Cumulative Frequency (GCSE Maths)

The first cumulative frequency is always exactly the same as the first frequency.

To find the missing frequency in the second row, subtract the previous cumulative frequency from the current cumulative frequency.

To find the missing cumulative frequency in the third row, add the current frequency to the previous cumulative frequency.

To find the missing frequency in the final row, subtract the previous cumulative frequency from the final cumulative frequency.

For the first interval $20 < d \leqslant 30$, the cumulative frequency is the same as the frequency.

Add the frequency of the second interval to the previous cumulative frequency.

Add the frequency of the third interval to the running total.

Add the frequency of the final interval to complete the calculation.

Combine all the calculated values into the final table.

To find the cumulative frequencies, add the frequency of each interval to the running total:

• For $0 < s \leqslant 10$, the cumulative frequency is the first frequency: $24$.
• For $10 < s \leqslant 20$, add the next frequency: $24 + 31 = 55$.
• For $20 < s \leqslant 30$, add the next frequency: $55 + 15 = 70$.
• For $30 < s \leqslant 40$, add the final frequency: $70 + 10 = 80$.

The first cumulative frequency is always exactly the same as the first frequency.

Subtract the previous cumulative frequency from the current cumulative frequency:

Add the current frequency to the previous cumulative frequency:

Subtract the previous cumulative frequency from the final cumulative frequency:

Insert the calculated missing values back into the table:

The first cumulative frequency is just the first frequency:

Add the second frequency to the running total:

Add the third frequency to the new total:

Add the fourth frequency to the total (this matches the total number of plants given in the question):

The completed cumulative frequency table is as follows:

Plot the cumulative frequency against the upper bound of each time interval:

• Coordinates: $(20, 16)$, $(30, 40)$, $(50, 59)$, and $(80, 67)$.
• Since the lowest possible time is $0$, the starting coordinate is $(0, 0)$.

Plot these points on the provided grid and join them starting from $(0, 0)$ using a smooth, continuous curve or straight line segments.

For a cumulative frequency graph, you always plot the cumulative frequency against the upper bound (the highest value) of each class interval.

The coordinates to plot are: $(50, 5)$, $(100, 16)$, $(150, 43)$, $(200, 67)$, $(250, 80)$, and $(300, 90)$.

Because an apple cannot have a weight of $0\text{ g}$ or less, the cumulative frequency at $0\text{ g}$ is $0$. Your starting coordinate is $(0, 0)$.

• Carefully plot each of the coordinates on the provided grid.
• Make sure to read the scale correctly (e.g., each small square on the y-axis represents $2$ units, as there are $5$ squares between $0$ and $10$).

Join the plotted points together starting from $(0,0)$ using a smooth, continuous curve (an "S" shaped ogive) or straight line segments.

• Locate $90$ on the horizontal speed axis.
• Draw a vertical line from $90$ straight up to where it intersects your drawn curve.
• From that point of intersection, draw a horizontal line across to the y-axis to read the cumulative frequency, which is approximately $88$.

The reading of $88$ represents the cars travelling up to $90\text{ km/h}$. Subtract this from the total number of cars ($100$):

The total number of people is $100$. The median is at the halfway point:

• Locate $50$ on the vertical $y$-axis (Cumulative frequency).
• Draw a horizontal line from $50$ across to where it intersects your drawn curve.
• From that point of intersection, draw a vertical line straight down to the horizontal $x$-axis (Time) to read the estimated median value.

The total cumulative frequency is $60$. The median is at the halfway point:

• Locate $30$ on the y-axis (cumulative frequency).
• Draw a horizontal line from $30$ across to where it intersects your drawn curve.
• From that point of intersection, draw a vertical line straight down to the x-axis (height) to read the median value.

• Plot points at the upper boundary of each time interval against its cumulative frequency: $(10, 16)$, $(20, 50)$, $(30, 82)$, $(40, 96)$, and $(50, 100)$.
• Start the graph at $(0, 0)$ and join the points with a smooth curve.

Calculate the position for the median:

Locate $50$ on the y-axis, draw a horizontal line to the curve, and a vertical line down to the x-axis. This gives a median of $20\text{ minutes}$.

Calculate the position for the lower quartile:

Locate $25$ on the y-axis, read across to the curve, and down to the x-axis. This gives a lower quartile of $13\text{ minutes}$.

Calculate the position for the upper quartile:

Locate $75$ on the y-axis, read across to the curve, and down to the x-axis. This gives an upper quartile of $28\text{ minutes}$.

The highest point on the cumulative frequency $y$-axis indicates a total frequency ($n$) of 12.

The median position is half of the total frequency:

• Locate 6 on the $y$-axis (Cumulative Frequency) and draw a horizontal line across to the graph.
• From where it intersects the line, draw a vertical line down to the $x$-axis (No. of Books Read).
• The exact reading is $3.2$ books.

The lower quartile position is one-quarter of the total frequency:

• Locate 3 on the $y$-axis, move horizontally to the graph, and then vertically down to the $x$-axis.
• The exact reading is $2.6$ books.

The upper quartile position is three-quarters of the total frequency:

• Locate 9 on the $y$-axis, move horizontally to the graph, and then vertically down to the $x$-axis.
• The exact reading is $4$ books.

Identify the median position, which is half of the total cumulative frequency:

• Locate $20$ on the y-axis (cumulative frequency) and draw a horizontal line across to the curve.
• From the intersection on the curve, drop a vertical line down to the x-axis (height) to read the estimated median value.

• The Upper Quartile (UQ) is at $\frac{3}{4}$ of the total: $40 \times \frac{3}{4} = 30$. Reading from the graph, this is approximately $70\text{ minutes}$.
• The Lower Quartile (LQ) is at $\frac{1}{4}$ of the total: $40 \times \frac{1}{4} = 10$. Reading from the graph, this is approximately $45\text{ minutes}$.

Subtract the Lower Quartile from the Upper Quartile:

• Read the cumulative frequency at $50\text{ minutes}$, which is approximately $15$.
• Read the cumulative frequency at $90\text{ minutes}$, which is approximately $37$.

The number of people in this interval is the difference:

Divide the number of people in the interval by the total number of people:

• Plot points at the upper boundary of each time interval against its cumulative frequency: $(20, 5)$, $(40, 35)$, $(60, 55)$, $(80, 70)$, $(100, 78)$, and $(120, 80)$.
• Include the starting point $(0, 0)$ and join the points with a smooth curve.

• Use the graph to find the number of people taking less than $90\text{ minutes}$ by reading the cumulative frequency at $t = 90$.
• The graph indicates approximately $75$ people.