Construct and interpret frequency histograms and polygons, and cumulative frequency graphs (ogives), including using an ogive to estimate the median and quartiles of a data set

Plot cumulative frequency (vertical) against the upper class boundary (horizontal): the points $(40, 2)$, $(50, 6)$, $(60, 14)$, $(70, 26)$, $(80, 34)$, $(90, 40)$, joined by a smooth rising curve, starting from $(30, 0)$.

Median: read across from a cumulative frequency of $\tfrac{n}{2} = 20$, then down to the axis, giving a median of about $65$ marks.

Quartiles: read across from $\tfrac{n}{4} = 10$ for $Q_1 \approx 55$ and from $\tfrac{3n}{4} = 30$ for $Q_3 \approx 75$, so the interquartile range is about $75 - 55 = 20$ marks.

Markers reward correct plotting (cumulative frequency against the upper boundary), reading across from $20$, $10$ and $30$, and an answer within a small reading tolerance of the true values. Reading from a frequency (not cumulative frequency) loses marks.

Part (a): the polygon joins the top-of-bar midpoints, at (class centre, frequency): $(10, 4)$, $(20, 11)$, $(30, 9)$, $(40, 6)$.

Part (b): the polygon is closed by joining each end down to the horizontal axis at the next class centre beyond the data with frequency $0$, that is to $(0, 0)$ on the left and $(50, 0)$ on the right, so the polygon meets the axis.

Markers reward the four (centre, frequency) coordinates and the idea that the ends are taken down to zero at the adjacent empty class centres. Plotting at the class boundaries instead of the class centres is the common error.

Part (a) - the gaps in a frequency histogram. A frequency histogram displays grouped numerical data, so the bars are drawn touching with no gaps between them. The gap is therefore $0$.

Part (b) - the tallest bar. Bar height equals the class frequency, so the tallest bar is the one with the largest frequency:

so the tallest bar has height $7$. (Check: the bars sit at the class centres $5$, $10$, $15$, $20$ and are each $5$ wide, so they meet edge to edge, confirming the zero gap.)

Add a running total down the table

The cumulative frequency of a class is the number of values at or below the upper end of that class, so you keep a running sum:

First class

$4$.

Up to $20$

$4 + 9 = 13$.

Up to $30$

$13 + 7 = 20$.

Up to $40$

$20 + 5 = 25$.

So the cumulative frequencies are $4$, $13$, $20$, $25$. (Check: the final cumulative frequency must equal the total number of values, and $4 + 9 + 7 + 5 = 25$, which matches the last running total.)

Part (a) - the median position. The median sits at the halfway point of the data, so you read across from a cumulative frequency of

then drop down to the horizontal axis to read the median value.

Part (b) - the lower quartile position. The lower quartile is one quarter of the way up, so read across from

then drop to the axis for $Q_1$. (Check: $Q_3$ would use $\tfrac{3n}{4} = 45$, which sits above the median position $30$, as it should.)

Part (a) - cumulative frequencies. Keep a running total down the classes:

so the cumulative frequencies are $6$, $20$, $32$, $40$.

Part (b) - the read-off heights. With $n = 40$ values:

• median at $\dfrac{n}{2} = \dfrac{40}{2} = 20$,
• lower quartile at $\dfrac{n}{4} = \dfrac{40}{4} = 10$,
• upper quartile at $\dfrac{3n}{4} = \dfrac{3 \times 40}{4} = 30$.

So you read across the ogive from cumulative frequencies $20$, $10$ and $30$ respectively, then drop to the time axis. (Check: the final cumulative frequency $40$ equals the number of commuters, confirming the running total is complete.)

Part (a) - the end bars. Each bar is centred on its class centre, so the leftmost bar is centred at a mark of $45$ and the rightmost bar at a mark of $85$.

Part (b) - the polygon vertices on the bars. A frequency polygon joins the midpoint of the top of each bar, and the midpoint of a bar's top sits at the class centre (horizontal) and the frequency (vertical). So the vertices are

reading (class centre, frequency) off each bar in turn. (To close the polygon onto the axis you would also add a point at the next empty class centre on each side, $(35, 0)$ and $(95, 0)$, both at frequency $0$.)

Part (a) - cumulative frequencies. Run a total down the upper class boundaries ($40, 50, 60, 70, 80, 90$):

so the cumulative frequencies are $2$, $6$, $14$, $26$, $34$, $40$. (The last value equals $n = 40$, the number of students, as a check.)

Part (b) - the median. With $n = 40$, read across from a cumulative frequency of $\tfrac{n}{2} = 20$. On the ogive this lands inside the $60$ to $70$ class, where the cumulative frequency climbs from $14$ (at $60$) to $26$ (at $70$). Reading down to the mark axis gives a median of about

so the median is approximately $65$ marks.

Part (c) - the quartiles. The lower quartile is at $\tfrac{n}{4} = 10$, which sits in the $50$ to $60$ class (cumulative frequency $6$ rising to $14$):

The upper quartile is at $\tfrac{3n}{4} = 30$, in the $70$ to $80$ class (cumulative frequency $26$ rising to $34$):

so $Q_1 \approx 55$ and $Q_3 \approx 75$ marks.

Part (d) - the interquartile range. The IQR is the spread of the middle half:

so the interquartile range is about $20$ marks. (Check: $Q_1 < \text{median} < Q_3$, that is $55 < 65 < 75$, exactly as the order of the three quartiles requires.)

Part (a) - interquartile range. The IQR is the upper quartile minus the lower quartile:

Part (b) - days at or below $7$ mm. The lower quartile $Q_1 = 7$ mm is the value with one quarter of the data at or below it, so

had $7$ mm or less (read off the ogive, an estimate, so rounding to a whole day is expected).

Part (c) - days above $19$ mm. The upper quartile $Q_3 = 19$ mm has three quarters of the data at or below it, so one quarter lies above it:

had more than $19$ mm.

Part (d) - days between the quartiles. Between $Q_1$ and $Q_3$ lies the middle half of the data:

had rainfall between $7$ mm and $19$ mm. (Check: the three regions $13 + 25 + 13 = 51$, which rounds back to $50$ within the rounding of half-days, confirming the split is consistent.)