PDF examples solution



S3.1 – Continuous random variablesPresenter example worked solutionSuppose you choose a real number X from the interval [0, 10] or 0≤x≤10 with a Probability Density Function f(x) = Cx(x - 10).Find ? and hence state the function f(x).-∞∞fxdx=1010Cxx-10dx=1C010x2-10xdx=1Cx33-5x2010=1C1033-5×102-0=1C10003-500=1-5003C=1C=-3500fx=-3x2500+3x50Find P(X≤4).PX≤x=axfxdx=04-3x2500+3x50dx=-x3500+3x210004=-1×43500+3×42100-0=44125 or 0.352Find the mode of the distribution.This is the x value corresponding the maximum value of the probability density function.This can be observed to be 5 from a graph of f(x).fx=-3x2500+3x50f'x=-3x250+350Solve f'x=00=-3x250+3503x250=3503x=15x=5Check x=5 is a maximum turning point.f''(x)=-3250f''5=-3250<0x=5 is a maximum turning point and is the mode of the distribution.Note: Ensure any maximum turning point is within the domain of the functions and consider the endpoints that the function is defined.Find the mean or expected value of x.?X=μ= abxfxdx= 010x-3x2500+3x50dx=010-3x3500+3x250dx=-3x42000+x350010=-3×1042000+10350-0=-300002000+100050=-15+20=5Note: This equals the median and mode because the distribution is symmetrical over the domain.Find Var(X) and hence the standard deviation.VarX=EX-μ2=EX2-μ2, ?X2= abx2fxdxVarX=010x2-3x2500+3x50dx-52=010-3x4500+3x350dx-25=-3x52500+3x4200010-25=-3×1052500+3×104200-0-25=-3000002500+30000200-25=5σ=Var(X)=5≈2.236 (3 dp)Find the median of the distribution.Method 1: From the cumulative distribution function (CDF):CDF=axfxdx=0x-3x2500+3x50dx=-x3500+3x21000x=-x3500+3x2100Graph of the cumulative distribution function:Median = 5Method 2: By integrationThe x value where the area under the curve from the lower bound is equal to 0.5.Solve axfxdx=0.50x-3x2500+3x50dx=0.5-x3500+3x21000x=0.5-x3500+3x2100=0.5-x3+15x2=250-x3+15x2-250=0(x-5)(-x2+10x-50)=0x=5Median = 5This can also be observed from the graph of fx=-x3+15x2-250, where fx=0.Find the 10th percentile of the distributionMethod 1: From the cumulative distribution function:14014452632075006324602635885010th percentile is approximately 2.Method 2: By integrationSolve axfxdx=0.10x-3x2500+3x50dx=0.1-x3500+3x21000x=0.1-x3500+3x2100=0.1-x3+15x2=50-x3+15x2-50=0x≈1.958This can also be observed from the graph of fx=-x3+15x2-50, where fx=0. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download