A mean when you only have the bands

Mathematics · Data & Statistics · ages 14-15

Name ______________________ Date ____________

Frequencies down a grouped table read 5, 7, 9 and 4 for the classes 0-5, 5-10, 10-15 and 15-20. The median lies in 5 ≤ x < 10.
Circle one: True False
Using only grouped counts and class midpoints, a calculated mean must be reported as an estimate because the exact values within the classes are unknown.
Circle one: True False
When estimating a mean from a grouped table that gives class counts but no exact observations, why do we use each class's midpoint?
- Because midpoints always come out as whole numbers
- Because every class in a table has to be the same width
- Because the table omits exact values and we use one representative value for each class
- Because the mean always lands inside the middle class of the table
A survey records how many minutes 20 students spent on homework.
Time (min) | Frequency 0 ≤ t < 10 | 6 10 ≤ t < 20 | 5 20 ≤ t < 30 | 7 30 ≤ t < 40 | 2
Use class midpoints to estimate the mean time.
Answer: ______________
A grouped table has frequencies 5, 22, 9 and 4. Dee estimates the mean by adding the four midpoints and dividing by 4. What has she done wrong?
- She should have added the class widths instead of the midpoints
- She should have used the upper boundary of each class
- Nothing, dividing by the number of classes is the right move
- She has averaged the four classes, not the people, so each midpoint needs multiplying by its frequency first
Sam uses each class's lower boundary instead of its midpoint. Compared with the midpoint estimate, what happens to his mean estimate?
- It is lower, because each stand-in is smaller
- It is higher, because every lower boundary lies above its class midpoint
- It stays the same, because the frequency column has not changed
- It becomes exact for any data set, because class boundaries are real numbers
From a grouped table you can say exactly how long the longest visit was.
Circle one: True False
A grouped table has a unique largest frequency of 26. Leo writes 26 as the mode. What can the table identify instead?
- The midpoint of that class, because it must be the most frequent exact value
- The total frequency, because that is the observed value a mode measures
- The class beside the 26, reported as the modal class
- Nothing, because a grouped table cannot identify any class with the most observations
If every value in every class equals that class's midpoint, the midpoint estimate equals the true mean.
Circle one: True False
A clinic records the ages of 30 patients.
Age (years) | Frequency 10 ≤ a < 15 | 2 15 ≤ a < 20 | 6 20 ≤ a < 25 | 10 25 ≤ a < 30 | 8 30 ≤ a < 35 | 4
Use class midpoints to estimate the mean age.
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

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False · There are 25 values, so the median is at position (25 + 1) / 2 = 13. The running totals are 5, 12, 21 and 25, and the 13th is past 12, so it sits in 10 ≤ x < 15 rather than 5 ≤ x < 10.
True · Midpoints replace unknown observations. The result could equal the true mean, but the grouped counts alone cannot confirm that. It should therefore be reported as an estimate.
Because the table omits exact values and we use one representative value for each class · The table supplies class counts but not the exact observations needed for a true total. Using each midpoint as a representative value gives an estimated total and mean. This does not mean every calculation with a grouped table needs midpoints.
17.5 · Midpoints 5, 15, 25 and 35 give 30 + 75 + 175 + 70 = 350 minutes, and 350 / 20 = 17.5.
She has averaged the four classes, not the people, so each midpoint needs multiplying by its frequency first · Dividing by 4 treats the class holding 22 people exactly like the class holding 4. The frequencies are where the people are, and they have to weight the midpoints.
It is lower, because each stand-in is smaller · Each lower boundary is below its midpoint. With the same frequencies, using smaller stand-ins lowers the estimated total and mean. This comparison is with the midpoint estimate, not a claim about the unknown true mean.
False · The top class gives a band the longest visit fell inside, not the visit itself. A table with a top class of 80 ≤ t < 100 tells you the longest was under 100 minutes and nothing sharper.
The class beside the 26, reported as the modal class · The 26 is a count. It could also happen to be an observed value, but the count does not establish that value's frequency. The table identifies the class containing the most observations, not an exact modal value.
True · Under this condition, every substituted midpoint equals the original observation. The estimated total and count therefore match the real total and count, giving the same mean. This is sufficient, but not the only way an estimate can be exact.
23.5 · Midpoints 12.5, 17.5, 22.5, 27.5 and 32.5 give 25 + 105 + 225 + 220 + 130 = 705, and 705 / 30 = 23.5 years.