Answer key · Unit 16

Statistics.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 5

Lesson 1 of 5 · About 25 minutes

Mean, median and mode

Three ways to give a long list of readings one number: the balance point, the value in the middle and the most common value, and what happens to each when a bad reading creeps in.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The balance point
Practice · Calculate
Question
Five friends counted the books they read in a year: $4, 7, 7, 9, 13$ ⁠. What is the mean?

Correct answer
8 ± 0.01
Setup, choices and feedback
The Scene at this step
Scatter
1
Bad reading
off
Readings / Mean and spread
Mean and spread
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 7 ± 0.01
  Misconception: Treats the mean, the median and the mode as one number, so any of them answers for the others.
  Feedback from the catalogue
  All three are “averages” in everyday speech, but each answers a different question. The mean is the total divided by the number of values, the balance point. The median is the value in the middle once the data are in order. The mode is the value that comes up most often. In symmetric data they are close, but with a tail to one side or an outlier they come apart, so it matters which one is asked for.
Feedback for a right answer
$\frac{4 + 7 + 7 + 9 + 13}{5} = \frac{40}{5} = 8$ ⁠. If each of them had read $8$ ⁠ books, together they would have read the same number of books.

Feedback for a wrong answer
Add the five numbers, then divide by $5$ ⁠.

Hint
Add up all the values and divide by how many there are.

Worked solution
1. The total: $4 + 7 + 7 + 9 + 13 = 40$ ⁠.
2. The number of values: $5$ ⁠.
3. The mean: $\bar{x} = \frac{40}{5} = 8$ ⁠.
If you answered $7$ ⁠, you took the value in the middle (the median) or the one that appears twice (the mode). Here both are $7$ ⁠, but the mean is $8$ ⁠: the $13$ ⁠ pulls it up.
Step 2 of 6
Open this step
The value in the middle
Practice · Calculate
Question
The wait for a bus, in minutes, on five mornings: $12, 3, 9, 5, 8$ ⁠. What is the median?

Correct answer
8 min ± 0.01 min
Setup, choices and feedback
The Scene at this step
Scatter
1
Bad reading
off
Readings / Mean and spread
Readings
Accepted within
± 0.01 min
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 9 min ± 0.01 min
  Misconception: Takes the middle value of the list as written, without putting the values in order first.
  Feedback from the catalogue
  The median is the middle of the data in order of size, not the middle of the list as it was written. First sort from smallest to largest, and only then take the value in the middle. In the list $12, 3, 9, 5, 8$ ⁠ the middle entry is $9$ ⁠, but once sorted, $3, 5, 8, 9, 12$ ⁠, the median is $8$ ⁠.
- 7.4 min ± 0.01 min
  Misconception: Treats the mean, the median and the mode as one number, so any of them answers for the others.
  Feedback from the catalogue
  All three are “averages” in everyday speech, but each answers a different question. The mean is the total divided by the number of values, the balance point. The median is the value in the middle once the data are in order. The mode is the value that comes up most often. In symmetric data they are close, but with a tail to one side or an outlier they come apart, so it matters which one is asked for.
Feedback for a right answer
In order: $3, 5, 8, 9, 12$ ⁠, and the middle one is $8$ ⁠. Two times are shorter and two are longer.

Feedback for a wrong answer
Sort the five times by size and take the third.

Hint
Put the times in order from shortest to longest before you look for the middle.

Worked solution
1. Sort from smallest to largest: $3, 5, 8, 9, 12$ ⁠.
2. There are $5$ ⁠ values, so the middle one is the third.
3. The median is $8$ ⁠.
If you answered $9$ ⁠, you took the middle of the list as written, without sorting. If you answered $7.4$ ⁠, you worked out the mean: $\frac{37}{5}$ ⁠.
Step 3 of 6
Open this step
An even count
Practice · Calculate
Question
Six children have $1, 4, 6, 10, 13, 20$ ⁠ marbles. What is the median?

Correct answer
8 ± 0.01
Setup, choices and feedback
The Scene at this step
Scatter
1
Bad reading
off
Readings / Mean and spread
Readings
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 6 ± 0.01
  Misconception: With an even number of values, gives one of the two middle values as the median instead of their mean.
  Feedback from the catalogue
  With an even number of values there is no single value in the middle: there are two. The median is their mean, the point exactly between them. In $1, 4, 6, 10, 13, 20$ ⁠ the middle two are $6$ ⁠ and $10$ ⁠, and the median is $\frac{6 + 10}{2} = 8$ ⁠. That way half the values lie below it and half above.
- 10 ± 0.01
  Misconception: With an even number of values, gives one of the two middle values as the median instead of their mean.
  Feedback from the catalogue
  With an even number of values there is no single value in the middle: there are two. The median is their mean, the point exactly between them. In $1, 4, 6, 10, 13, 20$ ⁠ the middle two are $6$ ⁠ and $10$ ⁠, and the median is $\frac{6 + 10}{2} = 8$ ⁠. That way half the values lie below it and half above.
- 9 ± 0.01
  Misconception: Treats the mean, the median and the mode as one number, so any of them answers for the others.
  Feedback from the catalogue
  All three are “averages” in everyday speech, but each answers a different question. The mean is the total divided by the number of values, the balance point. The median is the value in the middle once the data are in order. The mode is the value that comes up most often. In symmetric data they are close, but with a tail to one side or an outlier they come apart, so it matters which one is asked for.
Feedback for a right answer
The middle two are $6$ ⁠ and $10$ ⁠, and the median is their mean: $\frac{6 + 10}{2} = 8$ ⁠. Three children are below it and three above.

Hint
The values are already in order, but there are six. Which two are in the middle?

Worked solution
1. The values are in order: $1, 4, 6, 10, 13, 20$ ⁠.
2. There are $6$ ⁠ values, an even number, so two sit in the middle: the third and the fourth, $6$ ⁠ and $10$ ⁠.
3. The median: $\frac{6 + 10}{2} = 8$ ⁠.
If you answered $6$ ⁠ or $10$ ⁠, you took only one of the middle two. If you answered $9$ ⁠, you worked out the mean: $\frac{54}{6} = 9$ ⁠.
Step 4 of 6
Open this step
The most common value
Practice · Calculate
Question
One morning a shop sold shoes in sizes $36, 38, 38, 38, 40, 41, 43, 46$ ⁠. What is the mode?

Correct answer
38 ± 0.01
Setup, choices and feedback
The Scene at this step
Scatter
1
Bad reading
off
Readings / Mean and spread
Readings
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 3 ± 0.01
  Misconception: Gives how many times the most common value appears (or how tall the tallest bar is) instead of the value itself.
  Feedback from the catalogue
  The mode is the value that comes up most often, not the number of times it does. If size $38$ ⁠ appears three times, the mode is $38$ ⁠, and $3$ ⁠ is only how often it appears. On a histogram it is the difference between where the tallest bar stands (on the horizontal axis) and how tall it is.
- 40 ± 0.01
  Misconception: Treats the mean, the median and the mode as one number, so any of them answers for the others.
  Feedback from the catalogue
  All three are “averages” in everyday speech, but each answers a different question. The mean is the total divided by the number of values, the balance point. The median is the value in the middle once the data are in order. The mode is the value that comes up most often. In symmetric data they are close, but with a tail to one side or an outlier they come apart, so it matters which one is asked for.
- 39 ± 0.01
  Misconception: Treats the mean, the median and the mode as one number, so any of them answers for the others.
  Feedback from the catalogue
  All three are “averages” in everyday speech, but each answers a different question. The mean is the total divided by the number of values, the balance point. The median is the value in the middle once the data are in order. The mode is the value that comes up most often. In symmetric data they are close, but with a tail to one side or an outlier they come apart, so it matters which one is asked for.
Feedback for a right answer
Size $38$ ⁠ was sold three times, more than any other. For the shop it is the useful number: that is the size to order more of.

Hint
Which size was sold most often?

Worked solution
1. Count how often each size appears: $36$ ⁠ once, $38$ ⁠ three times, and every other size once.
2. The mode is the size that appears most often: $38$ ⁠.
If you answered $3$ ⁠, you gave the number of times, not the size. If you answered $40$ ⁠, you worked out the mean ( $\frac{320}{8}$ ⁠), and if $39$ ⁠, the median ( $\frac{38 + 40}{2}$ ⁠). Shoes in size $39$ ⁠ are no help to the shop: nobody bought one.
Step 5 of 6
Open this step
A bad reading
Practice · Predict, then look
Question
The twenty readings on the board have a mean of $5$ ⁠. Switch on Bad reading: one more reading, of $0$ ⁠. What happens to the mean and the median?

Correct answer
- Correct The mean drops clearly, and the median hardly moves
Setup, choices and feedback
The Scene at this step
Scatter
1fixed
Bad reading
offfixed
Readings / Mean and spread
Mean and spread
Choices
- Correct The mean drops clearly, and the median hardly moves
- Wrong Both drop by the same amount
  Misconception: Expects one extreme value to drag the median as far as it drags the mean.
  Feedback from the catalogue
  The mean uses the size of every value, so one value far away pulls it along. The median uses only the order: one extreme value moves the middle by at most one place in the list, however far away it is. That makes the median resistant to outliers, which is why it is used for things like pay, where a few huge salaries inflate the mean.
- Wrong The median drops more than the mean
  Misconception: Expects one extreme value to drag the median as far as it drags the mean.
  Feedback from the catalogue
  The mean uses the size of every value, so one value far away pulls it along. The median uses only the order: one extreme value moves the middle by at most one place in the list, however far away it is. That makes the median resistant to outliers, which is why it is used for things like pay, where a few huge salaries inflate the mean.
- Wrong Neither changes: one reading in $21$ ⁠ makes no difference
  Feedback
  One reading does not shift the order much, which is why the median barely moves. But the mean adds up the sizes, and $0$ ⁠ is $5$ ⁠ away from the mean.
Once the prediction is locked in, the Scene changes
Bad reading
onfixed
Readings / Mean and spread
Mean and spread
The Scene then shows: 4.762 · Ȳ, the mean of the y values/nodes/scatter/outputs/mean
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The mean drops from $5$ ⁠ to $4.76$ ⁠, the median only from $5.10$ ⁠ to $5.07$ ⁠. The new reading goes to the bottom of the list and shifts the middle by just half a place.

Hint
The mean adds up every value. The median only asks who is in the middle.

Worked solution
1. The mean: the total was $20 \times 5 = 100$ ⁠. With the new reading it is still $100$ ⁠, but divided by $21$ ⁠: $\frac{100}{21} \approx 4.76$ ⁠.
2. The median: with $20$ ⁠ readings it is the mean of the tenth and eleventh in order. With $21$ ⁠ it is the eleventh, and the new reading went in at the bottom, so the eleventh is the old tenth: $5.07$ ⁠ instead of $5.10$ ⁠.
3. So the mean moves about seven times as far as the median.
Even if the bad reading had been $-100$ ⁠, the median would have moved exactly as much. The mean would have plunged.
Step 6 of 6
Open this step
The new mean
Practice · Calculate
Question
Twenty readings with a mean of exactly $5$ ⁠, and one bad reading of $0$ ⁠. What is the mean of all $21$ ⁠ readings? Round to two decimal places.

Correct answer
4.762 ± 0.01
Setup, choices and feedback
The Scene at this step
Scatter
1fixed
Bad reading
onfixed
Readings / Mean and spread
Readingsfixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Ȳ, the mean of the y values/nodes/scatter/outputs/mean
Wrong answers it recognises
- 2.5 ± 0.01
  Misconception: Averages two group means as if the groups were the same size: 20 readings averaging 5 and 1 reading of 0 give 2.5.
  Feedback from the catalogue
  An average of two averages gives each group the same weight, even when one is far bigger. Here there are $20$ ⁠ readings averaging $5$ ⁠ and one reading of $0$ ⁠. The total is $20 \times 5 + 0 = 100$ ⁠ over $21$ ⁠ readings, so the mean is $\frac{100}{21} \approx 4.76$ ⁠, not $2.5$ ⁠. The single reading is only one in $21$ ⁠.
Feedback for a right answer
$\frac{20 \times 5 + 0}{21} = \frac{100}{21} \approx 4.76$ ⁠. The board shows the same value at Mean and spread.

Feedback for a wrong answer
The bad reading counts: the total stays $100$ ⁠, but now you divide by $21$ ⁠.

Hint
What do the twenty readings add up to? Then add the new one and divide by the new count.

Worked solution
1. From the mean back to the total: $20 \times 5 = 100$ ⁠.
2. Add the bad reading: $100 + 0 = 100$ ⁠, and there are now $21$ ⁠ readings.
3. The mean: $\frac{100}{21} \approx 4.76$ ⁠.
If you answered $2.5$ ⁠, you averaged the two means, $5$ ⁠ and $0$ ⁠, as if both groups were the same size. But one side has $20$ ⁠ readings and the other only one.

Lesson 2 of 5 · About 25 minutes

Range and standard deviation

Two groups with the same mean can be very different. How far the values spread: the range, the standard deviation and why it squares, and what share of the readings falls within one standard deviation.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
The range
Practice · Calculate
Question
The noon temperatures, in degrees Celsius, on five days: $14, 19, 11, 23, 17$ ⁠. What is the range?

Correct answer
12 ± 0.01
Setup, choices and feedback
The Scene at this step
Scatter
1
Bad reading
off
Readings / Mean and spread
Readings
Accepted within
± 0.01
A fixed value in the question
Feedback for a right answer
$23 - 11 = 12$ ⁠ degrees: every temperature lies inside a band $12$ ⁠ wide.

Feedback for a wrong answer
The range is the gap between the highest value and the lowest.

Hint
Find the highest and the lowest.

Worked solution
1. The highest: $23$ ⁠. The lowest: $11$ ⁠.
2. The range: $23 - 11 = 12$ ⁠.
The range is easy to work out, but it depends only on the two extreme values. One unusually hot day would change it, even if all the other days were alike.
Step 2 of 5
Open this step
The distance from the mean
Practice · Calculate
Question
Eight readings: $2, 4, 4, 4, 5, 5, 7, 9$ ⁠. The mean is $5$ ⁠. What is the standard deviation? (Divide by $n = 8$ ⁠.)

Correct answer
2 ± 0.01
Setup, choices and feedback
The Scene at this step
Scatter
1
Bad reading
off
Readings / Mean and spread
Mean and spread
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 7 ± 0.01
  Misconception: Takes the standard deviation to be the range (largest minus smallest) or half of it.
  Feedback from the catalogue
  The range depends on only two values, the largest and the smallest. The standard deviation uses all of them: it is roughly the typical distance of a value from the mean. Square every distance, average, and take the root: $s = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}$ ⁠. In $2, 4, 4, 4, 5, 5, 7, 9$ ⁠ the range is $7$ ⁠, but the standard deviation is only $2$ ⁠, because most values sit close to the mean of $5$ ⁠.
- 3.5 ± 0.01
  Misconception: Takes the standard deviation to be the range (largest minus smallest) or half of it.
  Feedback from the catalogue
  The range depends on only two values, the largest and the smallest. The standard deviation uses all of them: it is roughly the typical distance of a value from the mean. Square every distance, average, and take the root: $s = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}$ ⁠. In $2, 4, 4, 4, 5, 5, 7, 9$ ⁠ the range is $7$ ⁠, but the standard deviation is only $2$ ⁠, because most values sit close to the mean of $5$ ⁠.
Feedback for a right answer
The squares add to $32$ ⁠, their mean is $4$ ⁠, and the root is $2$ ⁠. A typical reading is about $2$ ⁠ from the mean.

Hint
For each reading find its distance from $5$ ⁠, square it, average, and take the root.

Worked solution
1. The distances from the mean: $-3, -1, -1, -1, 0, 0, 2, 4$ ⁠. They add to $0$ ⁠, as they always do, which is why we square them.
2. The squares: $9, 1, 1, 1, 0, 0, 4, 16$ ⁠, adding to $32$ ⁠.
3. The variance: $\frac{32}{8} = 4$ ⁠.
4. The standard deviation: $s = \sqrt{4} = 2$ ⁠.
If you answered $4$ ⁠, you stopped at the variance and forgot the root. If you answered $7$ ⁠ or $3.5$ ⁠, you worked out the range or half of it, which depend on only two values.
Step 3 of 5
Open this step
Setting the spread
Practice · Reach the target
Question
Change Scatter until the standard deviation on the board is $s = 1.5$ ⁠, then press Check.

Goal
- 1.5 ± 0.02
  At the start of the step: 0.5
Setup, choices and feedback
The Scene at this step
Scatter
0.5
Bad reading
offfixed
Readings / Mean and spread
Mean and spreadfixed
Read from the Scene at
S, the standard deviation of the y values (dividing by n)/nodes/scatter/outputs/spread
One setting that reaches it
Scatter
1.5
Feedback for a right answer
$s = 1.5$ ⁠. Every distance from the mean grew three times, so the average square grew nine times and its root three times.

Feedback for a wrong answer
Not yet. Watch the value of $s$ ⁠ the board prints as you move it.

Hint
The blue band is $\bar{y} \pm s$ ⁠. Widen it.

Worked solution
1. You start at $s = 0.5$ ⁠, with Scatter at $0.5$ ⁠.
2. Multiplying every distance from the mean by a number multiplies the standard deviation by the same number.
3. To get from $0.5$ ⁠ to $1.5$ ⁠ you need three times as much: set Scatter to $1.5$ ⁠.
Step 4 of 5
Open this step
Range against standard deviation
Practice · Predict, then look
Question
Switch Bad reading on again, a reading of $0$ ⁠. Which grows more, in proportion: the range or the standard deviation?

Correct answer
- Correct The range nearly doubles, and the standard deviation grows by less than half
Setup, choices and feedback
The Scene at this step
Scatter
1fixed
Bad reading
offfixed
Readings / Mean and spread
Mean and spreadfixed
Choices
- Correct The range nearly doubles, and the standard deviation grows by less than half
- Wrong Both grow by the same factor
  Misconception: Takes the standard deviation to be the range (largest minus smallest) or half of it.
  Feedback from the catalogue
  The range depends on only two values, the largest and the smallest. The standard deviation uses all of them: it is roughly the typical distance of a value from the mean. Square every distance, average, and take the root: $s = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}$ ⁠. In $2, 4, 4, 4, 5, 5, 7, 9$ ⁠ the range is $7$ ⁠, but the standard deviation is only $2$ ⁠, because most values sit close to the mean of $5$ ⁠.
- Wrong The standard deviation grows more than the range
  Feedback
  The standard deviation is sensitive to far values, since it squares distances. But it averages over $21$ ⁠ readings, while the range rests entirely on the extreme value.
- Wrong Neither changes
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Bad reading
onfixed
Readings / Mean and spread
Mean and spreadfixed
The Scene then shows: 7.109 · The range of the y values, largest − smallest/nodes/scatter/outputs/range
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The range jumps from $3.78$ ⁠ to $7.11$ ⁠, nearly double, because the lowest value is now $0$ ⁠. The standard deviation rises from $1$ ⁠ to $1.44$ ⁠.

Hint
The range depends on two values only. The standard deviation averages all of them.

Worked solution
1. The range: before, from $3.33$ ⁠ to $7.11$ ⁠, which is $3.78$ ⁠. With the bad reading the lowest is $0$ ⁠, so the range is $7.11 - 0 = 7.11$ ⁠.
2. The standard deviation: the new reading is about $5$ ⁠ from the mean, and its square, $25$ ⁠, joins twenty small squares. The mean square rises from $1$ ⁠ to about $2.09$ ⁠, and its root from $1$ ⁠ to $1.44$ ⁠.
3. So the range nearly doubles, and the standard deviation rises by about $44\%$ ⁠.
Both measures react to an outlier, but the range is built entirely from the extremes, while the standard deviation shares the work among all the readings.
Step 5 of 5
Open this step
Inside the band
Practice · Calculate
Question
The blue band on the board is $\bar{y} \pm s$ ⁠, from one standard deviation below the mean to one above it. How many of the $20$ ⁠ readings lie inside it?

Correct answer
13 ± 0.4
Setup, choices and feedback
The Scene at this step
Scatter
1fixed
Bad reading
offfixed
Readings / Mean and spread
Mean and spreadfixed
Accepted within
± 0.4
Computed from the Scene at this step’s setting
Readings inside the band ȳ ± s, edges included/nodes/scatter/outputs/inBand
Wrong answers it recognises
- 20 ± 0.4
  Misconception: Expects (nearly) all the data to lie within one standard deviation of the mean.
  Feedback from the catalogue
  The standard deviation is a typical distance, not a largest one, so plenty of values lie farther out. In a normal distribution about $68\%$ ⁠ of the values are within one standard deviation of the mean, about $95\%$ ⁠ within two, and $99.7\%$ ⁠ within three. So about a third of the values lie outside one standard deviation.
- 19 ± 0.4
  Misconception: Expects (nearly) all the data to lie within one standard deviation of the mean.
  Feedback from the catalogue
  The standard deviation is a typical distance, not a largest one, so plenty of values lie farther out. In a normal distribution about $68\%$ ⁠ of the values are within one standard deviation of the mean, about $95\%$ ⁠ within two, and $99.7\%$ ⁠ within three. So about a third of the values lie outside one standard deviation.
Feedback for a right answer
$13$ ⁠ out of $20$ ⁠, which is $65\%$ ⁠: about two thirds. The other seven readings are farther than $s$ ⁠ from the mean.

Feedback for a wrong answer
Count again the beads inside the band, between $\bar{y} - s$ ⁠ and $\bar{y} + s$ ⁠.

Hint
Count the beads between the two dashed lines.

Worked solution
1. Here $\bar{y} = 5$ ⁠ and $s = 1$ ⁠, so the band runs from $4$ ⁠ to $6$ ⁠.
2. Count the beads whose height is between $4$ ⁠ and $6$ ⁠: $13$ ⁠.
3. That is $\frac{13}{20} = 65\%$ ⁠ of the readings.
If you expected nearly all of them inside, remember that the standard deviation is a typical distance, not a largest one. Usually about a third of the readings lie outside it.

Lesson 3 of 5 · About 25 minutes

The normal distribution

A histogram shows how data spread out. On a Galton board: the mode is the tallest bar, the standard deviation grows like the square root of the number of rows, and the $68$–$95$ rule: how much of the data lies within one and two standard deviations.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A histogram
Practice · Calculate
Question
The board has $10$ ⁠ rows, and the chance of bouncing right is $25\%$ ⁠. A bin’s number is how many times the ball went right, from $0$ ⁠ on the left. What is the mode: which bin has the tallest bar?

Correct answer
2 ± 0.01
Setup, choices and feedback
The Scene at this step
Rows of pegs
10fixed
Chance of bouncing right
25 %fixed
Balls / Bell curve
Bell curve
Other settings
Probe/nodes/board/props/probe = 2
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The bin with the highest chance (the lower one on a tie)/nodes/board/outputs/likeliestBin
Wrong answers it recognises
- 2.5 ± 0.01
  Misconception: Treats the mean, the median and the mode as one number, so any of them answers for the others.
  Feedback from the catalogue
  All three are “averages” in everyday speech, but each answers a different question. The mean is the total divided by the number of values, the balance point. The median is the value in the middle once the data are in order. The mode is the value that comes up most often. In symmetric data they are close, but with a tail to one side or an outlier they come apart, so it matters which one is asked for.
- 140.8 ± 20
  Misconception: Gives how many times the most common value appears (or how tall the tallest bar is) instead of the value itself.
  Feedback from the catalogue
  The mode is the value that comes up most often, not the number of times it does. If size $38$ ⁠ appears three times, the mode is $38$ ⁠, and $3$ ⁠ is only how often it appears. On a histogram it is the difference between where the tallest bar stands (on the horizontal axis) and how tall it is.
Feedback for a right answer
Bin $2$ ⁠: two bounces right and eight left. That is the mode, the most common result, even though the mean is $10 \times 0.25 = 2.5$ ⁠.

Hint
Look at the amber bars: which is the tallest? Count bins from the left, starting at $0$ ⁠.

Worked solution
1. On the histogram each bar is a bin, and its height is how many balls are expected in it.
2. The tallest bar stands over bin $2$ ⁠, so the mode is $2$ ⁠.
3. The mean, $2.5$ ⁠, falls between two bins: no ball lands “in bin $2.5$ ⁠”. That is why the mode and the mean differ here.
If you answered a number around $140$ ⁠, you gave the bar’s height, how many balls are in the bin, not the bin itself.
Step 2 of 5
Open this step
The width of the bell
Practice · Reach the target
Question
On a fair board the standard deviation is $\sqrt{n \cdot p \cdot (1 - p)}$ ⁠ bins, where $n$ ⁠ is the number of rows and $p = 0.5$ ⁠. Change Rows of pegs so the standard deviation is exactly $2$ ⁠ bins, then press Check.

Goal
- 2 ± 0.01
  At the start of the step: 1.414
Setup, choices and feedback
The Scene at this step
Rows of pegs
8
Chance of bouncing right
50 %fixed
Balls / Bell curve
Bell curve
Read from the Scene at
The standard deviation in bins, √(rows · p · (1 − p))/nodes/board/outputs/spread
One setting that reaches it
Rows of pegs
16
Feedback for a right answer
$16$ ⁠ rows: $\sqrt{16 \times 0.25} = \sqrt{4} = 2$ ⁠. Doubling the standard deviation takes four times as many rows.

Feedback for a wrong answer
Not yet. How many rows give $\frac{n}{4} = 4$ ⁠?

Hint
$n \cdot 0.5 \cdot 0.5 = \frac{n}{4}$ ⁠ has to be $2^2$ ⁠.

Worked solution
1. You want $\sqrt{n \times 0.25} = 2$ ⁠.
2. Square both sides: $n \times 0.25 = 4$ ⁠.
3. So $n = 16$ ⁠: set Rows of pegs to $16$ ⁠.
With $8$ ⁠ rows the standard deviation is $\sqrt{2} \approx 1.41$ ⁠, not $1$ ⁠: it grows like the square root of the number of rows, not like the number itself.
Step 3 of 5
Open this step
One standard deviation
Practice · Predict, then look
Question
The board has $16$ ⁠ fair rows: the mean is bin $8$ ⁠ and the standard deviation $2$ ⁠ bins. What share of the balls lands between bin $6$ ⁠ and bin $10$ ⁠, within one standard deviation of the mean?

Correct answer
- Correct About two thirds
Setup, choices and feedback
The Scene at this step
Rows of pegs
16fixed
Chance of bouncing right
50 %fixed
Balls / Bell curve
Balls
Choices
- Correct About two thirds
- Wrong About a third
  Feedback
  A third is roughly the share outside one standard deviation, on both sides together.
- Wrong About $95\%$ ⁠
  Feedback
  $95\%$ ⁠ is roughly the share within two standard deviations.
- Wrong Nearly all of them
  Misconception: Expects (nearly) all the data to lie within one standard deviation of the mean.
  Feedback from the catalogue
  The standard deviation is a typical distance, not a largest one, so plenty of values lie farther out. In a normal distribution about $68\%$ ⁠ of the values are within one standard deviation of the mean, about $95\%$ ⁠ within two, and $99.7\%$ ⁠ within three. So about a third of the values lie outside one standard deviation.
Once the prediction is locked in, the Scene changes
Balls / Bell curve
Bell curve
The Scene then shows: 0.6677 · The share of the histogram's area, bars one bin wide, between mean − within × spread and mean + within × spread, 0 to 1/nodes/board/outputs/shareWithin
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About $0.67$ ⁠ of the histogram’s area lies between $6$ ⁠ and $10$ ⁠, very close to the normal curve’s $68\%$ ⁠.

Hint
The standard deviation is a typical distance from the mean, not the largest one.

Worked solution
1. The histogram of $16$ ⁠ rows is almost the same as the bell curve, the normal distribution.
2. In a normal distribution about $68\%$ ⁠ of the area lies within $\mu \pm \sigma$ ⁠, here between $8 - 2 = 6$ ⁠ and $8 + 2 = 10$ ⁠.
3. On the board itself: $0.668$ ⁠ of the bars’ area, with half of bars $6$ ⁠ and $10$ ⁠ falling inside the range.
About a third of the balls land more than one standard deviation from the mean, and that is perfectly ordinary.
Step 4 of 5
Open this step
Two standard deviations
Practice · Predict, then look
Question
And what share lies within two standard deviations, between bin $4$ ⁠ and bin $12$ ⁠?

Correct answer
- Correct About $95\%$ ⁠
Setup, choices and feedback
The Scene at this step
Rows of pegs
16fixed
Chance of bouncing right
50 %fixed
Balls / Bell curve
Balls
Choices
- Correct About $95\%$ ⁠
- Wrong Every ball
  Feedback
  Close, but not all: about one ball in twenty lands farther out.
- Wrong Twice the share for one standard deviation: more than all of them
  Feedback
  A share cannot pass $100\%$ ⁠. Each further standard deviation adds less, because the curve is low at the edges.
- Wrong About $75\%$ ⁠
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Balls / Bell curve
Bell curve
Other settings
Within/nodes/board/props/within = 2
The Scene then shows: 0.951 · The share of the histogram's area, bars one bin wide, between mean − within × spread and mean + within × spread, 0 to 1/nodes/board/outputs/shareWithin
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$0.951$ ⁠ of the area: almost exactly $95\%$ ⁠. Only one ball in twenty lands outside bins $4$ ⁠ to $12$ ⁠.

Hint
The curve falls fast away from the middle. How much is left beyond two standard deviations?

Worked solution
1. The $68$ ⁠– $95$ ⁠– $99.7$ ⁠ rule: in a normal distribution $68\%$ ⁠ lies within one standard deviation, $95\%$ ⁠ within two, $99.7\%$ ⁠ within three.
2. Here two standard deviations are $4$ ⁠ bins, so the range runs from bin $4$ ⁠ to bin $12$ ⁠.
3. The board gives $0.951$ ⁠, just as the rule says.
Step 5 of 5
Open this step
Real data
Practice · Calculate
Question
Adult women’s heights are roughly normal, with a mean of 162 cm and a standard deviation of 6 cm. About what per cent of women are between 150 cm and 174 cm tall? Write the number only.

Correct answer
95 ± 1.5
Setup, choices and feedback
The Scene at this step
Rows of pegs
16
Chance of bouncing right
50 %
Balls / Bell curve
Bell curve
Accepted within
± 1.5
A fixed value in the question
Wrong answers it recognises
- 100 ± 1.5
  Misconception: Expects (nearly) all the data to lie within one standard deviation of the mean.
  Feedback from the catalogue
  The standard deviation is a typical distance, not a largest one, so plenty of values lie farther out. In a normal distribution about $68\%$ ⁠ of the values are within one standard deviation of the mean, about $95\%$ ⁠ within two, and $99.7\%$ ⁠ within three. So about a third of the values lie outside one standard deviation.
Feedback for a right answer
$174 - 162 = 12 = 2 \times 6$ ⁠: two standard deviations each side, so about $95\%$ ⁠.

Feedback for a wrong answer
How many standard deviations lie between the mean and each end? $68\%$ ⁠ belongs to one standard deviation only.

Hint
How many standard deviations are there from 162 to 174?

Worked solution
1. The distance from the mean to each end: $174 - 162 = 12$ ⁠ and $162 - 150 = 12$ ⁠.
2. In standard deviations: $\frac{12}{6} = 2$ ⁠.
3. By the $68$ ⁠– $95$ ⁠– $99.7$ ⁠ rule, about $95\%$ ⁠ lie within two standard deviations.
Only about one woman in twenty is taller than 174 cm or shorter than 150 cm, about half of them on each side.

Lesson 4 of 5 · About 25 minutes

Correlation

Two variables together: the scatter plot, the direction and strength of a link, the correlation coefficient $r$ and what it does not measure: not the slope, not a link that is not a straight line, and not a cause.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Rising and falling
Practice · Predict, then look
Question
The readings on the board fall from left to right, and hug a line fairly closely. What will the correlation coefficient $r$ ⁠ be?

Correct answer
- Correct Close to $-1$ ⁠
Setup, choices and feedback
The Scene at this step
Slope
-0.6fixed
Noise
0.8fixed
Outlier
offfixed
Measurements / Best-fit line
Measurementsfixed
Choices
- Correct Close to $-1$ ⁠
- Wrong Close to $1$ ⁠, because the points hug a line
  Feedback
  The points do hug a line, so $r$ ⁠ is far from $0$ ⁠. But the line falls, so the sign is negative.
- Wrong Negative, so a weak link
  Misconception: Takes a negative correlation coefficient as a weak or missing link.
  Feedback from the catalogue
  The sign of $r$ ⁠ only tells the direction of the link: positive when $y$ ⁠ rises with $x$ ⁠, negative when it falls. The strength is the size of $r$ ⁠, its distance from $0$ ⁠. $r = -0.87$ ⁠ is a strong link, stronger than $r = 0.6$ ⁠: the points hug a falling line.
- Wrong Close to $0$ ⁠
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Measurements / Best-fit line
Best-fit linefixed
The Scene then shows: -0.872 · The correlation coefficient, NaN when every y is the same/nodes/scatter/outputs/r
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$r = -0.87$ ⁠: a strong negative link. As $x$ ⁠ grows, $y$ ⁠ shrinks, and the points stay close to the falling line.

Hint
The sign of $r$ ⁠ is the direction of the line. Its size is how tightly the cloud hugs it.

Worked solution
1. The direction: the points fall, so $r < 0$ ⁠.
2. The strength: the points are close to a line, so $|r|$ ⁠ is close to $1$ ⁠.
3. Together: $r$ ⁠ is close to $-1$ ⁠. The board gives $r = -0.87$ ⁠.
A negative $r$ ⁠ is not a weak link. $r = -0.87$ ⁠ is exactly as strong as $r = 0.87$ ⁠, just the other way.
Step 2 of 6
Open this step
Tighter
Practice · Predict, then look
Question
Right now $r = 0.70$ ⁠. Turn Noise down from $1.6$ ⁠ to $0.4$ ⁠, keeping the slope the same. What happens to $r$ ⁠?

Correct answer
- Correct $r$ ⁠ gets closer to $1$ ⁠
Setup, choices and feedback
The Scene at this step
Slope
0.6fixed
Noise
1.6fixed
Outlier
offfixed
Measurements / Best-fit line
Best-fit linefixed
Choices
- Correct $r$ ⁠ gets closer to $1$ ⁠
- Wrong $r$ ⁠ stays the same, since the slope did not change
  Misconception: Reads the strength of a correlation from how steep the points rise, rather than how tightly they hug a line.
  Feedback from the catalogue
  The correlation coefficient $r$ ⁠ does not measure the slope. It measures how tightly the points line up: $r$ ⁠ is near $1$ ⁠ or $-1$ ⁠ when they lie almost on a straight line, and near $0$ ⁠ when they are scattered. A gentle, tight cloud gives a higher $r$ ⁠ than a steep, scattered one. The slope answers a different question: how much $y$ ⁠ rises when $x$ ⁠ rises by $1$ ⁠.
- Wrong $r$ ⁠ gets smaller
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong $r$ ⁠ becomes $0$ ⁠, because there is hardly any noise
  Feedback
  With no noise the points sit exactly on the line, which is the strongest link, $r = 1$ ⁠.
Once the prediction is locked in, the Scene changes
Noise
0.4fixed
Measurements / Best-fit line
Best-fit linefixed
The Scene then shows: 0.9663 · The correlation coefficient, NaN when every y is the same/nodes/scatter/outputs/r
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$r$ ⁠ rises to $0.97$ ⁠. The same slope, but the points hug the line more tightly, so the link is stronger.

Hint
$r$ ⁠ measures how tightly the points hug a line.

Worked solution
1. The noise is the scatter of the points about the line. When it shrinks, the points move closer to the line.
2. That is exactly what $r$ ⁠ measures: $r^2$ ⁠ is the share of the scatter in $y$ ⁠ the line accounts for.
3. So $r$ ⁠ rises, from $0.70$ ⁠ to $0.97$ ⁠, even though the slope stays $0.6$ ⁠.
Step 3 of 6
Open this step
Steep or tight
Practice · Predict, then look
Question
Two clouds of points. Cloud A has slope $0.3$ ⁠ and noise $0.2$ ⁠: a gentle, tight rise. Cloud B has slope $1$ ⁠ and noise $2$ ⁠: a steep, scattered rise. Which has the larger $r$ ⁠?

Correct answer
- Correct Cloud A, the gentle, tight one
Setup, choices and feedback
The Scene at this step
Slope
1
Noise
2
Outlier
offfixed
Measurements / Best-fit line
Measurementsfixed
Choices
- Correct Cloud A, the gentle, tight one
- Wrong Cloud B, the steep one
  Misconception: Reads the strength of a correlation from how steep the points rise, rather than how tightly they hug a line.
  Feedback from the catalogue
  The correlation coefficient $r$ ⁠ does not measure the slope. It measures how tightly the points line up: $r$ ⁠ is near $1$ ⁠ or $-1$ ⁠ when they lie almost on a straight line, and near $0$ ⁠ when they are scattered. A gentle, tight cloud gives a higher $r$ ⁠ than a steep, scattered one. The slope answers a different question: how much $y$ ⁠ rises when $x$ ⁠ rises by $1$ ⁠.
- Wrong Both have the same $r$ ⁠, since both rise
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Slope
0.3
Noise
0.2
Measurements / Best-fit line
Best-fit linefixed
The Scene then shows: 0.9663 · The correlation coefficient, NaN when every y is the same/nodes/scatter/outputs/r
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Cloud A has $r = 0.97$ ⁠. Try cloud B ( $1$ ⁠ and $2$ ⁠): only $r = 0.79$ ⁠. Slope and strength are two different things.

Hint
Ask how far the points are from the line, not how steep it is.

Worked solution
1. $r$ ⁠ does not depend on the slope itself, but on how much the line explains compared with the scatter about it.
2. In cloud A the scatter, $0.2$ ⁠, is tiny next to the rise across the board, so $r = 0.97$ ⁠.
3. In cloud B the line is steep, but the scatter, $2$ ⁠, is large, so $r = 0.79$ ⁠.
If you chose cloud B, you read the strength from the slope. The slope says how much $y$ ⁠ rises; $r$ ⁠ says how confidently you can predict it.
Step 4 of 6
Open this step
Aiming for $r$
Practice · Reach the target
Question
Change Slope and Noise until the correlation coefficient is $r = -0.9$ ⁠, then press Check.

Goal
- -0.9 ± 0.02
  At the start of the step: 0.8857
Setup, choices and feedback
The Scene at this step
Slope
0.6
Noise
0.8
Outlier
offfixed
Measurements / Best-fit line
Best-fit linefixed
Read from the Scene at
The correlation coefficient, NaN when every y is the same/nodes/scatter/outputs/r
One setting that reaches it
Slope
-0.25
Noise
0.3
Feedback for a right answer
$r$ ⁠ is close to $-0.9$ ⁠: a strong falling link. Many settings get there, because $r$ ⁠ depends on the slope compared with the noise.

Feedback for a wrong answer
Not yet. The sign comes from the slope, the size from how much noise there is next to it.

Hint
First turn the line round, then tune the noise until $r$ ⁠ arrives.

Worked solution
1. A negative $r$ ⁠ needs a negative slope: take Slope below $0$ ⁠, say to $-0.6$ ⁠.
2. With $-0.6$ ⁠ and Noise $0.8$ ⁠ you get $r = -0.87$ ⁠, a little too weak. Turn Noise down a touch, to $0.7$ ⁠, and $r$ ⁠ reaches $-0.9$ ⁠.
3. A steeper slope with a little more noise works too: what matters is how they compare.
Step 5 of 6
Open this step
Not a straight line
Practice · Choose
Question
A ball is thrown up and its height measured every tenth of a second, from the throw until it is back in the hand. The scatter plot of height against time gives $r \approx 0$ ⁠. What does that tell you?

Correct answer
- Correct There is no straight-line link: the height depends on time, but along an arch, not a straight line
Setup, choices and feedback
The Scene at this step
Slope
0
Noise
0.8
Outlier
off
Measurements / Best-fit line
Best-fit line
Choices
Shown in a random order.
- Correct There is no straight-line link: the height depends on time, but along an arch, not a straight line
- Wrong The height has nothing to do with the time
  Misconception: Concludes from r near 0 that the two variables have no relationship at all.
  Feedback from the catalogue
  $r$ ⁠ measures only a straight-line link, how well a line describes the points. Points on a symmetric parabola, such as the height of a ball thrown up against time, give $r \approx 0$ ⁠, even though the height is fixed entirely by the time. Look at the scatter plot before you compute $r$ ⁠.
- Wrong The measurements must be wrong
  Feedback
  The measurements are fine. $r$ ⁠ simply is not built for an arch: half the points rise and half fall, and the two cancel.
- Wrong The link is weak: the height can be predicted only roughly
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Feedback for a right answer
The points sit on a parabola, and the height can be predicted exactly from the time. $r$ ⁠ measures only how well a straight line describes them.

Hint
Picture the points: what shape do they make?

Worked solution
1. The height rises and then falls, so the points form a symmetric arch.
2. The straight line that fits them best is almost level, so $r \approx 0$ ⁠.
3. But the link is perfect: $h = v_0 t - \frac{1}{2} g t^2$ ⁠ gives the height at every moment.
That is why you always look at the scatter plot, not just at $r$ ⁠.
Step 6 of 6
Open this step
Not a cause
Practice · Choose
Question
In a seaside town, over a year, weekly ice-cream sales and the number of drownings that week are strongly correlated: $r = 0.8$ ⁠. What can you conclude?

Correct answer
- Correct Probably a third thing, hot weather, raises both
Setup, choices and feedback
The Scene at this step
Slope
0.6
Noise
0.8
Outlier
off
Measurements / Best-fit line
Best-fit line
Choices
Shown in a random order.
- Correct Probably a third thing, hot weather, raises both
- Wrong Ice cream causes drowning, and its sale should be limited
  Misconception: Reads a strong correlation as proof that one variable causes the other.
  Feedback from the catalogue
  A correlation says that two variables rise and fall together, not why. Often a third factor moves both: in hot summers more ice cream is sold, and more people swim, and so more drown. Ice cream does not cause drowning. To know what causes what you need an experiment, where you change one thing and hold everything else fixed.
- Wrong Drownings make people buy ice cream
  Misconception: Reads a strong correlation as proof that one variable causes the other.
  Feedback from the catalogue
  A correlation says that two variables rise and fall together, not why. Often a third factor moves both: in hot summers more ice cream is sold, and more people swim, and so more drown. Ice cream does not cause drowning. To know what causes what you need an experiment, where you change one thing and hold everything else fixed.
- Wrong The correlation is a fluke and means nothing
  Feedback
  A correlation of $0.8$ ⁠ over a whole year is unlikely to be a fluke. It is real; it just does not say what causes what.
Feedback for a right answer
In hot weeks more ice cream is sold, and more people go into the sea. The heat moves both variables, so they rise and fall together.

Hint
When do people buy a lot of ice cream, and when do they swim in the sea?

Worked solution
1. A correlation says only that two variables change together.
2. Here a third variable affects both: the temperature. Summer brings both more ice cream and more swimming.
3. To test whether ice cream causes drowning you would need an experiment: change only the ice-cream sales and hold the weather fixed.
A correlation is a clue worth looking into, not proof of a cause.

Lesson 5 of 5 · About 25 minutes

The line of best fit

Which straight line fits a cloud of points best? Turn a line by hand until the squares add up to the least, and discover least squares, the balance point, predicting from the line, and what an outlier does to it.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
By hand
Practice · Reach the target
Question
Your line passes through $(\bar{x}, \bar{y})$ ⁠. Change Slope of your line until the squares’ area is at most $5\%$ ⁠ above the least possible, then press Check.

Goal
- 0 ± 5
  At the start of the step: 363.9
Setup, choices and feedback
The Scene at this step
Slope of your line
0
Noise
0.8fixed
Outlier
offfixed
Measurements / Best-fit line
Mean and spreadfixed
Read from the Scene at
Per cent by which the squares about the hand-turned line exceed the least possible (the best-fit line’s); 0 at the least-squares slope/nodes/scatter/outputs/guessExcess
One setting that reaches it
Slope of your line
0.55
Feedback for a right answer
You found the bottom of the parabola. At this slope the residuals’ squares add up to the least they can: this is the line of best fit, “least squares”.

Feedback for a wrong answer
Not yet. If the squares grow as you raise the slope, go the other way.

Hint
Watch the average square under the plot and the dot on the chart beside the Scene: look for the slope where they are smallest.

Worked solution
1. At slope $0$ ⁠ the line is the mean line, and its squares are the scatter of $y$ ⁠ about $\bar{y}$ ⁠.
2. As you raise the slope towards the points, the sticks shorten and the squares shrink.
3. Past a certain point the line is too steep and the squares grow again. The total area is a parabola in the slope, with its bottom at a slope of about $0.62$ ⁠.
4. Any slope from $0.55$ ⁠ to $0.65$ ⁠ passes the check.
Step 2 of 5
Open this step
The balance point
Practice · Choose
Question
Which point does the line of best fit always pass through, whatever the cloud of points?

Correct answer
- Correct The balance point $(\bar{x}, \bar{y})$ ⁠
Setup, choices and feedback
The Scene at this step
Slope of your line
0fixed
Noise
0.8fixed
Outlier
offfixed
Measurements / Best-fit line
Best-fit line
Choices
Shown in a random order.
- Correct The balance point $(\bar{x}, \bar{y})$ ⁠
- Wrong The first and the last reading
  Misconception: Draws the line of best fit through the first and last points, or through as many points as possible.
  Feedback from the catalogue
  The line of best fit need not pass through any reading. It is the line for which the total of the squared vertical distances from all the points is smallest, so every point has a say, not just two. It always passes through the balance point $(\bar{x}, \bar{y})$ ⁠, the means of both variables.
- Wrong The origin $(0, 0)$ ⁠
  Feedback
  Only if by chance $\bar{y} = b\bar{x}$ ⁠. Here the line crosses the $y$ ⁠-axis at about $1.90$ ⁠.
- Wrong Through as many readings as possible
  Misconception: Draws the line of best fit through the first and last points, or through as many points as possible.
  Feedback from the catalogue
  The line of best fit need not pass through any reading. It is the line for which the total of the squared vertical distances from all the points is smallest, so every point has a say, not just two. It always passes through the balance point $(\bar{x}, \bar{y})$ ⁠, the means of both variables.
Feedback for a right answer
The line of best fit always passes through $(\bar{x}, \bar{y})$ ⁠, so its equation is $\hat{y} - \bar{y} = b(x - \bar{x})$ ⁠.

Hint
On the board the line turns about one point. Which?

Worked solution
1. For any slope $b$ ⁠, the line that fits the points best passes through $(\bar{x}, \bar{y})$ ⁠: then the residuals balance and add to $0$ ⁠.
2. That leaves only the slope to choose, which is what you did by hand in the last step.
3. So $a = \bar{y} - b\bar{x}$ ⁠: on the board $5.01 - 0.62 \times 5.02 \approx 1.90$ ⁠.
The line need not pass through any reading. Every reading has a say through its square.
Step 3 of 5
Open this step
Predicting
Practice · Calculate
Question
The board prints the line of best fit, $\hat{y} = a + bx$ ⁠. By it, what is the predicted value of $y$ ⁠ at $x = 8$ ⁠? Round to two decimal places.

Correct answer
6.861 ± 0.05
Setup, choices and feedback
The Scene at this step
Slope of your line
0defaultfixed
Noise
0.8fixed
Outlier
offfixed
Measurements / Best-fit line
Best-fit linefixed
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Ŷ at predictX on the best-fit line, a + b·predictX/nodes/scatter/outputs/prediction
Feedback for a right answer
$\hat{y} = 1.90 + 0.62 \times 8 \approx 6.86$ ⁠. That is the prediction for a typical value at $x = 8$ ⁠; a single reading there would miss it by about $s_e \approx 0.8$ ⁠.

Feedback for a wrong answer
Put in $x = 8$ ⁠ and remember $a$ ⁠: the line does not pass through the origin.

Hint
Put $x = 8$ ⁠ into the equation on the board.

Worked solution
1. From the board: $\hat{y} = 1.90 + 0.62x$ ⁠.
2. Put in $x = 8$ ⁠: $0.62 \times 8 = 4.96$ ⁠.
3. Add the intercept: $1.90 + 4.96 = 6.86$ ⁠.
If you got $4.96$ ⁠, you left out $a$ ⁠. And a prediction inside the range of the readings, like this one, is far more trustworthy than one well outside it.
Step 4 of 5
Open this step
One far point
Practice · Predict, then look
Question
Switch on Outlier: one reading on the right, far below the rest. What happens to the slope of the line of best fit, now $0.62$ ⁠?

Correct answer
- Correct The line gets flatter
Setup, choices and feedback
The Scene at this step
Slope of your line
0defaultfixed
Noise
0.8fixed
Outlier
offfixed
Measurements / Best-fit line
Best-fit linefixed
Choices
- Correct The line gets flatter
- Wrong The line gets steeper
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong The line stays put: it is one reading in $21$ ⁠
  Feedback
  For the median one reading makes little difference, but not for the line: it shrinks squares, and a far reading has a huge square.
Once the prediction is locked in, the Scene changes
Outlier
onfixed
Measurements / Best-fit line
Best-fit linefixed
The Scene then shows: 0.4675 · B, the least-squares slope Sxy / Sxx/nodes/scatter/outputs/slope
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The slope drops to $0.47$ ⁠. The outlier’s square is so large that the line gives up some fit to the rest to get closer to it.

Hint
A far reading has a huge square. Which way will the line move to shrink it?

Worked solution
1. The outlier sits at $x = 9.4$ ⁠, about $5$ ⁠ below the line. Its square is about $25$ ⁠, more than all the other squares together.
2. To shrink the total, the line lowers its right-hand end towards it.
3. So the slope drops from $0.62$ ⁠ to $0.47$ ⁠, and $r$ ⁠ from $0.89$ ⁠ to $0.69$ ⁠.
Unlike the median, the line of best fit is very sensitive to outliers, because of the squares. So it pays to check any suspicious reading.
Step 5 of 5
Open this step
How much the line explains
Practice · Calculate
Question
A study finds $r = 0.8$ ⁠ between hours of study and marks. What per cent of the scatter in the marks (of the squares’ area about the mean) does the line of best fit account for? Write the number only.

Correct answer
64 ± 0.5
Setup, choices and feedback
The Scene at this step
Slope of your line
0defaultfixed
Noise
0.8
Outlier
off
Measurements / Best-fit line
Best-fit line
Accepted within
± 0.5
A fixed value in the question
Feedback for a right answer
$r^2 = 0.8^2 = 0.64$ ⁠: the line removes $64\%$ ⁠ of the squares’ area, and $36\%$ ⁠ remains as scatter about the line.

Feedback for a wrong answer
Not $r$ ⁠ itself: the share explained is $r^2$ ⁠, as the board prints it, $r^2 = 1 - \frac{s_e^2}{s^2}$ ⁠.

Hint
On the board, which number compares the two average squares?

Worked solution
1. The board shows that $r^2 = 1 - \frac{s_e^2}{s^2}$ ⁠: the share of the squares’ area about the mean that the line removes.
2. Here $r^2 = 0.8^2 = 0.64$ ⁠.
3. That is $64\%$ ⁠.
$r = 0.8$ ⁠ sounds like “ $80\%$ ⁠”, but the share explained is smaller. And $r = 0.5$ ⁠ accounts for only a quarter.

Across the unit · 13 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Mean, median and mode

The balance point

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The value in the middle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

An even count

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The most common value

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A bad reading

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

The new mean

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Range and standard deviation

The range

Question

Correct answer

The Scene at this step

Hint

Worked solution

The distance from the mean

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Setting the spread

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Range against standard deviation

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

Inside the band

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution