Data questions are the ones candidates rush. The arithmetic is easy, so the eyes move fast — and the two failures that follow are predictable: reading the wrong row of a table, and answering a question about the mean when it asked for the median. Both produce answers that look fine.
Probability questions share the same reading risk. They are almost never computational beyond a single fraction, but they are always asking a precise question — "at least one" and "exactly one" are different questions, and only one of them is on the paper.
Split the work into two steps that cannot be skipped.
For averages, remember that the three "measures of centre" — mean, median and mode — answer different questions and can differ by a lot. The mean of 1, 2, 3, 4 and 40 is 10; the median is 3. A question that mentions an outlier is usually a hint that the mean is not the measure being asked about.
Mean. Add the values and divide by how many there are. The mean of 4, 8, 10 and 14 is 36 / 4 = 9. Reversed, if the mean of five numbers is 12, their total is 12 x 5 = 60 — this reverse form answers many questions in one step.
Median. Sort the values first, then take the middle one. For 3, 7, 9, 11 and 15 the median is 9. With an even count, average the two middle values: for 3, 7, 9 and 11, the median is (7 + 9) / 2 = 8. Sorting first is not optional — the median of an unsorted list is the single most common careless error in this topic.
Mode and range. The mode is the most frequent value: in 2, 3, 3, 5, 7 it is 3. The range is the largest value minus the smallest: 15 - 3 = 12.
Variance and standard deviation. These measure spread, not centre. Steps: find the mean, square each deviation from the mean, average those squares (divide by how many values there are), and — for standard deviation — take the square root. For 1, 3, 5, 7, 9 the mean is 5; the squared deviations are 16, 4, 0, 4, 16, summing to 40; the variance is 40/5 = 8; the standard deviation is root 8, about 2.83. Variance is in square units of the data, SD is back in the original units — which is why SD is the one usually quoted.
Reading tables and charts. Percentages from a table are computed against the correct total. If 40 of 200 respondents chose an option, that is 40 / 200 = 20 percent. Watch for tables whose final row is a total row, because including it in an average double-counts everything.
Probability as a fraction. Work out the favourable outcomes and the total equally likely outcomes, then write the fraction.
| Situation | Favourable | Total | Probability |
|---|---|---|---|
| An even number on a fair die | 2, 4, 6 (three) | 6 | 3/6 = 1/2 |
| A heart from a full deck | 13 | 52 | 13/52 = 1/4 |
| Two heads from two fair coins | HH (one) | HH, HT, TH, TT (four) | 1/4 |
Independent events multiply. The probability of two independent things both happening is the product: a head then a head is 1/2 x 1/2 = 1/4.
A class of 40 students sits a test. The marks are grouped as follows.
| Marks | Number of students |
|---|---|
| 40 to 49 | 4 |
| 50 to 59 | 10 |
| 60 to 69 | 16 |
| 70 to 79 | 8 |
| 80 to 89 | 2 |
First, a count check: 4 + 10 + 16 + 8 + 2 = 40, which matches the stated class size. Doing this check takes three seconds and catches a misread cell before it becomes a wrong answer.
What percentage of students scored 60 or above? That is the 16 + 8 + 2 = 26 students in the top three bands, so 26 / 40 = 0.65, which is 65 percent.
What is the modal band? The highest frequency is 16, which sits in the 60 to 69 band. Note that the mode is the most frequent band; if the question instead asked for the median band, the median is the 20th and 21st student in order, and cumulative counts (4, then 14, then 30) place both inside the 60 to 69 band as well.
If a student is picked at random, what is the probability that they scored below 50? Four students of 40, so 4 / 40 = 1/10.
What is the median of 3, 7, 9 and 11?
A fair die is rolled once. What is the probability of an even number?
The mean of five numbers is 12. What can you say about them?
What is the variance of 1, 3, 5, 7, 9?
Without looking back: what must you do to a list before finding its median, and how do you compute "at least one" probability quickly?
The answer is: sort the list first, and for an even count average the two middle values. For "at least one", subtract the probability of none from 1.
Fill in the blanks from memory:
The mean of 4, 8, 10 and 14 is . If the mean of five numbers is 12, their total is . The probability of getting at least one head when two fair coins are tossed is . The variance of 1, 3, 5, 7, 9 is , and its standard deviation is the square root of that.
In a survey of 250 people, 60 percent said they read news online, and of those, one third also read a printed newspaper. How many people read both, and how many read neither online news nor print?
Online readership is 60 percent of 250, which is 150 people. One third of those, 150 / 3 = 50 people, also read print. So 50 people read both.
Now the harder half of the question, which needs a decision rather than a computation. Of the 250 people, 150 read online, so 100 do not. Among those 100, the question's wording — one third of those who read online read print too — says nothing about print-only readers, so the number who read neither cannot be determined from the information given. If an option says "cannot be determined", that is very often the correct answer to a question shaped like this one, and choosing it is a mark earned by reading precisely rather than by calculating.
That covers reading data from tables, charts and summaries. The next lesson returns to words: translating a sentence into an equation. Every quantitative technique you have met so far arrives on the paper wrapped in prose, so the translation step is the one that decides whether the technique is ever reached.