OMPT drill — Probability & statistics

Descriptive statistics

Mean, median, quartiles, standard deviation — and knowing which one a skewed dataset breaks. OMPT-E leans on these harder than most students expect. Write the formula down before plugging in numbers; on this strand the setup is the whole battle.

Tested inOMPT-E

Lesson

Descriptive statistics compresses a dataset into a few honest numbers: a centre (mean or median), a spread (range, interquartile range, or standard deviation), and, when needed, position markers like quartiles. The mean is the balance point: total divided by count. The median is the middle value once the data is sorted. On a tidy symmetric dataset they nearly agree, and the interesting questions start exactly where they do not.

That disagreement is the concept OMPT-E leans on hardest. One extreme value drags the mean toward it, sometimes violently, while the median barely notices: it only cares about the order of the values, not their size. So a single billionaire in the room wrecks the average income and leaves the median income intact. Knowing which summary to trust for skewed data, and being able to say why, is worth as many marks here as any computation.

Problem 1Numeric answer

Five houseplants were measured at heights 4, 7, 9, 10 and 15 cm. What is the mean height?

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Answer: 9

The sum is $4 + 7 + 9 + 10 + 15 = 45$ , and dividing by the 5 observations gives a mean of $9$ cm. As a check, the mean here equals the middle value, though that is coincidence, not a rule.

Problem 2Multiple choice

What is the median of the data set $3, 5, 8, 9, 12, 14$ ?

A $8$
B $8.5$
C $9$
D $8.75$

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Answer: B — $8.5$

With an even number of ordered observations, the median is the mean of the two middle ones: $\frac{8 + 9}{2} = 8.5$ .

Problem 3Numeric answer

Compute the population standard deviation of $2, 4, 6, 8$ . Round to two decimals.

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Answer: 2.24

The mean is $5$ . The squared deviations are $9, 1, 1, 9$ , summing to 20. Dividing by $n = 4$ gives a variance of 5, so the standard deviation is $5 \approx 2.24$ .

Problem 4Multiple choice

Every value in a data set is increased by 5. What happens to the mean and the standard deviation?

ABoth increase by 5
BMean increases by 5; standard deviation is unchanged
CMean is unchanged; standard deviation increases by 5
DBoth are unchanged

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Answer: B — Mean increases by 5; standard deviation is unchanged

Shifting all data by the same amount drags the mean along by exactly that amount. The spread, however, is about distances between values, and those distances are untouched by a shift — so the standard deviation stays the same. Multiplying by a constant is the operation that scales the spread.

Problem 5Spot the error

A student summarises the ordered data set $1, 4, 6, 7, 9, 11, 15$ . Step 1: the median is the fourth value, 7. Step 2: the first quartile is 4 and the third quartile is 11. Step 3: the interquartile range is $15 - 1 = 14$ . Which step contains the error?

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Answer: Step 3

Steps 1 and 2 are right: with seven values the median is the fourth, and the quartiles are the medians of the lower half $(1, 4, 6)$ and upper half $(9, 11, 15)$ . Step 3 computes the range, not the interquartile range. The IQR is $Q_{3} - Q_{1} = 11 - 4 = 7$ , the width of the middle half of the data.

Problem 6Numeric answer

A course grade is built from an exam counting 40% and homework counting 60%. A student scored 6.0 on the exam and 8.5 on homework. What is the final grade?

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Answer: 7.5

This is a weighted mean: $0.4 \cdot 6.0 + 0.6 \cdot 8.5 = 2.4 + 5.1 = 7.5$ . The result sits closer to 8.5 than to 6.0 because homework carries the heavier weight.