Answer:
- x = 12.5 metres
- x = 20 metres
- Weight = 67.8 kg
Explanation:
You want the median for each of the histograms shown.
Median
The median value is the value of the independent variable that divides the area of the histogram into two equal parts. It can be found as the variable value corresponding to half of the cumulative sum of areas.
To find it, we can find half of the total of the area of the histogram. This locates the bar containing the dividing line. Then we find the median as the variable value interpolated between the area before the bar and the area including the bar.
1.
This histogram is completely symmetrical, so the median is the value of x halfway between the limits of the middle bar:
(10 +15)/2 = 12.5
The median is x = 12.5.
2.
If we consider an interval of width 5 to be "1 unit", multiplying the width by the heights of the bars gives the series of areas ...
(0.2, 0, 0.9, 0.9, 0.5, 0.5, 0.8, 0.2}
The total is 4, so half that is 2. The calculator tells us the cumulative total area (working from the left) is 2 at x = 20.
The median is x = 20.
3.
Again, we choose "1 unit" horizontally to be 5 kg. Then the bar heights are the areas of the bars. The cumulative area is 20, so half the area is 10.
The highest bar, between 65 and 70, has a cumulative area of 6.8 at its left side and 12.6 at its right side. Then the cumulative area will be 10 at the point that is (10 -6.8)/(12.6 -6.8) = 3.2/5.8 = 16/29 of the distance between 65 and 70. The value of weight (kg) there is ...
(16/29)(70 -65) +65 ≈ 67.759 . . . . kg
The median weight is about 67.8 kg.
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Additional comment
The calculator image shows a calculation of median weight that is essentially the same, but executed differently.
If a1 and a2 are the cumulative areas on either side of the bar containing the median, and x1 and x2 are the x-values on either side of that same bar, then the median x is calculated from the median area m as ...
x = ((a2 -m)x1 +(m -a1)x2)/(a2 -a1)
We have chosen bars of equal width to form the cumulative sum of areas. That way, we only need to be concerned with the bar height in the interval. You could use actual area values, which requires you multiply height by width for each bar. The above formula still works for that case.
The quartile locations can be found in similar fashion. If you have very many of these to do, a spreadsheet or statistics application may be helpful.
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