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I know that to find median you have to set $$\int_{-\infty}^x f(x)\,dx= 0.5.$$
The median is the point of equal areas on either side. The mean is the point of balance, which is basically the center of mass if the probability density function was solid. Median = $\int_{-\infty}^M f(x) dx = \frac{1}{2}$ or the area equals 1/2 (since the total area is 1)
In order to calculate the median, we should first order the numbers from smallest to highest, as the middle value is the median. In this question we have 100 numbers (an even number), so the position of the median is located at the Y values corresponding to X values 50 and 51 and is found by averaging these two values.
Having trouble on something that should be really, really easy. I need to find the median of the following probability distribution...but according to the website I linked below...I'm doing it incorrectly. Anyways, I computed the following probability table along with its mean and variance.
But in the histogram the hint is confusing me. What does that mean 43 is the median of the frequencies, but it's not the median of the values. For the median of the values, if you sum up all the frequencies below the median, and all the frequencies above the median, you should get the same number. Please help.
So the median is the 23rd number in the sorted list. Example2: There are 66 numbers. 66 plus 1 is 67, then divide by 2 and you get 33.5 33 and a half? That means that the 33rd and 34th numbers in the sorted list are the two middle numbers. So to find the median: add the 33rd and 34th numbers together and divide by 2
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$\begingroup$ @PeterSheldrick: the mean and median are hardly ever equal, except for distributions that are symmetric. But that has nothing to do with what lord12 is trying to do. $\endgroup$ – Robert Israel
Continuous Random Variable - Uniform Median, Exponential Mode 0 What is the difference between a continuous distribution function and a cumulative distribution function?
I don't know, let's find out. Maybe the median is in the $[0,5]$ part. Maybe it is in the other part. To get some insight, let's find the probability that our random variable lands between $0$ and $5$. This is $$\int_0^5(0.04)x\,dx.$$ Integrate. We get $0.5$. What a lucky break! There is nothing more to do. The median is $5$. Well, it wasn't ...