Averages (Mean, Median, Mode and Range): Question 6

Worked solution

Setting up: reading a cumulative frequency diagram

There are $n = 80$ students in total, which is the cumulative frequency at the top of the curve. To estimate the median and the quartiles we read across from the correct cumulative-frequency values to the curve, then down to the time axis.

For estimates taken from a cumulative frequency curve we use:

• Median at cumulative frequency $\dfrac{n}{2} = \dfrac{80}{2} = 40$
• Lower quartile ($Q_1$) at cumulative frequency $\dfrac{n}{4} = \dfrac{80}{4} = 20$
• Upper quartile ($Q_3$) at cumulative frequency $\dfrac{3n}{4} = \dfrac{3 \times 80}{4} = 60$

(We use $\tfrac{n}{2}$, $\tfrac{n}{4}$ and $\tfrac{3n}{4}$ here, not $\tfrac{n+1}{2}$ etc., because we are reading an estimate off a continuous curve rather than picking a term from a short ordered list.)

(a) Estimating the median

Read across from cumulative frequency $40$ to the curve, then down to the time axis. This lands between the plotted points $(0.5,\,32)$ and $(0.6,\,52)$.

Interpolating along that segment:

$\text{median} \approx 0.5 + \frac{40 - 32}{52 - 32} \times (0.6 - 0.5) = 0.5 + \frac{8}{20} \times 0.1 = 0.5 + 0.04 = 0.54$

$\boxed{\text{median} \approx 0.54 \text{ seconds}}$

(b) Estimating the interquartile range

Lower quartile. Read across from cumulative frequency $20$; this lands between $(0.4,\,12)$ and $(0.5,\,32)$:

$Q_1 \approx 0.4 + \frac{20 - 12}{32 - 12} \times 0.1 = 0.4 + \frac{8}{20} \times 0.1 = 0.44 \text{ seconds}$

Upper quartile. Read across from cumulative frequency $60$; this lands between $(0.6,\,52)$ and $(0.7,\,72)$:

$Q_3 \approx 0.6 + \frac{60 - 52}{72 - 52} \times 0.1 = 0.6 + \frac{8}{20} \times 0.1 = 0.64 \text{ seconds}$

Interquartile range. The interquartile range is the upper quartile minus the lower quartile:

$\text{IQR} = Q_3 - Q_1 \approx 0.64 - 0.44 = 0.20$

$\boxed{\text{IQR} \approx 0.20 \text{ seconds}}$

Interpretation

The middle half of the students had reaction times spread over a range of about $0.20$ s, centred near the median of $0.54$ s. Because these are read from a curve, small differences (for example, $0.53$ s or $0.55$ s for the median) are perfectly acceptable; the method matters more than the last decimal place. Note that the interquartile range measures only the spread of the central $50$ per cent of the data, so it ignores the very fastest and very slowest students; this makes it less sensitive to extreme values than the full range.