Question

Suppose that the continuous random variable X has pdf given by: x <1 0.16x 15 x 33 f(x)= 0.06 3<x55 [124 x>5 • Find the corresponding cdf for X: You must determine the arbitrary constants. x <1 1<x3 Ex(x)={ 3< x <5 x>5 • Use the cdf to find P(2.4 <x< 10) = • Use the cdf to determine the following percentiles: the 50th percentile (median) the 80th percentile the 90th percentile

please show work and explain for my understanding.

Answer 1

I consider the above given that po xa1 - 0.16* Isx<3 0.06 34X25 6 12x3 x>5 at corresponding cdf ..x<1 Foc (s) = 0 15:33 F(x) = 56.162 doc 3-245 = 0.08[32-17= 0.08x?=0.08 Fla) Food de 30.06[ 20.06[–3] = 0.061-0-18 F (*) = ( 12x3 dx x25

= 12[ =-6[** *) = 6T 7 : the corresponding cdf four x Coy F GOCJ = 0.088?0.08 152? 3 0.063-0.18 3 <355 x>5 - 60 - opus x5 P(2.44<<10): at divided 3 parts by sing range = P(8.4< «< 3) + P(3<x55] + P(x>5) = 0.081227 +001435* 0.0

50.08[9-1) -0.08[5:76-1) + 0.06183)-006(33 +1 #7 - 0/02 - 7 0.644-0.381 +0.12 -0.18 = 0.2 P(2.44<x< 10) = 0.2 3) the soth percentile at range t<x<3 =0.08 (364) -0.08=0.5 0.58 0.08 x=0.5+0.08 0.08 x ² = 7.25 *=2.69 the roth percentile at range of<x<5

0.062 -0.18 = 0.8 X=0.8+0.18 - 0-06 16:33 at the 90th percentile at Range 3cxzs X=0.06% -0.18 = 9 x = 9+0.18 0.oo - 183