Question

The random variable B is normally distributed with mean zero and unit variance. Find the probability that the quadratic equation X2 +2BX + 1 = 0 has real roots. Given that the two roots X and X, are real, find, giving your answers to three significant figures: (i) the probability that both X and X, are greater than ; (ii) the expected value of X1 + X2l.

Answer 1

Solution:

Given that:- x2 +28x+1=0 Here, The solution of the quadratic is, X=-3+ B2-1 which has real roots if 1812): ket &(z) be the probability that a standard normally distributed variable 2 satisfies zsz. Then the required answeri's 06-1) +C1-501)) = 2-2001) =>0.3174 forcij: We need the smaller root to be greater than ş: The smaller Toot is -B-182-ī. Now Provided 182-4 is real, ule have -8-182-13 § 4 Bræcorg2-1 CB+'s)'s > B2-1 and CB+') <0 BB B + 25 st&B< >B> -13 * However, if B<- then the condition that v82-1 is real, icelBist, implies the stronger condition B<-I. * the condition that both roots are real and greater than Ş is therefore, - 13 <B<-1 * And the probability that both roots are real so greater than ķis, G1) - (-2.6) = 0 (2.6) - $ C1) → 0.9953 -0.8413 0.1540

* The conditional probability that both roots are greater thants given that they are real is, PC bort roots > § 1 both real) - pcbork roots > and both roots reals pcbolt roots real) 0.1540 0:3174 ->10.485 to doc forcip): * The sum of the roots i5'-9B, so we want the expectation of 1981 given that 18131, which is 5.Coxbeis sem og -re VI TITT VI ETT xze TESTS dxt 2xe dat ܐܫܐ e dx 2x 2T 261-0.00) 3.05