Topics › KS4 › Statistics · Higher only
Histograms
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The table shows the times some people waited. Hana is drawing a histogram. What is the frequency density for the class $10 < t \le 30$? Give answers to 2 decimal places where needed. $$\begin{array}{|l|c|}\hline \text{Time } t \text{ (minutes)} & \text{Frequency}\\\hline 0 < t \le 10 & 6\\\hline 10 < t \le 30 & 16\\\hline 30 < t \le 35 & 9\\\hline\end{array}$$
A
Option A:
$0.8$
B
Option B:
$320$
C
Option C:
$1.25$
D
Option D:
$0.53$
Answer and misconceptions
A
Option A:
$0.8$
✓ Correct
B
Option B:
$320$
✗
Multiplies frequency by class widthWorks out $16 \times 20$; frequency density is frequency ÷ class width, $16 \div 20 = 0.8$.
C
Option C:
$1.25$
✗
Divides class width by frequencyWorks out $20 \div 16$; the division is the other way round.
D
Option D:
$0.53$
✗
Divides by the upper class boundary, not the widthWorks out $16 \div 30$; the class width is $30 - 10 = 20$.
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The table gives the frequency densities from a histogram of masses. How many items are in the class $20 < m \le 35$? $$\begin{array}{|l|c|}\hline \text{Mass } m \text{ (g)} & \text{Frequency density}\\\hline 0 < m \le 20 & 1.2\\\hline 20 < m \le 35 & 3\\\hline 35 < m \le 50 & 2\\\hline\end{array}$$
A
Option A:
$3$
B
Option B:
$45$
C
Option C:
$0.2$
D
Option D:
$105$
Answer and misconceptions
A
Option A:
$3$
✗
Reads frequency density as frequencyThe height of the bar is frequency density, not frequency; frequency = $3 \times 15 = 45$.
B
Option B:
$45$
✓ Correct
C
Option C:
$0.2$
✗
Divides frequency density by class widthWorks out $3 \div 15$; frequency = frequency density × class width.
D
Option D:
$105$
✗
Multiplies by the upper class boundary, not the widthWorks out $3 \times 35$; the class width is $35 - 20 = 15$.
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A histogram is drawn for this data. Which class has the tallest bar? $$\begin{array}{|l|c|}\hline \text{Time } t \text{ (seconds)} & \text{Frequency}\\\hline 0 < t \le 10 & 14\\\hline 10 < t \le 15 & 9\\\hline 15 < t \le 30 & 24\\\hline 30 < t \le 60 & 21\\\hline\end{array}$$
A
Option A:
$0 < t \le 10$
B
Option B:
$15 < t \le 30$
C
Option C:
$30 < t \le 60$
D
Option D:
$10 < t \le 15$
Answer and misconceptions
A
Option A:
$0 < t \le 10$
✗
Divides by the upper class boundary, not the widthWorks out $14 \div 10$, $9 \div 15$, $24 \div 30$ and $21 \div 60$; the widths are $10, 5, 15, 30$, giving densities $1.4, 1.8, 1.6, 0.7$.
B
Option B:
$15 < t \le 30$
✗
Thinks the tallest bar has the highest frequencyIn a histogram, bar height is frequency density; $24 \div 15 = 1.6$ is less than $9 \div 5 = 1.8$.
C
Option C:
$30 < t \le 60$
✗
Multiplies frequency by class widthWorks out $21 \times 30 = 630$ as the height; frequency density is frequency ÷ class width.
D
Option D:
$10 < t \le 15$
✓ Correct
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The table gives the frequency densities from a histogram. What is the total frequency? $$\begin{array}{|l|c|}\hline \text{Length } x \text{ (cm)} & \text{Frequency density}\\\hline 0 < x \le 10 & 2\\\hline 10 < x \le 30 & 1.5\\\hline 30 < x \le 40 & 3\\\hline\end{array}$$
A
Option A:
$6.5$
B
Option B:
$185$
C
Option C:
$80$
D
Option D:
$65$
Answer and misconceptions
A
Option A:
$6.5$
✗
Adds the frequency densitiesWorks out $2 + 1.5 + 3$; each density must be multiplied by its class width: $20 + 30 + 30 = 80$.
B
Option B:
$185$
✗
Multiplies by the upper class boundaries, not the widthsWorks out $2 \times 10 + 1.5 \times 30 + 3 \times 40$; the widths are $10, 20, 10$.
C
Option C:
$80$
✓ Correct
D
Option D:
$65$
✗
Assumes all the classes have the same widthUses a width of $10$ for every class; the middle class is $20$ wide.
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The table gives the frequency densities from a histogram. Estimate how many values are in the interval $20 < t \le 25$. $$\begin{array}{|l|c|}\hline \text{Time } t \text{ (minutes)} & \text{Frequency density}\\\hline 0 < t \le 20 & 0.5\\\hline 20 < t \le 40 & 1.6\\\hline 40 < t \le 50 & 2.2\\\hline\end{array}$$
A
Option A:
$1.6$
B
Option B:
$8$
C
Option C:
$40$
D
Option D:
$32$
Answer and misconceptions
A
Option A:
$1.6$
✗
Reads frequency density as frequencyGives the bar height; the frequency is the area, $1.6 \times 5 = 8$.
B
Option B:
$8$
✓ Correct
C
Option C:
$40$
✗
Multiplies by the upper boundary, not the widthWorks out $1.6 \times 25$; the interval $20 < t \le 25$ is $5$ minutes wide.
D
Option D:
$32$
✗
Uses the whole class, not the part asked forWorks out $1.6 \times 20$, the frequency for all of $20 < t \le 40$; only $5$ minutes of that class is wanted.