Question 252226: Let the random variable X have the density function
f(x)= {kx for 0
{0 elsewhere
If the mode of this distribution is at x =sqrt(2)/4, then what is the median of X?
Answer by Greenfinch(383)   (Show Source):
You can put this solution on YOUR website! For a continuous random variable, the probability density function is
f(x) = { kx for 0 < x < sqrt (2/k)
{ else 0
The cumulative distribution function F(x) for this is therefore the integral from 0 to sqrt (2/k) of kx dx or [(k x^2)/2] from x = 0 to sqrt(2/k).
The graph of the function starts at 0 when x = 0 and has a slope of k. The highest point is therefore at the point sqrt(2/k) which is the mode. The shape is a triangle whose length is sqrt(2/k) and whose height is k x sqrt(2/k). This area is the total probability and is therefore equal to 1. Since the k appears in both parts, this cannot be used to determine k, but knowing that the mode is at sqrt (2)/4 gives sqrt( k) = 4 so k = 16. The pdf now becomes
f(x) = 16x for {0 < x < sqrt (1/8)
{ else
The median occurs when the cdf F(x) = 1/2
so F(x)= 8 x^2 = 1/2
x^2 = 1/16
x= 1/4
This fits in with expectations, The median is 1/sqrt2 = .7071 of the distance along the x axis.
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