Q6 Let X be a random variable with mean μ and suppose the expected value E[(X – μ)²k] exists for some k ≥ 1.Show for any d > 0, thatP(|X-μ ≥ d) ≤E[(X − µ)²]d2k(HINT: Use an argument similar to that used to prove Chebyshev's inequality.)

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Answered: Q6 Let X be a random variable with mean…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Show that for any random variable E[X2]≥ (E[X])² and When does one have equality?
Let X be a random variable taking positive values and assume that E(X) exists. Which of the following statements are true? • (i) E(X*) > (E(X))*. • (ii) E(1/X²) > 1/E(X²). • (ii) E(e-3X) > e°
(5) If Xis a r.v. with mean u and variance o?, using Chebyshev's-Inequality, AX-2 20) S (a) 0.5 (b) 0.05 (c)0.657 (d) 0.25.
A random sample of size n₁ = 14 is selected from a normal population with a mean of 76 and a standard deviation of 7. A second random sample of size n₂ = 9 is taken from another normal population with mean 71 and standard deviation 11. Let X₁ and X₂ be the two sample means. Find: (a) The probability that X₁ – X₂ exceeds 4. 1 2 (b) The probability that 4.3 ≤ X₁ – X2 ≤ 5.6. Round your answers to two decimal places (e.g. 98.76). (a) i (b) i
8. The time (in minutes) that it takes a car mechanic to replace a car battery is a random variable X that follows an exponential distribution with mean 15 minutes. a) Find P(14 < X < 22). Round off your answer to two decimal places. x 22 1 P(14 < X <22) = √₂/1² 15 e 15dx = P(x < 22) — P(x < 14) x 15 122 = e 15 22 1 - e 14 or 0.8 = = 22 14 = −e 15 + e15 = 0.16254≈ 0.16 - (1 – b) Find the 80th percentile. Round off to the nearest whole number. 1 X S 15 14 e 15 t e 15 dt X 0.8 = 1 e 15 X e 15 = = 0.2 14 22 = e 15 e 15 = 0.16 x — = ln (0.2). This gives x = 24.14156 15
My Find the standard deviation of the discret probability dist x P(x) xp(x)) (x-M)² | (x-mipul 2 σ = √(x-M)² pxx) ટે О 0.21 10*0.21=0 /10-112 =. M= Exp(x) 1 0.30 1*0.3 = 0.3 2 0=49 EXP(X) =M
4. Assume that X is an exponential random variable. Suppose further that Var(X) = 5. E(X). (a) Find the parameter 0> 0. (b) Compute Var(X + 4). (c) Compute P(X > 15[X > 11).

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Q6 Let X be a random variable with mean μ and suppose the expected value E[(X - μ)²] exists for some k > 1. Show for any d > 0, that P(|X-μ ≥ d) ≤ E[(x-μ)2k] d2k (HINT: Use an argument similar to that used to prove Chebyshev's inequality.)