Mixed practice: graphs and averages

5.1 One page, several kinds of data

This page mixes the chapter's two skills: summarising a data set with mean and median, and reading a graph carefully. Every question is new. The data is small enough to work by hand, so write the ordered list or the totals before choosing the measure.

5.1.1 Mean, median and missing values

Question 1. Find the mean, median and range of 12, 15, 15, 18, 20 and 24.

Answer. The total is 12 + 15 + 15 + 18 + 20 + 24 = 104 and there are 6 values, so the mean is 104 ÷ 6 ≈ 17.33. The ordered list is 12, 15, 15, 18, 20, 24; the two middle values are 15 and 18, so the median is (15 + 18) ÷ 2 = 16.5. The range is 24 − 12 = 12. The mean is above the median because the two largest values pull it up.

Question 2. A data set has 8 values and mean 14. Seven of the values are 9, 11, 13, 15, 16, 17 and 15. Find the eighth value.

Answer. The total is 8 × 14 = 112. The seven known values add to 9 + 11 + 13 + 15 + 16 + 17 + 15 = 96. The eighth value is 112 − 96 = 16. Checking, the total 96 + 16 = 112 and 112 ÷ 8 = 14.

Question 3. A frequency table gives the value 2 with frequency 5, the value 3 with frequency 8 and the value 4 with frequency 7. Find the total frequency, the mean and the median.

Answer. The total frequency is 5 + 8 + 7 = 20 values. The total of value × frequency is 2×5 + 3×8 + 4×7 = 10 + 24 + 28 = 62, so the mean is 62 ÷ 20 = 3.1. For the median, positions 10 and 11 are needed. The cumulative frequencies are 5 after the value 2 and 13 after the value 3, so both positions hold 3 and the median is 3.

5.1.2 Dot plots and line graphs

Question 4. A dot plot shows the value 1 twice, the value 3 four times, the value 5 three times and the value 8 once. Find the number of values, the mean and the median.

Answer. The number of values is 2 + 4 + 3 + 1 = 10. The total is 2×1 + 4×3 + 3×5 + 1×8 = 2 + 12 + 15 + 8 = 37, so the mean is 37 ÷ 10 = 3.7. In order the values are 1, 1, 3, 3, 3, 3, 5, 5, 5, 8; the 5th and 6th values are both 3, so the median is 3. The single 8 raises the mean but does not move the median.

Question 5. A shop records visitors and buyers for a week: visitors 20, 25, 22, 30, 28, 40 and 45; buyers 12, 15, 14, 20, 18, 26 and 30. Find the mean number of visitors and the mean number of buyers, and describe the pattern.

Answer. The visitors total 20 + 25 + 22 + 30 + 28 + 40 + 45 = 210 over 7 days, so the mean is 30 visitors a day. The buyers total 12 + 15 + 14 + 20 + 18 + 26 + 30 = 135, so the mean is 135 ÷ 7 ≈ 19.3 buyers a day. Both counts peak on Sunday (45 visitors, 30 buyers) and the visitors line stays above the buyers line all week. The gap is smallest on the first and third days (8 each) and largest on Sunday (15).

Question 6. Which of these statements are true? (i) Adding 5 to every value raises the mean and the median by 5. (ii) Doubling every value doubles the mean and the median. (iii) Removing a value below the mean always increases the mean. (iv) Adding a value below the median always leaves the median unchanged.

Answer. (i) True: shifting every value by the same amount shifts both measures by that amount. (ii) True: scaling every value scales both measures by the same factor. (iii) True for any data set with more than one value, because the new total loses less than the old mean from the sum. (iv) False in general: 1, 2, 3, 4, 5 has median 3, but adding 0 gives 0, 1, 2, 3, 4, 5, whose median is (2 + 3) ÷ 2 = 2.5, so the median can change.

Question 7. A student says the mean of 4, 6, 6, 8 and 11 is 7. Show the calculation and explain how one wrong value would change the answer.

Answer. The total is 4 + 6 + 6 + 8 + 11 = 35, and 35 ÷ 5 = 7, so the mean is 7. If the 11 were misread as 1, the total would be 25 and the mean would fall to 5; if the 4 were misread as 14, the total would be 45 and the mean would rise to 9. A single misread value moves the mean by its error divided by the count.