If you want the total, like the total cookies to buy, the mean is the right tool, because mean × count = total. If you want a typical value, like a typical home price in a Chicago neighborhood, the median is safer, because one mansion sale would distort the mean. A common mistake is to call the mean "more accurate." Both are exact; they just answer different questions.

Eight students count their books at home: 4, 6, 7, 9, 10, 12, 15, 18. The range is the largest value minus the smallest: 18 − 4 = 14. The range is quick, but it depends only on two values. One student with 200 books would push the range to 196, and it would say nothing about the other seven.

The interquartile range, or IQR, measures the spread of the middle half of the data. First find the median: the two middle values are 9 and 10, so the median is 9.5. The lower half is 4, 6, 7, 9; its median is (6 + 7) ÷ 2 = 6.5. That is the first quartile, Q1. The upper half is 10, 12, 15, 18; its median is (12 + 15) ÷ 2 = 13.5. That is the third quartile, Q3. IQR = Q3 − Q1 = 13.5 − 6.5 = 7. The middle half of the students have roughly 6.5 to 13.5 books.

With an odd count, leave the median out of both halves. For 2, 4, 5, 7, 8, 10, 11, the median is 7. The lower half is 2, 4, 5, so Q1 = 4. The upper half is 8, 10, 11, so Q3 = 10. IQR = 10 − 4 = 6. Common mistakes: forgetting to order the data first, and subtracting the wrong pair. Range uses max and min; IQR uses Q3 and Q1. If one value becomes an outlier, the range jumps but the IQR barely moves.

Ten quiz scores, in order: 5, 6, 6, 7, 8, 8, 9, 9, 10, 10. Find five numbers. Minimum: 5. Maximum: 10. Median: the middle two are 8 and 8, so 8. The lower half 5, 6, 6, 7, 8 has median 6, so Q1 = 6. The upper half 8, 9, 9, 10, 10 has median 9, so Q3 = 9. The five-number summary is 5, 6, 8, 9, 10.

A box plot draws these on a number line. Draw a box from Q1 = 6 to Q3 = 9. Draw a line inside the box at the median, 8. Draw lines called whiskers from the box out to the minimum 5 and the maximum 10. The box covers the middle half of the scores. Its width is the IQR, 9 − 6 = 3. Each of the four pieces (whisker, half-box, half-box, whisker) holds about 25% of the data.

Reading a box plot takes care. A long whisker means the values in that quarter are spread out. It does not mean there are more of them. A common mistake is to think a longer box or whisker holds more data. It holds the same quarter, spread across more space. Box plots hide the exact values and the count. They are best for comparing groups side by side, like two classes’ quiz scores on one number line.

Four friends scored 2, 4, 6 and 8 in a game. The mean is (2 + 4 + 6 + 8) ÷ 4 = 20 ÷ 4 = 5. How far is each score from the mean? 2 is 3 away, 4 is 1 away, 6 is 1 away, 8 is 3 away. These distances are absolute deviations; distance is never negative, so ignore the sign. Average the distances: (3 + 1 + 1 + 3) ÷ 4 = 8 ÷ 4 = 2. The mean absolute deviation, or MAD, is 2. On average, a score sits 2 points from the mean.

Try 10, 12, 14, 20. Mean: 56 ÷ 4 = 14. Distances from 14: 4, 2, 0, 6. MAD: (4 + 2 + 0 + 6) ÷ 4 = 12 ÷ 4 = 3. A data set where every value is the same, like 6, 6, 6, 6, has MAD 0, because nothing deviates. The larger the MAD, the more spread out the values are around their mean.

Common mistakes: using signed differences, which always add to zero and give a MAD of 0 by accident; dividing by the wrong count; and measuring from the median instead of the mean. MAD goes with the mean, the way IQR goes with the median. When you report a mean as the center, report MAD as the spread. When you report a median, report the IQR.

A distribution is the whole pattern of the data: which values appear and how often. When you look at a dot plot or histogram, describe its shape first. If the left and right sides are close to mirror images, the shape is symmetric. Heights of students in one grade are usually close to symmetric, with a hump in the middle and fewer very short or very tall students.

If one side stretches out much farther than the other, the shape is skewed. Number of pets is skewed to the right: most students have 0, 1 or 2, and a few have 5 or 6, so the tail points right toward the large values. In a right-skewed set, the mean is usually larger than the median, because the tail pulls the mean. In a left-skewed set, the mean is usually smaller. Also look for peaks. Two peaks might mean two different groups are mixed together, like the heights of sixth graders and eighth graders on one plot.

Finally, look for clusters, gaps and outliers. A cluster is a bunch of values close together. A gap is a stretch of the number line with no values. An outlier sits far from the rest. A complete description of homework minutes might be: "The data are skewed right, with a cluster from 20 to 40 minutes. There is a gap from 60 to 74 and one outlier at 120 minutes." A common mistake is to describe the bars instead of the data, like calling a histogram "tall." Say where the values pile up and where they thin out.

Two seventh-grade classes measure their heights. Class A has a mean of 152 cm with a MAD of 4 cm. Class B has a mean of 160 cm with a MAD of 4 cm. The difference in means is 160 − 152 = 8 cm. How big is that? Compare it to the spread: 8 ÷ 4 = 2, so the difference is 2 MADs. When the gap between centers is about 2 or more MADs, the groups clearly sit in different places; most students in B are taller than most in A.

Now compare two classes’ quiz scores. Class C has mean 78 with MAD 8; Class D has mean 80 with MAD 8. The difference is 2 points, only 2 ÷ 8 = 0.25 of a MAD. The dot plots would overlap almost completely. You should not conclude that Class D does better. A small difference in the center is not meaningful when the spread is large.

You can do the same with box plots, using medians and IQRs. Draw both box plots on one number line. If the boxes barely overlap, the groups differ. If one box sits mostly inside the other, they are similar. A common mistake is to compare only the maximum values: "Class A has the tallest student, so Class A is taller." One student is not a group. Compare centers, then spreads, then shapes.

Will it snow in Chicago in January? Very likely. Will it snow in July? Almost impossible. Probability turns words like "likely" into numbers. The probability of an event is a number from 0 to 1. A probability of 0 means the event cannot happen. A probability of 1 means it is certain. A probability of 1/2, or 0.5, means the event is as likely to happen as not, like a fair coin landing heads.

Probabilities can be written as fractions, decimals or percents. 1/4 = 0.25 = 25% all describe the same chance. Events near 0, like 0.05, are unlikely. Events near 1, like 0.9, are likely. Rolling a 7 on a regular six-sided die has probability 0, since no face shows 7. Rolling a number less than 7 has probability 1, since every face does.

A common mistake is to write a probability bigger than 1 or less than 0. If you get 7/6 or 120%, something went wrong, usually a count on top that is too large. Another mistake is to think "unlikely" means "will not happen." An event with probability 0.1 happens about 1 time in 10 in the long run. It still happens.

A spinner has 8 equal sections; 3 are red and 5 are blue. Each section is an outcome, and because the sections are the same size, the outcomes are equally likely. The theoretical probability of red is the number of red sections divided by the total: 3/8 = 0.375. You do not need to spin at all. This is what should happen in the long run.

A regular die has six equally likely outcomes: 1, 2, 3, 4, 5, 6. The probability of an even number is 3/6 = 1/2, because 2, 4 and 6 are the three favorable outcomes. A bag holds 4 red and 6 blue marbles. P(red) = 4/10 = 2/5 = 0.4. P(blue) = 6/10 = 3/5 = 0.6. The two add to 1, because a marble must be red or blue. The probability that an event does not happen is 1 minus the probability that it does: P(not red) = 1 − 0.4 = 0.6.

You can predict counts too. If you roll a die 60 times, you expect a 3 about 1/6 of the time: 60 × 1/6 = 10 times. Expect, not guarantee. Common mistakes: treating outcomes as equally likely when they are not (a spinner with one big section and three small ones), and putting the wrong number on top. P(red) uses the red count on top and the total, 10, on the bottom, not the blue count.

Devon’s class spins the 3-red, 5-blue spinner 50 times and records 18 reds. The experimental probability of red is the number of reds divided by the number of trials: 18/50 = 0.36. The theoretical probability was 3/8 = 0.375. They are close but not identical. That is normal. Fifty spins are not enough to land exactly on 0.375; in fact, 0.375 × 50 = 18.75 is not even a whole number.

The more trials you run, the closer the experimental probability tends to get to the theoretical one. Ten flips of a coin might give 7 heads, an experimental probability of 0.7. A hundred flips gave 56 heads, or 0.56. Devon’s class combined 2,500 flips and got 1,261 heads, which is 1,261 ÷ 2,500 = 0.5044. Each bigger set of trials sits closer to 0.5.

Experimental probability is the only choice when you cannot count equally likely outcomes. What is the chance a thumbtack lands point up? A tack is not symmetric, so you drop it 200 times and count. If it lands point up 130 times, the experimental probability is 130/200 = 0.65. Common mistakes: dividing by the number of successes instead of the number of trials, and trusting a tiny experiment. Three flips that all land heads do not mean P(heads) = 1.

Flip two coins. What are the possible results? Students often say three: two heads, two tails, or one of each. But list them carefully, first coin then second: HH, HT, TH, TT. There are four outcomes, and they are equally likely. The full list is the sample space. "One head and one tail" happens in two of the four, HT and TH, so its probability is 2/4 = 1/2, not 1/3. This is why the organized list matters.

Now flip a coin and roll a die. For each of the 2 coin results there are 6 die results, so the sample space has 2 × 6 = 12 outcomes. They are H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6. P(heads and an even number) = 3/12 = 1/4, since H2, H4 and H6 are favorable. A quick check: multiply the number of outcomes for each part. That product is the size of the sample space for a compound event.

A table works well for two dice. The rows are the first die (1 to 6) and the columns are the second (1 to 6), giving 6 × 6 = 36 cells. Count the cells whose sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That is 6, so P(sum of 7) = 6/36 = 1/6. A sum of 2 appears in only one cell, (1,1), so P(sum of 2) = 1/36. Common mistake: treating (2,5) and (5,2) as one outcome. They are different rolls.

A cafe sells sandwiches with 2 breads (white, wheat) and 3 fillings (turkey, cheese, veggie). A tree diagram starts with 2 branches for bread. From each bread, draw 3 branches for filling. Count the ends: 2 × 3 = 6 sandwiches: white-turkey, white-cheese, white-veggie, wheat-turkey, wheat-cheese, wheat-veggie. Each path from the start to a tip is one outcome.

Add a third choice, 2 drinks (water, juice). From each of the 6 sandwich tips draw 2 more branches. Now there are 2 × 3 × 2 = 12 complete meals. If a customer picks each choice at random, every path is equally likely, so the probability of any one meal, such as wheat-cheese-juice, is 1/12. The probability of "a wheat sandwich with juice" counts 3 paths, one for each filling, so it is 3/12 = 1/4.

Tree diagrams shine when the branches are not the same at every step. Suppose a game: flip a coin; if heads, you roll a die; if tails, you spin a 4-section spinner. The tree has 6 tips under heads and 4 under tails, 10 total, but the tips are not equally likely, because heads and tails each get half the probability. Common mistake: multiplying counts when the choices depend on each other. Draw the tree and count paths instead of assuming.

Spin a 4-section spinner (1, 2, 3, 4) and flip a coin. The sample space has 4 × 2 = 8 equally likely outcomes. What is P(3 and heads)? Only one outcome, 3H, so 1/8. What is P(even number and tails)? Two outcomes, 2T and 4T, so 2/8 = 1/4. Counting always works when outcomes are equally likely.

There is a shortcut when the parts do not affect each other, which we call independent events. The spinner does not change the coin. So P(3 and heads) = P(3) × P(heads) = 1/4 × 1/2 = 1/8, the same answer. For two dice, P(both show 6) = 1/6 × 1/6 = 1/36. For three coin flips, P(all heads) = 1/2 × 1/2 × 1/2 = 1/8.

"At least one" events are easier through the opposite, called the complement. P(at least one head in two flips) = 1 − P(no heads) = 1 − P(TT) = 1 − 1/4 = 3/4. Check by listing: HH, HT, TH are three of the four outcomes. One common mistake is adding when you should multiply. For "3 and heads," 1/4 + 1/2 = 3/4 is far too large. Another is multiplying when the events are not independent, like drawing two marbles without putting the first back.

The forecast says a 30% chance of rain each day this week. What is the probability of rain on at least 2 of the 5 school days? Computing this by hand is messy. Instead, run a simulation. Use random digits 0 to 9. Let 0, 1 and 2 stand for rain; that is 3 out of 10 digits, matching 30%. Digits 3 to 9 mean no rain. Take 5 random digits for one week: 8 1 4 2 7 means rain on Tuesday and Thursday, 2 rainy days.

One week is one trial. Run 50 trials, count how many have 2 or more rainy days, and divide by 50. If 24 trials qualify, the estimate is 24/50 = 0.48. The key rule: the tool must match the real probability. For a basketball player who makes 80% of free throws, let digits 0 to 7 mean make and 8, 9 mean miss. For a 1/6 chance, roll a die and pick one face. For 1/4, use a spinner with four equal sections, or two coin flips where HH counts.

Common mistakes: matching the wrong number of digits, since 0 to 3 gives 40%, not 30%. Reusing the same digits every trial. Stopping after a handful of trials. A simulation is an experiment, so its answer is an experimental probability. It gets more trustworthy with more trials, but it is still an estimate.

You want to know how many hours students at your school sleep, but the school has 800 students. Asking all 800 is a census. Asking a smaller group is taking a sample. The whole group you care about is the population. A sample only helps if it looks like the population, and the best way to get that is to choose at random: every student has the same chance of being picked.

Random does not mean careless. Put every student’s name or ID number in a list, then use a random number generator, or draw numbers from a hat, to pick 50. That is a random sample. Compare that to asking the 50 students in the cafeteria line at 7:30 a.m., who are early risers, or asking the basketball team, who are taller than average. Those samples are biased: some students had a much better chance of being chosen than others, so the sample leans one way.

Random samples still vary. Two random samples of 50 students might give mean sleep of 7.1 and 7.4 hours. Larger samples vary less. The U.S. Census counts every person once every ten years; between counts, the government estimates by sampling. Common mistake: thinking a big biased sample beats a small random one. A million responses to an online poll that only night owls saw still tell you about night owls.

A random sample of 50 students at an 800-student school is asked their favorite lunch. 20 pick pizza. The sample proportion is 20/50 = 0.4, or 40%. If the sample looks like the school, about 40% of all 800 students prefer pizza: 0.4 × 800 = 320 students. This is an inference: a conclusion about the population drawn from the sample.

Another: a random sample of 40 students shows 12 walk to school, so 12/40 = 0.3. In a district of 1,200 students, estimate 0.3 × 1,200 = 360 walkers. The estimate is not exact. A second sample might show 10 or 14 walkers, giving 300 or 420. Report it as "about 360," and remember that larger samples give tighter estimates.

The same idea works for quality checks. A factory pulls a random sample of 100 light bulbs and finds 3 that fail. The proportion is 3/100 = 0.03. In a shipment of 20,000 bulbs, expect about 0.03 × 20,000 = 600 failures. Common mistakes: multiplying the sample count by the population instead of using the proportion (20 × 800 is nonsense), and making inferences from a biased sample. A sample of the pizza line will always say pizza.

Do seventh graders and eighth graders spend different amounts of time on homework? You cannot ask everyone, so take a random sample of 30 from each grade. Seventh-grade sample: mean 45 minutes, MAD 10. Eighth-grade sample: mean 65 minutes, MAD 10. The difference in means is 65 − 45 = 20 minutes, which is 20 ÷ 10 = 2 MADs. That is a large, visible separation, so you can infer that eighth graders in this school generally spend more time.

Change the numbers: seventh-grade mean 52, eighth-grade mean 55, both MAD 12. The difference is 3 minutes, or 3 ÷ 12 = 0.25 MAD. Two random samples from the same population could easily differ by that much. Do not claim a real difference. Say the samples overlap too much to tell. A meaningful difference is one that is large compared with the spread.

Repeating the sampling shows why. If you drew many pairs of samples of 30, the difference in sample means would bounce around by a few minutes even when the populations were identical. That bounce is sampling variability. A difference must be big compared to that bounce, and compared to the spread within each group, to mean something. Common mistakes: comparing two biased samples, and comparing samples of very different sizes without noticing. Sizes of 30 and 3 do not carry equal weight.

Five students report hours of sleep and quiz score: (5, 68), (6, 72), (7, 80), (8, 85), (9, 88). Each pair is one student. A scatter plot puts sleep on the horizontal axis and score on the vertical axis and marks one point per student. The student who slept 7 hours and scored 80 is the point 7 units to the right and 80 units up. Five students, five points. The data now have two variables, sleep and score, and the plot lets you see them together.

Choose axes carefully. The variable you think might explain the other usually goes across (sleep), and the one being explained goes up (score). Pick scales that fit the data: sleep from 4 to 10 hours, score from 50 to 100. A scale that starts at 0 on both axes would squeeze the points into one corner. A plot with 24 students has 24 points, one for each pair of measurements.

Watch for three mistakes. Drawing two separate dot plots loses the pairing. Swapping the coordinates puts (5, 68) at 68 across and 5 up. Connecting the points with lines suggests a path. Points in a scatter plot are separate people, not a path. Leave them as dots and look for a pattern in the cloud.

Look at the sleep-and-score plot. Start at the left: students with little sleep have lower scores. Move right: scores climb. The cloud of points rises from lower left to upper right. This is an association between sleep and score: knowing one tells you something about the other. Now ask how tight the pattern is. If the points hug a line, the association is strong. If they spread widely, it is weak.

You can read individual facts too. Find the student with 8 hours of sleep: the point at 8 across sits at 85 up, so that student scored 85. Find the highest scorer: the top point, at 88, belongs to the 9-hour sleeper. Find how many students slept under 7 hours: count the points left of 7. In the five-student data, that is 2.

Be careful about what an association does not prove. More sleep goes with higher scores, but sleep might not be the cause. Students who sleep more may also finish homework earlier, or worry less. A scatter plot shows that two variables follow a trend together; it cannot by itself show that one causes the other. Common mistake: reading the plot as if the points were in time order. The leftmost point is the least sleep, not the first day.

Sleep and score rise together: that is a positive association. Now picture a plot of a car’s age (across) and its price (up). Older cars generally cost less, so the points fall from upper left to lower right. That is a negative association. Both can be strong or weak. A third possibility: a plot of shoe size and quiz score shows a shapeless cloud, because the two have nothing to do with each other. That is no association.

A quick test: slide your hand across the plot from left to right. If the points under your hand move up, the association is positive. If they move down, it is negative. If they stay level or jump around, there is none. Words matter: negative does not mean bad. A negative association between minutes of exercise and resting heart rate is good news.

Examples to sort. Hours of TV and hours of homework: likely negative. Height and arm span: positive and strong. Day of the month and high temperature in Chicago: no clear association within one month. Common mistake: calling a plot negative because the values are low, or positive because the values are high. Direction comes from the slope of the cloud, not from where it sits.

On the sleep plot, most points bunch between 6.5 and 8 hours. That bunch is a cluster. A cluster can hint at a group: maybe most students have a similar bedtime. Two clusters might mean two kinds of students, like those with after-school jobs and those without. Name clusters by where they sit on both axes: "a cluster around 7 hours and 80 points."

One student slept 9 hours but scored 55. That point sits far below the rising band. It is an outlier in the scatter plot: it does not follow the pattern the others follow. An outlier can be a data-entry error, a special case (Leo was sick), or a real surprise. Do not delete it silently. Note it, find out why if you can, and describe the pattern "with one outlier at (9, 55)."

Not every pattern is a line. Plot the height of a thrown ball against time and the points rise, then fall, in an arch. That is a nonlinear association. Plot a bacteria count against hours and the points bend upward faster and faster. When the points bend, a straight line will fit badly, and you should say the association is nonlinear. Common mistake: forcing a line through a curved pattern and reporting its slope as if it held for all values.

The five sleep points (5, 68), (6, 72), (7, 80), (8, 85), (9, 88) rise in a nearly straight band. Lay a ruler on the plot and draw a line of fit. It should follow the direction of the cloud and pass close to as many points as you can. A good line has about as many points above it as below it. The points above should not all sit at one end. The line does not have to touch any point.

One easy line goes through (5, 68) and (9, 88). Its steepness is (88 − 68) ÷ (9 − 5) = 20 ÷ 4 = 5. Check the middle: at 7 hours, this line gives 68 + 2 × 5 = 78, close to the actual 80. At 8 hours it gives 83, close to 85. It sits a little low in the middle, so a slightly better line is score = 5 × hours + 43, which gives 68, 73, 78, 83 and 88 at 5 through 9 hours. It is above the data twice, below twice, and exact twice.

Use the line to make a prediction. For 6.5 hours, the line gives 5 × 6.5 + 43 = 32.5 + 43 = 75.5, so predict about 75 or 76. Predictions inside the range of the data are reasonable. Predictions far outside, like 20 hours of sleep, are not; the line would say 143, which is not even a possible score. Common mistakes: drawing the line from the lowest point to the highest point no matter where the rest sit, and running it through two outliers.

A line of fit through plant-growth data passes through (1, 4) and (4, 13): after 1 week the plant was 4 cm, after 4 weeks it was 13 cm. Slope is rise over run: (13 − 4) ÷ (4 − 1) = 9 ÷ 3 = 3. The slope is 3 cm per week. It means: for each extra week, the line predicts about 3 cm more height. Always attach both units, the vertical unit per one horizontal unit.

Slope can be negative. A line through used-car data passes through (1, 20) and (5, 12), with age in years and price in thousands of dollars. Slope = (12 − 20) ÷ (5 − 1) = −8 ÷ 4 = −2. Each extra year of age goes with about $2,000 less in price. The sign tells the direction of the association, and the size tells how steep it is.

In the sleep line, score = 5 × hours + 43, the slope is 5 points per hour. Each extra hour of sleep goes with about 5 more quiz points, on average. "On average" matters; individual students sit above or below the line. Common mistakes: dividing run by rise, which gives 1/3 instead of 3. Subtracting in different orders on top and bottom, which gives −3. Forgetting that slope is a rate of change, not a total.

In the line score = 5 × hours + 43, set hours to 0 and the score is 43. That number is the y-intercept, where the line crosses the vertical axis. Does it mean a student with no sleep would score 43? No. No student in the data slept 0 hours; the least was 5. The intercept here is a mathematical anchor, not a fact about zero-sleep students.

Sometimes the intercept does mean something. A line for taxi fares, fare = 2.5 × miles + 3, has intercept 3. At 0 miles the fare is $3, which is the flat fee for getting in the cab. Here 0 miles is a real starting point. To predict, plug in: 8 miles gives 2.5 × 8 + 3 = 20 + 3 = 23 dollars. A line with slope −2 and intercept 50, y = −2x + 50, gives y = −2 × 15 + 50 = −30 + 50 = 20 when x = 15.

Always ask two questions about a prediction. Is the input inside the range of the data? Does the answer make sense? A line for ice cream sales, cones = 4 × temp − 100, predicts 4 × 80 − 100 = 220 cones at 80°F, which is sensible. At 20°F it predicts 4 × 20 − 100 = −20 cones, which is impossible; that temperature is far outside the summer data. Common mistake: reading the intercept as "the starting value" in every situation.

Some data are categories, not numbers: grade level, cat or dog, yes or no. This is categorical data. A two-way table organizes counts by two categories at once. Survey 100 students: do you prefer cats or dogs? Rows are grade (7th, 8th), columns are the pet. Seventh grade: 20 cats, 30 dogs. Eighth grade: 15 cats, 35 dogs. Add each row: 7th total 50, 8th total 50. Add each column: cats 35, dogs 65. The grand total is 100 either way, which confirms the table.

Each cell answers one how-many question. How many eighth graders prefer dogs? 35. How many students prefer cats? Read the column total: 35. How many seventh graders were surveyed? Read the row total: 50. Given some cells, you can find missing ones by subtraction. If 80 students are surveyed, 50 play a sport, and 18 of those also play an instrument, then 50 − 18 = 32 play a sport but no instrument.

Common mistakes: mixing up row and column when reading a cell, and double-counting when categories overlap. Each person belongs in exactly one cell. Also, do not use a two-way table when a variable is numerical, like exact height; use a scatter plot for two numerical variables. Two categories, two-way table; two numbers, scatter plot.

Counts alone can mislead when groups differ in size. Use relative frequency: a count divided by a total. In the pet table, 30 of the 50 seventh graders prefer dogs: 30/50 = 0.6, or 60%. For eighth graders, 35/50 = 0.7, or 70%. Dividing by the row total gives row relative frequencies. Because both grades had 50 students, the counts and the percents tell the same story here.

Now a table with unequal groups. 80 students: 50 play a sport and 30 do not. Of the sport players, 18 play an instrument. Of the non-players, 12 do. Raw counts, 18 versus 12, make it look like sport players play instruments more. But 18/50 = 0.36, or 36%, and 12/30 = 0.4, or 40%. The non-players actually have the higher share. Compare percents, not counts, when the groups are different sizes.

You can also divide by the grand total to get the share of everyone: 18/80 = 0.225, so 22.5% of all students play both a sport and an instrument. Choose the total that matches the question. "What percent of sport players" means divide by the sport row. "What percent of all students" means divide by 80. Common mistake: dividing by the wrong total, or mixing a row percent with a column percent in one comparison.

Is there an association between playing a sport and playing an instrument? Compare the row percents: 36% of sport players play an instrument, and 40% of non-players do. Those are close, only 4 percentage points apart. Knowing whether a student plays a sport tells you little about instruments. That is weak or no association.

Now a different survey of 200 students: studied with a group or alone, and passed or did not. Group: 60 passed, 20 did not, total 80. Alone: 60 passed, 60 did not, total 120. Row percents: group 60/80 = 0.75, or 75% passed. Alone 60/120 = 0.5, or 50% passed. Notice the raw pass counts are the same, 60 and 60, but the percents differ by 25 percentage points. That is a clear association: students who studied in groups passed at a higher rate.

Association again is not proof of cause. Group studiers might already be stronger students. Also, a difference of a couple of percentage points in small groups is not convincing; 5 of 10 versus 6 of 10 is one student. Common mistakes: comparing counts across rows of different size, and comparing a row percent to a column percent. Pick one direction, compute both rows, then compare.