EE 193 - Department of Electrical & Computer Engineering - Tufts ...

• 3.5.5 We have T ∼ N(µ, 15). From the problem statement we know[ t − µP [T > 10] = 0.5 → P15 > 10 − µ ] ( ) 10 − µ= 1 − Φ = 1 1515 2which means Φ((10 − µ)/15) = 1/2. Since Φ −1 (0.5) = 0 we conclude that µ = 10. Hence[ ]T − 10 32 − 10P [T > 32] = P >15 15= 1 − Φ(22/15) = 1 − Φ(1.47) ≈ 1 − 0.9292 = .0708[ T − 10P [T < 0] = P < 0 − 10 ]15 15= Φ(−0.66) = 1 − Φ(0.66) ≈ 1 − 0.7454 = 0.2546[ ]T − 10 60 − 10P [T > 60] = P 84] =23, 000100, 000, 000 .[ ] [ ]X − 70 84 − 7014> = 1 − Φ =σ X σ X σ X23100, 000 .σ X =14) ≈ 4.00 using norminv in MatlabΦ(1 −1 − 23100,000(b) We have[ X − 70P [X > 96] = P >4]96 − 70= 1 − Φ (6.5)4(c) The probability that any individual is over 90 inches is( ) 90 − 70P [X > 90] = 1 − Φ= 1 − Φ(5) ≈ 2.87e − 7 ≡ p.4Now, for this problem, there are 100,000,000 men in the U.S. If each man is a Bernoullitrial with p = 2.87e − 7, then the probability that there are no men over 90 inches inheight is (1 − p) 100,000,000 ≈ 3.56e − 13. This is a very small number indeed.(d) Here we are asking for the expected number of “successes” in 100,000,000 trials of abinomial RV with p = 2.87e − 7. This is just 100, 000, 000 × p ≈ 28.7 or roughly 30 men.2