Year 10 Extended · Unit 6 Statistics and the bell curve · Lesson 1 of 7

What a table of classes can and cannot tell you

When data arrive in classes, the individual values are gone. This lesson asks what a table of classes can still tell you about the mean, the mode and the median, and how closely.

Statistics and ProbabilityGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 8: find the mean of 12, 15, 15, 18 and 20.
    Answer16. The total is 80 and 80 ÷ 5 = 16.
  2. From Year 9: a table gives 3 values of 4, 5 values of 6 and 2 values of 9. How many values are there?
    Answer10. Add the frequencies: 3 + 5 + 2 = 10.
  3. From Year 9: what number is halfway between 20 and 30?
    Answer25. The midpoint is (20 + 30) ÷ 2 = 25.
Words used in this lesson
class
a group of values between two boundaries. For example: 0 ≤ x < 8.
mid-interval value
the value halfway between the class boundaries. For example: 6 for the class 4 ≤ x < 8.
modal class
the class with the greatest frequency (equal widths). For example: the tallest bar of a histogram.
median class
the class that holds the middle value. For example: the class where the running total passes n ÷ 2.
estimate
a value found from incomplete information. For example: x̄ ≈ Σfx ÷ n.
class width
the upper boundary minus the lower boundary. For example: 8 − 4 = 4.

The clinic that publishes classes, not times

A clinic in Nairobi publishes how long its patients waited, but only in ten-minute classes: so many waited 0 to under 10 minutes, so many 10 to under 20, and so on. You are asked for the mean waiting time.

Predict first: can you find the exact mean from the table? If not, what is the least it could be and the greatest, and where would you put your single best estimate? Then slide the values across their classes in the figure.

5
25.25mean at this position
20.25least the mean can be
30.25mean if every value were at an upper end

Values sit 5 tenths of the way across their classes: the mean is 25.25. It cannot be below 20.25 and it stays under 30.25.

Show me the steps
  1. Slide to 0. Every value sits at the lower end of its class, so the mean is (4 × 0 + 9 × 10 + 14 × 20 + 8 × 30 + 5 × 40) ÷ 40 = 20.25, the least it can be.
  2. Slide to 10. Every value sits at the upper end of its class and the mean is 30.25. The true mean is below this, because a value under 10 is not 10 itself.
  3. Slide to 5. The rings are at the mid-interval values 5, 15, 25, 35 and 45, and the mean is 25.25: exactly halfway between the two extremes, because every class is 10 wide.
  4. Slide to 3. The mean is 23.25. Each tenth of a class width moves the mean by one minute, whatever the frequencies are: 40 values, each moved the same amount.

Estimated mean. Replace every value in a class by the mid-interval value (halfway between the class boundaries), multiply each by its frequency, add, and divide by n, the total frequency: x̄ ≈ Σfx ÷ n. It is an estimate, because the real values inside each class are unknown.

Modal class is the class with the greatest frequency (when the classes are equal in width). The median class is the class that holds the middle value, found by counting up the frequencies.

The true mean lies between the mean with every value at a lower boundary and the mean with every value at an upper boundary. With equal class width w the estimate is exactly halfway, so it is never more than w ÷ 2 from the truth.

Divide by n, the number of values, not by the number of classes. With mid-interval values 5 and 15 and frequencies 1 and 9, Σfx = 140. Dividing by n = 10 gives 14, a value between 5 and 15. Dividing by 2 classes gives 70, which is far above every value in the table.

Worked examples

1. Parcel masses (kg): 0 to under 2 has 5 parcels; 2 to under 4 has 12; 4 to under 6 has 9; 6 to under 8 has 4. Estimate the mean.

  1. The mid-interval values are 1, 3, 5 and 7, and n = 5 + 12 + 9 + 4 = 30.
  2. Σfx = 5 × 1 + 12 × 3 + 9 × 5 + 4 × 7 = 5 + 36 + 45 + 28 = 114.
  3. The estimated mean is 114 ÷ 30 = 3.8 kg.

2. For the same table, find the modal class and the median class.

  1. The greatest frequency is 12, so the modal class is 2 to under 4 kg.
  2. With n = 30 the middle values are the 15th and 16th. The running totals are 5, then 17, so both lie in the second class: the median class is also 2 to under 4 kg.

3. Between what two values must the true mean lie?

  1. With every value at a lower boundary, the mean is (5 × 0 + 12 × 2 + 9 × 4 + 4 × 6) ÷ 30 = 84 ÷ 30 = 2.8.
  2. With every value at an upper boundary, every value is 2 more, so the mean is 4.8.
  3. The true mean is between 2.8 kg and 4.8 kg, and the estimate 3.8 kg is exactly halfway.

Your go. The first step is given. A table of journey times (minutes) has 0 to under 5: 4 journeys; 5 to under 10: 8; 10 to under 15: 5; 15 to under 20: 3. Estimate the mean.

  1. The mid-interval values are 2.5, 7.5, 12.5 and 17.5, and n = 20.
  2. Now work out each fx, add them, and divide by n.
Check the answer

fx = 10, 60, 62.5 and 52.5, which add to 185. Then 185 ÷ 20 = 9.25 minutes. The true mean is within 2.5 minutes of this, because every class is 5 wide.

With a GDC (optional)

A GDC can do the sum for you. Put the mid-interval values in one list and the frequencies in a second list, then ask for the one-variable statistics with the frequency list named. The mean it reports is the estimated mean, and n is the total frequency.

The GDC treats each mid-interval value as if it were a real data value, so the result is still an estimate. Check one product fx by hand before you trust the list. Everything here can be done by hand.

Check before you go on

1. The class 20 ≤ x < 30 has frequency 6 and the class 30 ≤ x < 40 has frequency 4. What is the estimated mean?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Times (minutes) are grouped: 0 to under 10 has 6 values; 10 to under 20 has 11; 20 to under 30 has 8; 30 to under 40 has 5. What is the estimated mean?

3. Ages are grouped: 10 to under 20 has 3 people; 20 to under 30 has 14; 30 to under 40 has 6; 40 to under 50 has 7; 50 to under 60 has 5. The median lies in a class. What is the lower boundary of that class?

Level 2 · Building

4. A table has classes that are all 10 wide. Its mean estimated from the mid-interval values is 26.4. The true mean must be less than what number?

5. Four classes have mid-interval values 5, 15, 25 and 35 and frequencies 4, 10, f and 6. The estimated mean is 20. Find f.

Level 3 · Stretch

6. A table of 50 values has classes that are 5 wide. Its estimated mean is 42.6. What is the greatest whole number the true mean could be?

7. Group A has 12 values with an estimated mean of 30. Group B has 28 values with an estimated mean of 50. What is the estimated mean of all 40 values together?

Something to think about

In the figure every class is 10 wide, and moving every value by one tenth of a class moved the mean by exactly 1. Suppose the classes were 5 wide instead. How far would the same slide move the mean, and what does that say about why narrower classes give a better estimate?

One way to think about it

Each tenth of a class width is 0.5 when the width is 5, so the same slide moves the mean by 0.5, and the whole range of possible means is only 5 wide instead of 10. The estimate is then within 2.5 of the true mean instead of within 5. Narrower classes shrink the uncertainty because each value is pinned down more tightly, at the price of a longer table.

Exit ticket

8. A table groups 60 exam marks into equal classes. Which of these can be read from the table with no estimating at all?