11.3 Diagrams for single-variable data and the histogram — Practise

Work through these on your own. For every histogram question, use the area, not the height. Check that the frequencies add to the total.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A class of width 5 contains 12 values. Find the frequency density.
A histogram bar has frequency density 3 for a class of width 8. Find the frequency in the class.
A class runs from 20 to 35. Find its class width.
A histogram has three classes: $0$ – $10$ with frequency density 2, $10$ – $20$ with frequency density 5, and $20$ – $40$ with frequency density 1. Find the total frequency.
A histogram has frequency density 4 for the class $5$ – $15$ . Find the number of values between $5$ and $10$ .
A grouped frequency table has classes with frequencies 12, 20 and 18. Find the total frequency.
A grouped table has classes $0$ – $10$ with frequency 12, $10$ – $20$ with frequency 20, and $20$ – $30$ with frequency 18. Using midpoints $5$ , $15$ and $25$ , estimate the mean.
A class of width 10 contains 25 values. Find its frequency density.
A histogram bar has frequency density 1.4 for the class $0$ – $20$ . Find the frequency in the class.
One class $0$ – $10$ contains 30 values, and another class $10$ – $30$ contains 40 values. Find the frequency density of each class.
A histogram of 200 values has frequency density 6 in the class $10$ – $20$ . Estimate the probability that a value lies in that class.
A histogram has frequency density 6 in the class $0$ – $5$ and frequency density 4 in the class $5$ – $15$ . Find the number of values in the class $5$ – $15$ . State whether you used the height or the area.

Answers and concise solutions

1. The frequency density

Answer: Divide the frequency by the class width: $\dfrac{12}{5} = 2.4$ .

2. The frequency

Answer: Frequency is frequency density multiplied by class width: $3 \times 8 = 24$ values.

3. The class width

Answer: Subtract the lower boundary from the upper boundary: $35 - 20 = 15$ .

4. The total frequency

Answer: Each frequency is density times width: $20 + 50 + 20 = 90$ values. Adding the areas of the bars gives the total frequency.

5. The number of values

Answer: The part from $5$ to $10$ has width $5$ , so the frequency is $4 \times 5 = 20$ . The area over part of a bar gives the frequency in that part.

6. The total frequency

Answer: Add the frequencies: $12 + 20 + 18 = 50$ values.

7. The estimated mean

Answer: Multiply each midpoint by its frequency: $60 + 300 + 450 = 810$ . Divide by the total frequency: $\dfrac{810}{50} = 16.2$ . This is an estimate, because the midpoints stand for the values in each class.

8. The frequency density

Answer: The frequency density is $\dfrac{25}{10} = 2.5$ . The bar is taller than a class of width 10 with 20 values, whose density is 2.

9. The frequency

Answer: The frequency is $1.4 \times 20 = 28$ values.

10. The two frequency densities

Answer: The first density is $\dfrac{30}{10} = 3$ and the second is $\dfrac{40}{20} = 2$ . The second class holds more values, but its bar is shorter, because the class is three times as wide as the first.

11. The estimated probability

Answer: The frequency is $6 \times 10 = 60$ , so the probability is $\dfrac{60}{200} = \dfrac{3}{10} = 0.3$ . This is an estimate from the sample.

12. The number of values

Answer: Use the area: $4 \times 10 = 40$ values. The height alone is 4, which is a frequency density, not a frequency.