Question

This is Probability and Statistics in Engineering and Science

Please show your work! especially for part B

A Poisson distribution with λ=2 X~Pois(2)

A binomial distribution with n=10 and π=0.45. X~binom(10,0.45)

Question 4. An inequality developed by Russian mathematician Chebyshev gives the minimum percentage of values in ANY sample that can be found within some number (k21) standard deviations from the mean. Let P be the percentage of values within k standard deviations of the mean value. Chebyshev's inequality states that for ANY distribution, P2100 (1-1/). a) For any distribution, what does Chebyshev's inequality say about the percentage of values that are within 2, 3, 5, 10 standard deviations of the mean? b) For each of the distributions below, determine the percentage of observations within 2 and 3 standard deviations of the mean value. Comment on how these percentages compare to the percentage found using Chebyshev's inequality. A standard normal distribution (XN(0,1)) An exponential distribution with A-2 (xexp(2)) A Poisson distribution with N-2 X*Pois(2) A binomial distribution with n-10 and π=0.45. X~binom(10,0.45)

Answer 1

(a) 100 * P(X-μ| < ko) > 100 * (1-1/k2) Hence when k 2 i.e.percentage value that are within 2 standard deviation the mean is at least 100 * (1-1/22)-75% When k-3 i.e. percentage value that are within 3 standard deviation the mean is at least 100 * (1-1/32) 88.89% When k - 5 i.e. percentage value that are within 5 standard deviation the mean is at least 100 * (1-1/52-96% When k 10 i.e. percentage value that are within 10 standard deviation the mean is at least 100 * (1-1/102)-99% 2.e. percentage value that are wntian eviaton te mean

(b) P(-2 < X < 21X ~ Ņ(0, 1)) × 100-24(2)一1) * 100 Using Chebychev's inequality, 100P(-2 < X < 2) > 75% 95.45%. P(-3 < X < 3X ~ Ņ(0, 1)) 100-24(3)-1) * 100-99.73% Using Chebuchev's inequality. 100P(-3 < X < 3) > 88.89%

Suppose X ~ Exp(2) then E(X) = mean = 1 /2 = 0.5 Var(X)= 1/22=0.25: SD(X)= 0.5 then 100 * P(05-2 * 0.5 < X < 0.5 + 2 * 0.5) 100 * P(0 < X < 1.5) 1.5 = 100 × 2 exp (-2z) dz = 100 × (1-exp(-3)) = 95.02% 100 * P(0.5-3 * 0.5 < X < 0.5 + 3 * 0.5) 100 * P(0 < X < 2) 2 exp (-2x) dx = 100 × (1-exp(-4)) 100 × = 981 7% Suppose X ~ Pois(2) then E(X) = mean = 2 Var(X) = 2 , SD= V2 then 100 * P(2 _ 2 * V2 < X < 2+2+V2) 100 * P(0 < X < 4.83) = 94.73% 100 * P(2 _ 3 * V2 < X < 2 3 * V2-100 * P(0 < X < 5.46) = 98.34% Suppose X ~ Binomial(10.0.45) then E(X) = mean = 4.5 Var(X) 10 * 0.45 * (1-0.45) , SD = VVar(X) = 1.57 then 100 * P(45-2 * 1.57 < X < 4.5-2 * 1.57) = 100 * P(2 < X < 7) = 87.30% 100 * P(4.5-3 * 1.57 < X < 4.5-3 * 1.57) = 100 * P(0 < X < 9) 99.97%