Question

3(a). f (x)=ce^(-0.1x) , if 0≤x≤10 and 0, otherwise Based on your computations below, explain and discuss the various probabilities with respect to the given exponential function. (i)deduce the constant c, AN[5]

(ii)compute the probabilities (α) P(x≥5) AP[4] (β) P(x<4) AP[4] (γ) P(3≤x≤11) AP[4]

(b) A random variable, X, is defined as ‘the absolute difference and sum of Numbers appearing when two fair dice are tossed’. Construct the Probability distribution of the random variable, X for (i)sum AP[4]

(ii)difference AP[4] of the above random variable. Outline a clear cut analysis of these two components with respect to the given random variable.

Answer 1

Solution: 3(Q) The paf of x as a f(x) = sc =Sce/1005x510 n. Otherwise Here f(x) is a pag than 10 S f(x) da El 10 0 > *10 dx=1 C 10 se 107 > C10(1-é) = 1 L xx10 е -1/10 c = Tole) 3 i: Ci e loces

410 Hence the pdf of a as, f(x)= i Otherwise е osx<10 e lole-1) 0 The caf of a as F(x) = I foco da X 0 10 da e e X YO 10 (6-1) = Toledo 40 -1/10 e 2/10) (i- e- F(x) Sell- ē 10) e-/ i o<x<10 i Olturwise 0

10 la p(x75) - |- P( 1<s) = 1- P(S) El- ell-ě 8/10) (P-1) El- ell- e/2) (e-1) = 0.3775 Hencer P(x>52 = 0.3775

P(x4) - f(u) ell-410) こ e-l ell-e 45) 11 - 0.5216 Hence P(x<4) = 0:5216

P( 35x311) = P(3<x=10) = P(x+10) - P(x+3) Flo) - A) ell- -3110) T-1) = 0.5899 pencer PC 3 = x311) = 0.5899

Solulidu: 3(6) 0 Kef a random variable x deuotes the sum of members then ? The probability distribuhon x as of i x=8 x=3 ' x=4 x=5 x=6 p(X=X) = X-7 (1/36 436 3/36 4/36 5136 6/36 5/36 4/36 3/36 2136 436 x=8 x=9 x=10 x=1l x=12

71 Let a random variable a aendes et absolule difference X Then Probability distribution of x as ) Ć X=0 36 x=1 p(x-2) = 10 36 8 36 x=2 6 36 X-3 x= 4 36 X-5 2 36

Iteve la probabili lq aistribulas of xas yo i X=0 X=1 5118 49 a=2 p(x=2= x-3 16 x=4 49 x=5 Yla