Question

2. The random variable X, representing the number of errors per 100 lines of software code, has the following probability distribution: 2 f(x) 2 0.11 5 7 8 10 0.27 0.16 0.14 0.32 (c) Suppose g(X) = (3X – 1)2. Find E[9(X)] (a) Find E(X). (b) Find E(X). 3. Use the distribution from Problem 2. (a) Find the variance of X, V(X). (b) Find the standard deviation of X, SD(X). (c) Find V(-3X). (d) Explain why V(X) and V(X – 2) give the same answer. 4. Use the distribution from Problem 2. Find the CDF, F(x). Write your answer as a piecewise function.

Please help Stats

Answer 1

Here the random The number of errors Software Code with variable x representing per 100 lines of a following probabilit x 25 17 Sto 0.11 0.27 0.16 0.14 0.3 (@) E(X) = { xf (w). (by defination cont) (22)P(X= 2) + bu = 5) x P(x = 5) + (2-7) *P(X= 7) + (K= 8) XP(128) + (n=10) P(x=10) 2x o.ll & 5x0.27 + 7 x 0.16 +1 8x014+ 10x 0.3 50.22 & 1.35 + 1.12 + 1.12 + 3. 6.81 ( E(+2) = I x2f (2) (by defination) = 2² 0.11 +52 x 0.27 & 72x0.16 . & 82 x 0.14 & 10² x 0.3 = 0.44 + 6+75+ 7.84 + 8 + 96+ 30 = 53.99

form (0) ELX) = 6-81 (e) suppose g(x)= (3x-1)2 Expanding g(x) = 9x² - 6x +1 so E{g(x)} = E(922) - E (6x) + 1 = 9. E (882) - 6 E (x) + 1 (1) E(X) = 53-99 = 94.53.99 - 6x6-81+1 = 485.91 40.86 + 1. = 446.05 (3 (2) var (x) = E(X2) - Ex) (by defination - 53.99- (-81) - - 53.99-46.3761 = 7.6139 (Appux) SD (*): + Vas(x). = + J7.6139 = 2.75933 (approx) (c) V(-3x) = 32 V (X) = 9x 7.6139 [van (x) = 706132 68.5251 (Approx) At firet V(x) = 7.6139 V(x-2) = V(x) to = 7.6139. is a This are same because because in in x-2: x-.2 ; 2 constant term; whose variance always become . Zero. so v(x) = V(x-2).