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I have been struggling with this topic for ages, and my exam is on Wednesday. I understand how to construct and read a histogram, but I cannot wrap my head around questions where you are asked to find the number of objects within a range that differs from the class widths in the question.

For example, the table might be as follows:

$$ \begin{array}{c|r} Class & \text{Frequency} \\ \hline 4 - 6 & 6 \\ 7 - 8 & 12 \\ 9 & 21 \\ 10 - 12 & 45 \\ 13 - 15 & 9 \\ 16 - 20 & 5 \\ \end{array} $$

The question given is to estimate the number of values between 8.5 and 13.5. I'm not looking for a solution to the problem, I just need to understand how to answer it. I've tried looking for an explanation, but what I came across only explained how to draw the histogram.

Gladiator Kittens
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1 Answers1

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It seems the data must be integers. (Otherwise, bins would be labeled something like $[4, 8)$, $[7, 9),$ and so on. So between 8.5 and 13.5, the integers are 9, 10, 11, 12, and 13. The total frequency must be $21 + 45.$

A trickier problem would be to estimate the number of values between 11 and 14 (inclusive), that is $[11, 14].$ It would be impossible to say for sure, but the answer may be something like $(2/3)45 + (2/3)9,$ assuming that interval $[10, 12]$ has three equally frequent values, and similarly for $[13,15].$

Notes: (a) I am using standard interval notation, in which square brackets include the endpoint and parentheses don't. (b) It is common practice for histogram intervals to be of equal width. In your histogram, I wonder why there is a separate interval that includes only 9. Intervals of unequal width are sometimes used, but the purpose of a histogram is to convey information clearly, and people are often confused when intervals are of unequal width.

BruceET
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