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Help me with this....After drawing a histogram....How do u find the class in which the median lies in a histogram....​

User Dany M
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2 Answers

12 votes
12 votes

Answer:

Get all the values of the histogram and put them in order:

For example 2,5,7,1,7,3 becomes 1,2,3,5,7,7 and then you get the middle number, which is in this case (3+5)/2=3.5

Then see which of the readings lie on 3.5.

Step-by-step explanation:

User Dannyyy
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21 votes
21 votes

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Step-by-step explanation:

The histogram will have a count associated with each class. You may have to read the count from the vertical scale of the graph. If it is a histogram of relative frequencies, each class will have a fraction or percentage instead of a count.

Find the total of all counts. Determine what half that number is. If the total is odd, then round up to the nearest integer. This is the index number of the median value.

Add up the counts in the classes starting at one side of the graph, working toward the other side. (Left-to-right is the usual direction for doing this.) When you find the class that increases the total to a value equal or greater than the index number, this is the class that contains the median.

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If the histogram is a relative frequency histogram, the class that brings the total to 50% is the one containing the median.

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There may be cases where the total count is even, and the total of all of the classes up to a point is exactly equal to the index number you're looking for. In that case, the median is between that class and the next. Here's an example:

class 1: count 3

class 2: count 5

class 3: count 4

class 4: count 4

Total count: 3+5+4+4 = 16. Median index = 16/2 = 8

Total of classes 1 and 2 = 3+5 = 8. This means exactly half of the represented values are in class 2 or below, and half are in class 3 or above. If you had access to the actual data values, you would find the median as the average of the 8th and 9th data values (when they are sorted into increasing order). Here, the median might be estimated as the value halfway between the upper bound of class 2 and the lower bound of class 3.

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2nd example

Consider the same histogram as above, but with a count of 3 in class 4. This brings the total count to 15, and the index of the median is 15/2 = 7.5, rounds to 8. Again, the 8th data value (median) is in class 2. There is no ambiguity in this case.

User Rfpdl
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