Histograms (GCSE Maths)

The frequency density is defined as the ratio of the frequency of a class to its class width. This is used to determine the height of the bars when drawing a histogram.

To draw a histogram, you must first calculate the frequency density for each class interval. The formula is:

For the interval $0 < l \le 8$:

For the interval $8 < l \le 10$:

For the interval $10 < l \le 12$:

For the interval $12 < l \le 16$:

For the interval $16 < l \le 20$:

Plot these values on the provided grid. Label the $x$-axis as "Length ($l$ cm)" and the $y$-axis as "Frequency Density":

• Draw a bar from $x = 0$ to $8$ with a height of $2$.
• Draw a bar from $x = 8$ to $10$ with a height of $3.5$.
• Draw a bar from $x = 10$ to $12$ with a height of $4.5$.
• Draw a bar from $x = 12$ to $16$ with a height of $1.5$.
• Draw a bar from $x = 16$ to $20$ with a height of $0.5$.

Divide the frequency by the class width to find the frequency density (FD):

• $150 < \text{cm} \le 155$: $\text{Width} = 5$, so $\text{FD} = 10 \div 5 = 2$
• $155 < \text{cm} \le 165$: $\text{Width} = 10$, so $\text{FD} = 25 \div 10 = 2.5$
• $165 < \text{cm} \le 170$: $\text{Width} = 5$, so $\text{FD} = 12 \div 5 = 2.4$
• $170 < \text{cm} \le 180$: $\text{Width} = 10$, so $\text{FD} = 33 \div 10 = 3.3$
• $180 < \text{cm} \le 190$: $\text{Width} = 10$, so $\text{FD} = 22 \div 10 = 2.2$
• $190 < \text{cm} \le 200$: $\text{Width} = 10$, so $\text{FD} = 5 \div 10 = 0.5$

Plot the heights on the $x$-axis and the calculated frequency densities on the $y$-axis, drawing bars for each respective class width.

• First, calculate the total "area" of the histogram, which represents the total frequency (the 266 people).
• Let 1 major vertical grid line equal 1 unit of frequency density.

• Next, calculate the area representing the eligible employees (those who commute 15 minutes or less).
• This includes the entire first bar (0 to 10 minutes) and half of the second bar (10 to 15 minutes).

• Calculate the percentage of eligible people relative to the total area.
• Note: using the 266 total people is unnecessary since the ratio of the areas directly gives the percentage.

To draw a histogram, you first need to calculate the frequency density for each class interval. The formula is:

Class Width = $30 - 0 = 30$

Class Width = $40 - 30 = 10$

Class Width = $50 - 40 = 10$

Class Width = $70 - 50 = 20$

Class Width = $100 - 70 = 30$

Plot these values on the provided grid to draw the histogram:

• Draw a bar from $x = 0$ to $30$ with a height of $0.5$.
• Draw a bar from $x = 30$ to $40$ with a height of $2.2$.
• Draw a bar from $x = 40$ to $50$ with a height of $2.8$.
• Draw a bar from $x = 50$ to $70$ with a height of $1.5$.
• Draw a bar from $x = 70$ to $100$ with a height of $0.3$.

Use the formula $\text{Frequency Density} = \dfrac{\text{Frequency}}{\text{Class Width}}$ for each interval:

• For $0 - 100$: $\text{FD} = \dfrac{20}{100} = 0.2$
• For $100 - 200$: $\text{FD} = \dfrac{60}{100} = 0.6$
• For $200 - 250$: $\text{FD} = \dfrac{25}{50} = 0.5$
• For $400 - 450$: $\text{FD} = \dfrac{35}{50} = 0.7$
• For $450 - 500$: $\text{FD} = \dfrac{5}{50} = 0.1$

The histogram shows the bar for 250 < £ ≤ 400

has a frequency density of $0.3$. Multiply by the class width ($150$) to get the frequency:

Draw the remaining bars on the histogram using the frequency densities calculated in Step 1.

For each bar on the histogram, identify its width and height (frequency density). The area of the bar gives the frequency:

• For $0 < t \le 5$: width is $5$ and $\text{FD} = 1$. Frequency is $5 \times 1 = 5$.
• For $5 < t \le 10$: width is $5$ and $\text{FD} = 1.8$. Frequency is $5 \times 1.8 = 9$.
• For $10 < t \le 15$: width is $5$ and $\text{FD} = 2$. Frequency is $5 \times 2 = 10$.
• For $15 < t \le 20$: width is $5$ and $\text{FD} = 3$. Frequency is $5 \times 3 = 15$.
• For $20 < t \le 30$: width is $10$ and $\text{FD} = 0.8$. Frequency is $10 \times 0.8 = 8$.
• For $30 < t \le 45$: width is $15$ and $\text{FD} = 0.6$. Frequency is $15 \times 0.6 = 9$.

Use the calculated values to draw the histogram bars on the provided grid:

• Draw the first bar from $x = 0$ to $5$ with a height of $1.2$.
• Draw the second bar from $x = 5$ to $10$ with a height of $2.6$.
• Draw the third bar from $x = 10$ to $20$ with a height of $1.4$.
• Draw the fourth bar from $x = 20$ to $30$ with a height of $1.5$.
• Draw the fifth bar from $x = 30$ to $50$ with a height of $0.6$.

The modal class is the tallest bar, which corresponds to the highest frequency density.

• A score of $75\%$ or more is exactly halfway through the $70-80$ interval.
• Assuming an even spread, half of the $12$ students in this interval is $6$.
• Add the $8$ students from the $80-100$ interval to find the total:

• The calculation is an estimate because the data is grouped.
• We do not know the exact distribution of students within the $70-80$ interval, so we must assume they are spread evenly.

The total frequency is $37$. The median position is:

• Using cumulative frequencies ($3, 8, 20, \dots$), the $19\text{th}$ value falls in the $60 < m \le 75$ interval.
• It is $19 - 8 = 11$ places into this interval.

Divide the interval width by the frequency:

Multiply the position by this step size:

Add this to the lower bound of the interval: