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A made-up survey of 200 households records the weekly hours spent watching streamed videos…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A made-up survey of 200 households records the weekly hours spent watching streamed videos. The data are strongly skewed to the right, with mean 40 hours and standard deviation 10 hours. At least how many households must lie between 20 and 60 hours? If the data had been mound-shaped instead, about how many would you expect in that range? Why can you not use the mound figure here?
Need a reminder? The key ideaShow
A z-score counts standard deviations from the mean and has no units. For mound-shaped data about , and lie within 1, 2 and 3 SDs.
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 band. gives 20 to 60 hours: within 2 SDs.
Step 2. Any shape. Chebyshev's rule: at least lie within 2 SDs. households, at least.
Step 3. If mound-shaped. About : households.
Answer: at least 150 households; about 190 if the data were mound-shaped.
Why not 190? The 68-95-99.7 idea is only for mound-shaped data. These data are skewed, with a long tail of heavy viewers. Chebyshev's rule promises less, but it holds for every shape.
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