Given the probability distribution of the random variable X X 2 4 6 f(x) 1/6 1/2 1/3 Calculate u_{g(x)} where g(x)=9x+108 (round to 2 decimal places)
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Given the probability distribution of the random variable X
| X | 2 | 4 | 6 |
| f(x) | 1/6 | 1/2 | 1/3 |
Calculate u_{g(x)} where g(x)=9x+108
(round to 2 decimal places)
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- Find the mean of the random variable X with PDF S 3x² _if 0Suppose that f(x)=k/(4^x), x=1,2,3,... is the probability function for a random variable X. Find the value of 1/k.Show that the variance of a geometric random variable X; is given by V ar(X) = q/p^2 .the random variable Y is such that : E(2Y+3)=6 and Var(2-3Y)=11 find E(Y^2)The probability of an event depends on the random variable X and can be expressed by the following formula: P(^|x) = {0 1-e for x>0, for x 0). Find the total probability of the event A if X is a normally distributed random variable with expectation & and variance oSuppose that f(x)=2/(3^x), x=1,2,3,... is the probability function for a random variable X. Find P(X>3). Use 4 decimal places.Suppose a random variable X has range {−3, −1, 1, 2}. Its proba- bility distribution (probability mass function) is given by P (X = −3) = 0.25 P (X = −1) = 0.25 P (X = 1) = 0.45 P (X = 2) = 0.05 a) Find E(X). b) Find E(X^2). b) Find V (X). b) Find V (X^2).Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter > (a) the probability that a repair time exceeds 6 hours? ☐ (b) the conditional probability that a repair takes at least 9 hours, given that it takes more than 8 hours? = 0.2. What isI have a probability distribution consisting of (x, p(x)): (-3, 0.1), (-2, 0.2), (-1, 0.3), (0, 0.05). The probabilities obviously do not sum to 1. How do I find <x> (average), <x2> (the second moment), <sigmax2> (the variance)?Identify the distributions of the random variables with the moment- generating functions shown below. For each random variable also indicate what the mean and the variance are. a) m(t) = e^2.2(e^t−1)b) m(t) = 1/(1 − 2t)^2 c) m(t) =( e^5t−e^t )/4tThe spinner is spun one fime and a fair coin is tossed. Find the probabilities. P(R) = P(T) = %3D Yellow Red P(B U T)= P(G^) = Blue Green Sample Space: P(G N H)= P(Y U H)C = P(BC)= P(G NT)=An 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.SEE MORE QUESTIONS