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A museum records the ages of 60 visitors on one made-up morning, in years…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A museum records the ages of 60 visitors on one made-up morning, in years. The groups are 0 to 10 (9 visitors), 10 to 20 (12), 20 to 40 (16), 40 to 60 (14) and 60 to 90 (9). The bins have different widths. Find the frequency density of the bin 20 to 40 and of the bin 60 to 90. A first drawing used the frequencies as heights. Which bin looked tallest in it, and which bin is really the most crowded per year of age?
Need a reminder? The key ideaShow
A histogram counts numerical values into bins of equal width, with touching bars. An edge value goes into the bin above. With unequal bins the height is the frequency density:
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The widths. The five widths are 10, 10, 20, 20 and 30 years.
Step 2. The two densities asked for. For 20 to 40: visitors per year. For 60 to 90: visitors per year.
Step 3. All densities. The others are , and visitors per year.
Step 4. Compare. With frequency as height, the bin 20 to 40 looked tallest, with 16 visitors. By density, the bin 10 to 20 is the most crowded, with 1.2 visitors per year of age.
Answer: and visitors per year. The frequency picture made 20 to 40 look tallest; 10 to 20 is really the most crowded.
Watch out. The wide bin 60 to 90 holds 9 visitors, as many as 0 to 10, but it spreads them over three times as many years.
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