A mean with one frequency missing
A group of students takes a short quiz, and every student scores 1, 2 or 3 points. Four students score 1 point and six score 3 points. The number who scored 2 points has been lost, so call it f. The teacher remembers one more fact: the mean score was 2.1 points.
The mean depends on every frequency in the table, including f. So knowing the mean is enough to find f, as long as f is kept in every place it belongs.
Four students scored 1 and six scored 3. The number who scored 2 is the unknown frequency f.
The mean fixes the total
The mean is the total of the values divided by how many values there are. Turned around, that says the total is the mean times the number of values.
The number of values is the total frequency, 4 + f + 6 = 10 + f. The students who scored 2 are students like any others, so f is part of the count. The total of all the scores is therefore 2.1 × (10 + f), written 2.1(10 + f).
The same total, added from the table
The total can also be added up from the table, row by row, as for any frequency table: each score times its frequency. The 4 students who scored 1 give 1 × 4 = 4 points, the f students who scored 2 give 2 × f = 2f points, and the 6 students who scored 3 give 3 × 6 = 18 points. Altogether that is 4 + 2f + 18 = 22 + 2f points.
The two expressions are the same total, found two ways, so they are equal:
22 + 2f = 2.1(10 + f)
Expand the bracket: 22 + 2f = 21 + 2.1f. Subtract 2f from both sides: 22 = 21 + 0.1f. Subtract 21: 1 = 0.1f. Divide by 0.1: f = 10. Ten students scored 2 points.
Both totals contain f: there are 10 + f students, and their scores add up to 22 + 2f. The mean, 2.1, is the second total divided by the first.
The same equation, written with
The mean from a frequency table is : the total of each value times its frequency, divided by the total frequency. Here and , so the mean says .
Multiply both sides by 10 + f and this is the equation above, 22 + 2f = 2.1(10 + f). Whichever way it is written, the unknown frequency appears twice: once in the total of the values and once in the number of values.
Check by working out the mean again
Put f = 10 back into the table. There are 4 + 10 + 6 = 20 students, and their scores add up to 4 + 20 + 18 = 42 points. The mean is 42 ÷ 20 = 2.1 points, as stated, so f = 10 is right.
The size of the answer makes sense too. With no scores of 2 at all, the mean would be 22 ÷ 10 = 2.2. Every score of 2 pulls the mean down toward 2, so a mean of 2.1, part of the way from 2.2 toward 2, needs some 2s. A mean even closer to 2 would need more of them.
Two mistakes to avoid
The first mistake is to leave f out of the count and write the total as 2.1 × 10 = 21. Then 22 + 2f = 21 gives f = −0.5, and a frequency cannot be negative or a fraction. An answer like that is the signal to check that the unknown frequency is in the number of values as well as in the total.
The second mistake is to treat the mean as the mean of the three scores, (1 + 2 + 3) ÷ 3 = 2. That ignores the frequencies, which are the whole point: the mean of the students’ scores depends on how many students got each score.
Worked example: A Missing Frequency in a Shop's Records, Recovered from a Stated Mean
Question A shop recorded the number of bicycles it sold each day. It sold 0 bicycles on 4 days, 1 bicycle on 6 days, 2 bicycles on 9 days, 3 bicycles on k days and 4 bicycles on 5 days. The mean number sold in a day was exactly 2 bicycles. (a) Find k. (b) Find the median number sold in a day, and say on how many days the shop sold more than the mean.
1.Count the days in terms of k: 4 + 6 + 9 + k + 5 = 24 + k days.
The unknown k stands in the f column, so it counts in the days as well as in the bicycles. 2.Count the bicycles in the same way: 0 × 4 + 1 × 6 + 2 × 9 + 3 × k + 4 × 5 = 6 + 18 + 3k + 20 = 44 + 3k bicycles.
The fx column: 3 bicycles on k days is 3k bicycles. 3.The mean is the bicycles divided by the days, and that is 2, so 44 + 3k24 + k = 2. Multiply both sides by 24 + k: 44 + 3k = 2(24 + k) = 48 + 2k.
The mean is the bicycles over the days: 44 + 3k24 + k = 2, so 44 + 3k = 48 + 2k. 4.(a) Subtract 2k from both sides and then 44: k = 4. Check: the shop traded on 24 + 4 = 28 days and sold 44 + 12 = 56 bicycles, and 56 ÷ 28 = 2 as stated.
(a) k = 4. The table is complete: 28 days and 56 bicycles, and 56 ÷ 28 = 2 as stated. 5.For the median there are 28 days, so it lies halfway between the 14th and the 15th. Counting along the frequencies, 4 days sold none, 4 + 6 = 10 days sold 1 or fewer and 10 + 9 = 19 days sold 2 or fewer. The 11th to the 19th days all sold 2, so the 14th and the 15th are both 2 and the median is 2 bicycles.
Counting along the frequencies, 4, 10 and then 19 days, the 14th and the 15th days both sold 2 bicycles, so the median is 2. 6.(b) More than the mean of 2 means 3 or 4 bicycles, which is 4 + 5 = 9 days. The median is 2 bicycles and the shop beat its mean on 9 of the 28 days.
(b) More than the mean means 3 or 4 bicycles: 4 + 5 = 9 days.
Answer: (a) k = 4, so the shop traded on 28 days and sold 56 bicycles; (b) the median is 2 bicycles and the shop sold more than the mean on 9 days
Common mistakes
- Forgetting that k appears in the divisor as well as in the total, and solving 44 + 3k = 2 × 24. The unknown days are days like any other: they count in the number of days the mean is taken over, so 24 + k is the divisor.
- Answering the median as 0 + 42 = 2 from the values in the first column. That the answer agrees here is a coincidence of this table. The median is the middle of the 28 days, so the frequencies have to be counted along.