Start with the observed group
Before calculating any average, ask what each row represents. One value per person, one value per group, and a frequency are not combined in the same way. To find the overall mean, multiply each group mean by its size, add the products, then divide by the total number of observations. If groups are the same size, this is a simple average of their means; otherwise, that simple average is generally not the overall mean.
Descriptive statistics summarize data; they do not automatically explain the underlying causes. A high mean may result from an extreme value. The median describes the center of an ordered data set without reflecting how far extreme values are from it. ETS includes several measures and basic probability in its program; the following exercises cover some of them, without inferential statistics.
Exercise 1: mean, median, and range
Here are fictional delivery times, in minutes, for five comparable deliveries. Every observation has the same weight; none is excluded.
| Delivery | Time, minutes |
|---|---|
| A | 12 |
| B | 14 |
| C | 14 |
| D | 16 |
| E | 34 |
For delivery times of 12, 14, 14, 16, and 34 minutes, find the mean, median, and range. If the 34-minute delivery is replaced by a 19-minute delivery, which of these measures remain unchanged?
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The solution, step by step
Mean 18, median 14, range 22; only the median remains unchanged.
The method in action
- The sum is 90, so the mean is 90/5 = 18. The third value in the ordered data set is 14. The range is 34 − 12 = 22.
- After the replacement, the sum is 75, so the mean is 15; the median remains 14 and the range becomes 19 − 12 = 7.
- An unchanged median does not mean every measure stays the same. Also, do not confuse range with standard deviation; they are different measures of variability.
Exercise 2: explicit dependence
For successive draws, write down the conditions at each step. Without replacement, the contents of the bag change. With replacement, the bag returns to its original composition.
A bag contains 3 red and 5 blue counters. Two counters are selected at random without replacement. What is the probability that both are red?
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The solution, step by step
3/28.
The method in action
- Each counter is equally likely to be drawn. The first probability is 3/8. After drawing a red counter, 2 red counters remain out of 7, or 2/7. Multiply the probability of drawing red first by the probability of drawing red second given that the first was red: 6/56 = 3/28. Independence is not required for this conditional product.
- 9/64 would use 3/8 twice, as if the counter were replaced. 6/15 adds numbers without defining an event. The result is less than 3/8, as expected: drawing two red counters is more restrictive than drawing one.
Exercise 3: at least one event
“At least one” combines several possibilities. Its complement, “none,” may be easier to calculate. Independence must be given or justified; it does not follow just because two different events are named.
A device succeeds on each of two independent attempts with probability 0.7 per attempt. What is the probability of at least one success?
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The solution, step by step
0.91.
The method in action
- Failure has a probability of 0.3. Two independent failures have a probability of 0.3 × 0.3 = 0.09. The complement is 1 − 0.09 = 0.91.
- 0.49 corresponds to two successes, not at least one. 1.4 adds probabilities without subtracting the overlap and exceeds 1, which is impossible for a probability.
Exercise 4: conditional probability
A condition narrows the group you choose from. Recheck the denominator after phrases such as “given that” or “selected from.”
A club has 20 members. Twelve cycle to meetings, and 5 of those cyclists also run. A member is selected at random from the cyclists. What is the probability that the selected member also runs?
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The solution, step by step
5/12.
The method in action
- The selection is from the 12 cyclists, 5 of whom run. Therefore, the denominator is 12.
- 5/20 would be the probability of choosing a member who both cycles and runs from the entire membership. Here, being a cyclist is already a condition.
The pitfall
The pitfall to avoid
Applying a formula without defining the population or event is a mistake. “The probability is 5/20” may be correct for one question and wrong for another. Write down the starting group, the condition, and what you are looking for.
Three consistency checks
The mean of positive observations must fall between their minimum and maximum. A probability is between zero and one. Changing the data may leave one measure unchanged while changing others. These checks do not replace a solution, but they can quickly eliminate some errors.
During review, restate the reasoning in words. For a draw, say what remains after the first event; for a median, identify the central position; for a condition, name the new group. This helps you identify what to revisit instead of accumulating isolated formulas.
Key takeaways
- The reference group determines the denominator.
- The median and mean respond differently to extreme values.
- Multiply by a conditional probability; independence lets you use the unconditional probability.