Suppose that X is a N(3,4) random variable and suppose that Y=2X+1. Determine P(Y>8).
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Suppose that X is a N(3,4) random variable and suppose that Y=2X+1.
Determine P(Y>8).
![Suppose that X is a N(3,4) random variable and suppose that Y=2X+1.
Determine P(Y>8).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b7f0249-62bd-4daa-aff9-56ee7c4d29e5%2Faa99de74-201a-4544-b3b5-4e999dc9159f%2F5ytx5on_processed.png&w=3840&q=75)
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- A QR code photographed in poor lighting, so that it can be difficult to distinguish black and white pixels. The gray color (X) in each pixel is therefore coded on a scale from 0 (white) to 100 (black). The true pixel value (without shadow) the code is Y = 0 for white, and Y = 1 for black. We treat X and Y as random variables. For the highlighted pixel in the figure is the gray color X = 20 and the true pixel value is white, i.e. Y = 0. We assume that QR codes are designed so that, on average, there are as many white as black pixels, which means that pY (0) = pY (1) = 1/2. In this situation, X is continuously distributed (0 ≤ X ≤ 100) and Y is discretely distributed, but we can still think about the simultaneous distribution of X and Y. We start by defining the conditional density of X, given the value of Y : fX|Y(x|0) = "Pixel is really white" fX|Y(x|1) =" Pixel is really balck " Use Bayes formula as given in the picture and find the probability for x = 20 like in the picture.20An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (*) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is ttt, then N (ttt) = 0. Suppose that the random variable X is defined in terms of N as follows: X=2N -2. The values of X are given in the table below. Outcome ttt hth tht htt thh hhh hht tth Value of X -2 2 0 0 2 4 2 0 Calculate the probabilities P(X=*) of the probability distribution of X. First, fill in the first row with the valuesof X. Then fill in the appropriate probabilities in the second row. Value x of X ___ ___ ___ ___ P(x=x) ___ ___ ___ ___
- Let X,, X2, and X3 be independent random variables, each are binomially distributed with n = 100 and p = 0.2. Let A = X,– 2X2 and B = X3 + 3X1. Find PAB- %3D %3DAn ordinary (fair) coin is tossed 3 times. Outcomes are thus triple of “heads” (h) and tails (t) which we write hth, ttt, etc. For each outcome, let R be the random variable counting the number of tails in each outcome. For example, if the outcome is hht, then R (hht)=1. Suppose that the random variable X is defined in terms of R as follows X=6R-2R^2-1. The values of X are given in the table below. A) Calculate the values of the probability distribution function of X, i.e. the function Px. First, fill in the first row with the values X. Then fill in the appropriate probability in the second row.16) number of medical tests that a patient will have on entering a hospital is a random variable X which can take on the values 0, 1, 2, 3, and 4. If P(X 2)=0.43, find P(X = 3).
- Suppose that X is a random variable * ? having the following m.g.f., Find Var(x) 7t_e4t M, (t) = eT 173 O 5 O 413 O 3148 O 55 O3.Suppose that X and Y are independent random variables for which Var(X)=Var(Y)=3. Find the values of (a) Var(X-Y); (b) Var(2X-3Y+1).4 Suppose that Alejandro is planting a garden of tulips. Let X be the Bernoulli random variable that returns 1 if the tulip is red, and 0 otherwise. Suppose Y is another Bernoulli random variable, independent of X, that returns 1 if the tulip survives longer than 30 days, and 0 otherwise. Both X and Y have parameters 1/2. Let Z be the random variable that returns the remainder of the division of X +Y by 2. For example, if X = 1 and Y = 0, Z = remainder() = 1 (a) Prove that Z is also a Bernoulli random variable, also with parameter 1/2. (b) Prove that X, Y, Z are pairwise independent but not mutually independent. (c) By computing Var[X+Y+Z] according to the alternative formula for variance and using the variance of Bernoulli r.v.'s, verify that Var[X+Y+Z] = Var[X]+Var[Y] +Var[Z] %3D (observe that this also follows from the proposition on slide 5 of the lecture segment entitled "Binomial distribution").
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