As I read this, the intervals are probably meant as continuous, the first one being 60 ≤ x < 70; if so, then 80 is actually the *first value in the class starting with 80, not the last in the class before that. But you may be meaning something different.*

There are 10 values below 80 and 10 values 80 or greater, so I would call the median 80, just by inspection. (With 20 values, I would take the median to be the 10.5th value, that is, the average of the 10th and 11th, not the 10th; since we don’t have access to the individual values, we can’t do that.)

Since 80 is “on the edge” between two classes, it could make sense to take either class as the “median class” in the formula.

If we take 70-80 as the median class, the formula gives 70 + [(20/2 – 4)/6]*10 = 80.

If we take 80-90 as the median class, the formula gives 80 + [(20/2 – 10)/7]*10 = 80.

So it doesn’t seem to make a difference. And if you look at my discussion of the derivation of the formula, you can see why.

]]>class freq

—– —-

60-70 4

70-80 6

80-90 7

90-100 3

Now I know that the median here is in the class 70-80, and I also know that the median would be the 10th value. In this case, would the frequency of the previous class interval still be considered – which is 4? (even though we know that the median value is the last value in the second class interval – 70-80)!

I mean, can I consider the frequency of 6 instead?

Please advise.

Thank you,

Vijay Gupta

This particular problem doesn’t have the same issues as the one in this post; there are no “holes”, and the answer is very simple. If you use the discriminant, you will end up with a quadratic inequality that is not pretty, but has a simple solution. If you find the inverse, no factors will cancel as they did in my work.

But it may be well worth discussing your work, wherever you are having trouble.

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