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- 11. Assume that X is a uniform random variable on the interval [-22, 14. (a) tion of its mean. In other words, compute Find the probability that X is within two standard devia- P(µ – 20 < X < µ +2o). -Give the formula for p(x) for a binomial random variable with n= 6 and p = 0.4. -.. - O'O (x=0, 1, 2 ...D p(x) = (Simplify your answers. Use integers or decimals for any numbers in the equation.)Let the following simple random sample X1, X2, · · , X11 following: 1. Binomial pmf. (11, ¾); 2. Uniform pmf; 3. Uniform pdf (0, a); 4. Exponential pdf with (µ). Find the corresponding pmf/pdf of Y1 , Y4, Y7 and F(Y;) where Yi < Y½ < .. < Y6 < ..< Yı1.
- [8.3]Find P(SN = 4) where N is a Binomial random variable with parameter r = 6 (number of times the experiment is performed) and p = .6.da1.25 31 Ju Suppose that the number of accidents occuring on a highway each day is a Poisson random variable with parameter A =3. Then the probability that at least 1 accident occurs today is P({X21})=1-e3o P({X21})=e3 P({Xs1})=1-e3 P({Xs1})=e¯3 .Microso | a Google Chrome 1-EV ENG 4) O
- Suppose the CDF of a random variable x is given by x2)?Q 6.1. Suppose Z = (Z₁, Z2, Z3) is a standard multi-variate Gaussian random variable i.e., for i ≤ 3, Zi~ N(0, 1) are i.i.d. random variables. Each of the random variables, (a)–(d), on the left is equal in distribution to exactly one random variables, (1)–(4), on the right. Pair up according to "equal in distribution" and explain briefly your reasoning. X₁ (a) X₂ (b) (x₂) - (2) (c) (x²) - (¹/1² 1/√²) (2₁) (3/√2 = √2 = (V2) Z₁ (d) (x) - (1) (²) X2 Y₁ (9)-(-) (2) (1=1) Y₂ (1) Y₁ (²) (4) - (1/² (3) (1/√√2 1/√2-1/√2) (x₁) = (²₁ ² ✓/³²) (3) 2 2 √2 1 Z3 Y₁ → ()-()) (4) = Y₂ 1/2) (2) 1) (2)Q 6.1. Suppose Z = (Z1, Z2, Z3) is a standard multi-variate Gaussian random variable i.e., for i ≤ 3, Zi~ N(0, 1) are i.i.d. random variables. Each of the random variables, (a)-(d), on the left is equal in distribution to exactly one random variables, (1)-(4), on the right. Pair up according to "equal in distribution" and explain briefly your reasoning. (a) (X₂) (¹) (X) = (2) 1/√2 (Ⓒ) (x₂) - (3/1² ¹1/1²) (2) (c = 2 0 = 2 Z₁ 1) (²) 3 (4¹) (X) = (²¹) (1) (2) (3) Y₁ 1 (29) - ()) (2) = Y₂ 1 Y₁ (121) = (1/² √₂) (2) (1/√2 1/√2 /2 −1/√√2, Y₂ X₁ X₂ = 22 1 1 2 0 Y₁ 2 1 (4) (2)) = (²) (²) 1 1 Z₁ Z₂ Z3