Open Math›Year 11›Mathematics Advanced›Probability and data›Data and random variables

Data and random variables

Classify random variables as discrete or continuous, construct frequency and cumulative relative frequency tables, and use relative frequency as an estimate of probability.

Worked examples

Classifying a variable and constructing a frequency table

Straightforward

Problem

The scores for 15 students on a maths quiz (out of 5) are listed below.

3,\; 5,\; 4,\; 2,\; 3,\; 5,\; 4,\; 3,\; 5,\; 2,\; 4,\; 3,\; 5,\; 4,\; 3

(a) Is quiz score a discrete or continuous variable? Give a reason.
(b) Construct a frequency table showing the frequency and relative frequency of each score.

Decide whether quiz score is discrete or continuous.

Quiz scores can only be the whole numbers 0, 1, 2, 3, 4 or 5 — they cannot take values such as 3.7. They are counted, not measured. Therefore quiz score is a discrete variable.

Tally the frequency of each score from the list.

\text{Score 2: } 2 \text{ times} \qquad \text{Score 3: } 5 \text{ times} \qquad \text{Score 4: } 4 \text{ times} \qquad \text{Score 5: } 4 \text{ times}

\text{Total: } 2 + 5 + 4 + 4 = 15 \checkmark

Divide each frequency by the total (15) to find the relative frequency.

| Score | Frequency | Relative frequency |
|---|---|---|
| 2 | 2 |

\dfrac{2}{15}

|
| 3 | 5 |

\dfrac{5}{15} = \dfrac{1}{3}

|
| 4 | 4 |

\dfrac{4}{15}

|
| 5 | 4 |

\dfrac{4}{15}

Answer

(a) Discrete, because scores can only take whole-number values and are counted, not measured. (b) See table above: relative frequencies are

\dfrac{2}{15}

\dfrac{1}{3}

\dfrac{4}{15}

and

\dfrac{4}{15}

for scores 2, 3, 4 and 5 respectively.

Reading the median and quartiles from a cumulative relative frequency table

Moderate

Problem

Twenty-five students recorded the number of hours they spent studying in a week. The results are in the frequency table below.

| Hours | Frequency |
|---|---|
| 0 | 2 |
| 1 | 5 |
| 2 | 8 |
| 3 | 6 |
| 4 | 4 |

(a) Complete the relative frequency and cumulative relative frequency columns.
(b) Estimate the median.
(c) Estimate the lower quartile

Q_1

Calculate the relative frequency for each row by dividing by the total of 25.

\frac{2}{25} = 0.08, \quad \frac{5}{25} = 0.20, \quad \frac{8}{25} = 0.32, \quad \frac{6}{25} = 0.24, \quad \frac{4}{25} = 0.16

Build the cumulative relative frequency column by accumulating the relative frequencies.

| Hours | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 0 | 2 | 0.08 | 0.08 |
| 1 | 5 | 0.20 | 0.28 |
| 2 | 8 | 0.32 | 0.60 |
| 3 | 6 | 0.24 | 0.84 |
| 4 | 4 | 0.16 | 1.00 |

Estimate the median: find the first value at which the cumulative relative frequency reaches or exceeds 0.50.

h = 1

: cumulative relative frequency

= 0.28 < 0.50

.

At

h = 2

: cumulative relative frequency

= 0.60 \ge 0.50

.

The median is

2

hours.

Estimate

Q_1

: find the first value at which the cumulative relative frequency reaches or exceeds 0.25.

h = 0

: cumulative relative frequency

= 0.08 < 0.25

.

At

h = 1

: cumulative relative frequency

= 0.28 \ge 0.25

.

So

Q_1 = 1

hour.

Answer

(a) See table in step 2. (b) Median

= 2

hours. (c)

Q_1 = 1

hour.

Using relative frequency as a probability estimate

Challenging

Problem

A fair six-sided die is rolled 90 times. The frequency of each face is shown in the table.

| Face | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 17 | 14 | 16 | 13 | 15 | 15 |

(a) Find the relative frequency of rolling an even number.
(b) The theoretical probability of rolling an even number on a fair die is

\dfrac{1}{2}

. Compare this with the relative frequency from part (a) and explain any discrepancy.

Identify the even-numbered faces and sum their frequencies.

Even faces: 2, 4 and 6.

f(\text{even}) = 14 + 13 + 15 = 42

Divide by the total number of rolls to find the relative frequency.

\text{Relative frequency of even} = \frac{42}{90} = \frac{7}{15} \approx 0.467

Compare the relative frequency to the theoretical probability and explain the difference.

Theoretical probability:

P(\text{even}) = \dfrac{1}{2} = 0.500

.

Relative frequency:

\dfrac{7}{15} \approx 0.467

.

The discrepancy of approximately 0.033 is expected. With only 90 trials, random variation causes the relative frequency to differ from the theoretical value. With a much larger number of rolls, the relative frequency would be expected to converge to

\dfrac{1}{2}

Answer

(a)

\dfrac{7}{15} \approx 0.467

. (b) The relative frequency (0.467) is close to but not equal to the theoretical probability (0.500). This is expected with 90 trials — random variation in a finite sample causes the relative frequency to differ from the theoretical value; larger samples produce closer agreement.

Practise

Q1·Straightforward

Which of the following is a discrete random variable?

The height of a randomly selected studentThe time taken to run 400 mThe number of cars passing through an intersection in one hourThe volume of water in a tank

Explanation

The number of cars is a count and can only take non-negative integer values (0, 1, 2, 3, ...), so it is a discrete random variable. Height, time and volume are measured quantities that can take any value in a range, making them continuous.

Q2·Straightforward

Which of the following is a continuous random variable?

The number of goals scored in a matchThe number of students absent from school on a given dayThe mass (in kg) of a randomly selected orangeThe number of text messages sent in a day

Explanation

Mass is a measured quantity that can take any value within a range (e.g. 0.142 kg, 0.1423 kg), so it is continuous. Goals, absent students and text messages are all counts and can only take non-negative integer values, making them discrete.

Q3·Straightforward

The number of siblings reported by 10 students is listed below.

0,\; 2,\; 1,\; 2,\; 0,\; 3,\; 2,\; 1,\; 0,\; 1

How many students reported having 2 siblings?

Explanation

Going through the list:

0, \mathbf{2}, 1, \mathbf{2}, 0, 3, \mathbf{2}, 1, 0, 1

. The value 2 appears at positions 2, 4 and 7 — a total of 3 times.

Q4·Straightforward

The number of siblings reported by 10 students is listed below.

0,\; 2,\; 1,\; 2,\; 0,\; 3,\; 2,\; 1,\; 0,\; 1

Express the relative frequency of reporting 0 siblings as a fraction in simplest form.

Explanation

The value 0 appears at positions 1, 5 and 9 — a total of 3 times out of 10 students. The relative frequency is

\dfrac{3}{10}

, which is already in simplest form.

Q5·Moderate

The number of books read last month by 25 students is summarised in the table below.

| Books | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 0 | 3 | 0.12 | 0.12 |
| 1 | 7 | 0.28 | 0.40 |
| 2 | 9 | 0.36 | 0.76 |
| 3 | 4 | 0.16 | 0.92 |
| 4 | 2 | 0.08 | 1.00 |

What is the cumulative relative frequency for 2 books or fewer?

Explanation

The cumulative relative frequency for 2 books or fewer is

0.12 + 0.28 + 0.36 = 0.76

. This means 76% of students read 2 or fewer books.

Q6·Moderate

The number of books read last month by 25 students is summarised in the table below.

| Books | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 0 | 3 | 0.12 | 0.12 |
| 1 | 7 | 0.28 | 0.40 |
| 2 | 9 | 0.36 | 0.76 |
| 3 | 4 | 0.16 | 0.92 |
| 4 | 2 | 0.08 | 1.00 |

Using the cumulative relative frequency, estimate the median number of books read. The median is the first value at which the cumulative relative frequency reaches or exceeds 0.50.

Explanation

The cumulative relative frequency is 0.40 at 1 book (below 0.50) and 0.76 at 2 books (at or above 0.50). The median is therefore estimated as 2 books.

Q7·Moderate

Twenty students sat a quiz scored out of 5. Their scores are summarised below.

| Score | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 1 | 2 | 0.10 | 0.10 |
| 2 | 4 | 0.20 | 0.30 |
| 3 | 8 | 0.40 | 0.70 |
| 4 | 5 | 0.25 | 0.95 |
| 5 | 1 | 0.05 | 1.00 |

Estimate the lower quartile

Q_1

. The lower quartile is the first value at which the cumulative relative frequency reaches or exceeds 0.25.

Explanation

The cumulative relative frequency is 0.10 at score 1 (below 0.25) and 0.30 at score 2 (at or above 0.25). So

Q_1 = 2

Q8·Moderate

Q_3

. The upper quartile is the first value at which the cumulative relative frequency reaches or exceeds 0.75.

Explanation

The cumulative relative frequency is 0.70 at score 3 (below 0.75) and 0.95 at score 4 (at or above 0.75). So

Q_3 = 4

Q9·Challenging

A four-sector spinner labelled 1, 2, 3 and 4 is spun 80 times. Sector 3 lands face up 16 times. Using relative frequency as an estimate of probability, find

P(\text{sector 3})

from this experiment. Give your answer as a fraction in simplest form.

Explanation

The relative frequency estimate is

\dfrac{16}{80} = \dfrac{1}{5}

. This is used as an estimate for the probability of landing on sector 3 based on this experiment.

Q10·Challenging

For a fair four-sector spinner, the theoretical probability of landing on any one sector is

\dfrac{1}{4} = 0.25

. In 80 spins, sector 3 appeared 16 times, giving a relative frequency of

\dfrac{1}{5} = 0.20

. Which statement best explains the discrepancy between these two values?

The spinner must be biased, since the relative frequency differs from the theoretical probability.With a larger number of spins, the relative frequency would be expected to be closer to $\dfrac{1}{4}$.Relative frequency always equals the theoretical probability when the spinner is fair.The theoretical probability should be recalculated using the experimental data.

Explanation

A discrepancy between relative frequency and theoretical probability is expected with a small sample. As the number of trials increases, the relative frequency tends to approach the theoretical probability. A single experiment with 80 trials is not sufficient to conclude that the spinner is biased.

Q11·Challenging

A factory tests 150 light bulbs from a production run and finds 9 defective bulbs. Use relative frequency to estimate the probability that a randomly selected bulb from this production run is defective. Give your answer as a fraction in simplest form.

Explanation

The relative frequency estimate is

\dfrac{9}{150} = \dfrac{3}{50}

. This serves as an estimate of the probability that a randomly selected bulb is defective.

Q12·Challenging

A fair die is rolled in two experiments.

- Experiment A: 30 rolls, a 6 appears 3 times. Relative frequency

= \dfrac{3}{30} = 0.100

.
- Experiment B: 300 rolls, a 6 appears 42 times. Relative frequency

= \dfrac{42}{300} = 0.140

.

The theoretical probability of rolling a 6 is

\dfrac{1}{6} \approx 0.167

. Which experiment gives a relative frequency closer to the theoretical probability?

Experiment A, since fewer rolls means the result is more reliable.Experiment B, since the larger sample gives a relative frequency closer to $\dfrac{1}{6}$.Both experiments are equally reliable.Neither experiment is reliable because neither result equals $\dfrac{1}{6}$ exactly.

Explanation

The deviation of Experiment A from the theoretical probability is

|0.100 - 0.167| = 0.067

. The deviation of Experiment B is

|0.140 - 0.167| = 0.027

. Experiment B is closer. A larger sample generally produces a relative frequency closer to the theoretical probability.