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The CDF is the integral of the PDF and in this case is. F(n)=n00.25xdx=0.25×22|n0=0.25n20.25(02)2=0.125n2, where 0n8. Thus, if we wanted to determine the probability of being less than or equal to 2 we can use the CDF function. F(x)=0.125×2F(2)=0.125(22)=0.5.
Basically CDF gives P(X x), where X is a continuous random variable, i.e. it is the area under the curve of the distribution function below the point x. PDF of a continuous random variable gives the value P(X=x) and area at a point (say x) is 0. Suppose, a continuous random variable X follows Normal Distribution.
If X is a continuous random variable and Y=g(X) is a function of X, then Y itself is a random variable. Thus, we should be able to find the CDF and PDF of Y. It is usually more straightforward to start from the CDF and then to find the PDF by taking the derivative of the CDF.
Find the CDF of Y. Find the PDF of Y. Find EY.
0:21 2:48 Suggested clip Sketching a Probability Density Function : Edexcel S2 June 2012 YouTubeStart of suggested client of suggested clip Sketching a Probability Density Function : Edexcel S2 June 2012
Declare a variable u and substitute it into the integral: Differentiate u = 4x + 1 and isolate the x term. This gives you the differential, Du = 4dx. Substitute Du/4 for DX in the integral: Evaluate the integral: Substitute back 4x + 1 for u:
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