number 2 2. Suppose the random variable X has the cdf F(x) = 0, x < −1, = (x+2)/4, −1 ≤ x < 1, = 1, 1 ≤ x. (a) Does thedensity function f(x) of X exist? If yes, find f(x), if not, why not? Compute: (b) P(-1/2 < X ≤1/2 ); (c) P(X ≥ 0);(d) What is the median of X? (e) Is the median of X unique?

From the given information, FX=0, x<-1x+24, -1≤x<11, x≥1 From the above distribution function, The random variable X takes the values -1, 0 and 1. That is, FX=-1+24=14, X=-10+24=12, X=01, X=1a) Density function: For X=-1, PX=-1=14 For X=0, PX=0=FX=0-FX=-1 =12-14 =14 For X=1, PX=1=FX=1-FX=0 =1-12 =12 Therefore, X -1 0 1 f(X) 1/4 1/4 1/2 Here, the f(x) values greater than 0 and sum of f(x) values is equal to 1. Hence, f(x) is valid density function. b) P-12<X≤12=F12-F-12 =12+24--12+24 =58-38 =28 =14 c) PX≥0=PX=0+PX=1 =14+12 =34 d) Median of X: FX=12x+24=12x+22=1x+2=2 x=0 Hence, the median is 0. e) Here, the median of X is 0. That is, there is only one median for the variable X. Hence, the median of X is unique.