En anglais

When the bars are different widths

Mathematics · Data & Statistics · ages 15-16

Name ______________________ Date ____________

In a histogram with frequency density on the vertical axis, each bar's height always equals its class frequency.
Circle one: True False
A density histogram has nonempty classes 6 ≤ x < 10 and 10 ≤ x < 17. Why do their bars meet at 10?
- Both classes have the same frequency.
- Their numerical intervals share boundary 10.
- Both classes have the same density height.
- Their class midpoints occupy the same numerical position.
A histogram class has boundaries 12.75 and 17.75. What is its class width?
Answer: ______________
A class holds 21 values and is 7 units wide. What is its frequency density?
Answer: ______________
Class A has width 8 and density 3. Class B has width 4 and density 6. How do their frequencies compare?
- A's frequency is twice B's.
- B's frequency is twice A's.
- B's frequency is four times A's.
- Their frequencies are equal.
A histogram class has frequency 24 and frequency density 3. What is its width?
- 8 units
- 72 units
- 21 units
- 3 units
A class holds 42 values and runs from 6 to 20. What is its frequency density?
Answer: ______________
A bar covering 20 ≤ x < 45 is drawn at a frequency density of 0.8. How many values are in that class?
Answer: ______________
The area method estimates the count for 12 ≤ x < 14 inside the class 10 ≤ x < 20. What assumption does this method use?
- All values lie at the class midpoint.
- Values are spread evenly within the class.
- Every unit interval contains exactly one value.
- The class frequency equals the sum of its boundaries.
A histogram's classes are all 1 unit wide. What happens to the frequency densities?
- They all become 1
- They equal the frequencies, since dividing by 1 changes nothing
- They cannot be worked out
- They become the class widths, because dividing a frequency by 1 leaves the width behind

Answer key

For grown-ups. Fold this page away before handing over the rest.

When the bars are different widths W1-mt_0QAOgzD-Qw-s1

False · Height is frequency / width, so it does not always equal frequency. Width 1 guarantees numerical equality. An empty class also has height 0 and frequency 0 at any positive width.
Their numerical intervals share boundary 10. · The first class ends at 10 and the next begins there, so their numerical widths meet at that boundary. Their frequencies and density heights need not match.
5 · Subtract the lower boundary from the upper boundary: 17.75 - 12.75 = 5.
3 · Frequency density is frequency divided by class width: 21 / 7 = 3. The average frequency is 3 values per unit of width.
Their frequencies are equal. · A contains 8 x 3 = 24 values. B contains 4 x 6 = 24 values. Their frequencies are equal despite their different widths and heights.
8 units · Frequency = density x width, so width = frequency / density = 24 / 3 = 8.
3 · The class width is 20 - 6 = 14, so the density is 42 / 14 = 3 values per unit.
20 · The class is 45 - 20 = 25 wide, so the area is 0.8 x 25 = 20 values.
Values are spread evenly within the class. · The bar is drawn at one height across the whole class, which treats every part of it as equally crowded. Taking a slice of the area only works if that is roughly true.
They equal the frequencies, since dividing by 1 changes nothing · Density = frequency / width. When every width is 1, division leaves each frequency unchanged, so density and frequency have equal numerical values.