Question

The continuous random variable x takes values in the interval 0 SXS 3. For 0 sxs 3 the graph of pdf consists of two straight line segment meeting at the point (1,k). And the random variable Y is given by Y = X? Find the median value of Y in 3 s.f.)

Answer 1

fa) 27/4/20 CEFTA 11 CEFI FIXY TAS CE The continuous randon variable takes values in the internal os x L3 as shown in the graph above For os x 4 3 the graph of the pdf consists of a straight dine segments meeting at the point (1,k) Fist we find the aquation of the line going from (1,k) to (3,0). The equation of this line is given by (4-8) = 6 - 2 1 - x) [where end ) = (K) 1 (2242) = (3,0) ] (4 - 8- 9-K (-). & yak= (x-1) a 2(4-k) = k(1-3) on 24-2k=k-ok or 24 = 2k+k-ak y = { (3k ->ck) Hence the pdf of Xis of the 4= K (3-2)..... ® form f (x) = sk o sasi From graph and = (3-2) 15853 I using IS253 Mes jeucada = 1 > Judz +33 (0.33d2 = 4 * k[+]' + [3 ] ** + [-4) -(3-4) =

FIXIN TARA CEFT ktk x7 21 ) ak=1 onk=t fx(x) = { osasi 1 Ź (3-2) 15x23 Nour y=x². We need to find distribution of Y & then its median 1. Os x63 hence y takes values from o to (3²= 9 Tie osyça First we find the adf of x Fx (x) = P(xsx) = 1 do ie Ex(x) when osasi (3-udu when 15x43 Ex(x) = 2 when osast. ( + † [3x-2 when Isasa {a Šta (3-0) du = ☆ [30-- - + [3*-*)-(3-2)] = + [3x-21 - ] } Nous Fuly) = cdf. of Y = Plysy) : - P(xºsy) = P(-145 x 558) = P(x sg) - P(x=57) but osysa - Fx (17) - F (55) hence yao alausy Fy(y) = Ex lif) [:Ex(+7) 20 kosysat • Fuly) = when osysi -l2 + [314 - ] 4459

. Fyly) = ( 4 when osys! 2 + [354 - I f] when Isusa Nou let Essen be the median of Y then we must have P(YLEx) = ± Now we see that P(Y51) 2 1 1 2 = 1 . Elle = 1, hence median of Y is 1 LUODO