Statistics & Probability

Find the mean of the data set 4, 8, 10, 14, 14.

• A12
• B9
• C14
• D10✓

Answer: The mean is (4 + 8 + 10 + 14 + 14)/5 = 50/5 = 10.

A fair six-sided die is rolled once. Find the probability of obtaining a prime number.

• A1/6
• B2/3
• C1/2✓
• D1/3

Answer: The primes on a die are 2, 3 and 5, giving 3 favourable outcomes out of 6, so the probability is 3/6 = 1/2.

The random variable X follows a binomial distribution B(10, 0.3). Find E(X), the expected value.

• A7
• B3✓
• C3.3
• D0.3

Answer: For X ~ B(n, p), E(X) = np = 10 × 0.3 = 3.

The heights of a population are normally distributed with mean 170 cm. What proportion of the population is taller than 170 cm?

• A0.34
• B1
• C0.68
• D0.5✓

Answer: The normal distribution is symmetric about its mean, so exactly half (0.5) of the population lies above the mean of 170 cm.

A school has 1200 students split into Year 12 (700 students) and Year 13 (500 students). A researcher wants a sample of 60 students that reflects this split, taking 35 from Year 12 and 25 from Year 13, choosing randomly within each year group. Which sampling method is this?

• ASystematic sampling
• BStratified sampling✓
• CQuota sampling
• DOpportunity sampling

Answer: The population is divided into groups (strata) and a number is taken from each in proportion to its size (700:500 = 35:25), with random selection within each group. That is stratified sampling. Quota sampling would not use random selection within the groups.

Find the population variance of the data set 5, 7, 9, 11, 13.

• A8✓
• B10
• C2√2
• D40

Answer: The mean is (5+7+9+11+13)/5 = 9. The squared deviations are 16, 4, 0, 4, 16, summing to 40. The variance is 40/5 = 8. (Option C is the standard deviation √8, not the variance; option D forgets to divide by n.)

The ordered data set is 4, 6, 7, 9, 11, 12, 14, 15, 18, 21, 25 (11 values). Using the positions (n+1)/4 and 3(n+1)/4 for the quartiles, find the interquartile range.

• A9
• B12
• C6
• D11✓

Answer: With n = 11, the lower quartile is the 3rd value Q1 = 7 and the upper quartile is the 9th value Q3 = 18. The interquartile range is Q3 − Q1 = 18 − 7 = 11.

For a data set the lower quartile is 15 and the upper quartile is 27. Using the 1.5 × IQR rule, which of the following values would be classed as an outlier?

• A44
• B10
• C46✓
• D5

Answer: The IQR is 27 − 15 = 12, so 1.5 × IQR = 18. The fences are Q1 − 18 = −3 (lower) and Q3 + 18 = 45 (upper). A value is an outlier if it is below −3 or above 45. Only 46 lies outside this range; 44, 10 and 5 all lie within the fences.

On any day the probability it rains is 0.3. If it rains, the probability a train is late is 0.6; if it does not rain, the probability the train is late is 0.1. Given that the train was late, find the probability that it had rained.

• A0.18
• B0.60
• C0.25
• D0.72✓

Answer: P(late) = P(rain)·P(late|rain) + P(no rain)·P(late|no rain) = 0.3×0.6 + 0.7×0.1 = 0.18 + 0.07 = 0.25. By conditional probability, P(rain|late) = P(rain and late)/P(late) = 0.18/0.25 = 0.72.

In a group of 30 students, 18 study French, 15 study German and 7 study both subjects. How many students study neither French nor German?

• A4✓
• B0
• C11
• D7

Answer: By the addition rule, the number studying at least one subject is 18 + 15 − 7 = 26. So the number studying neither is 30 − 26 = 4. (Subtracting the overlap of 7 once avoids double-counting.)

A scatter diagram of two variables gives a product-moment correlation coefficient of r = −0.92. Which statement best interprets this value?

• AThere is a strong positive linear correlation
• BThere is a strong negative linear correlation✓
• CThere is no linear correlation
• DOne variable causes the other to decrease

Answer: The PMCC lies between −1 and +1. A value of −0.92 is close to −1, indicating a strong negative linear correlation: as one variable increases the other tends to decrease. Correlation does not by itself establish causation, so option D is not justified.

Two events A and B are independent with P(A) = 0.6 and P(B) = 0.3. Find P(A ∪ B).

• A0.90
• B0.18
• C0.72✓
• D0.78

Answer: For independent events P(A ∩ B) = P(A)·P(B) = 0.6 × 0.3 = 0.18. Then P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0.18 = 0.72. (Option A wrongly adds without subtracting the intersection.)