The probability distribution of random variable X is given below. What is E[X]?
X | 4 | 2 | 6 |
P(x) | 0.6 | 0.2 | 0.2 |
The probability distribution of random variable X is given below. What is σ2x?
X | 4 | 2 | 6 |
P(x) | 0.6 | 0.2 | 0.2 |
The probability distribution of random variable X is given below. Let Y = 4X − 5 be a new random variable. What is σ2y?
X | 4 | 2 | 6 |
P(x) | 0.6 | 0.2 | 0.2 |
The probability distribution of random variable X is given below. Let W = X2 − X be a new random variable. What is σ2w?
X | 4 | 2 | 6 |
P(x) | 0.6 | 0.2 | 0.2 |
The probability distribution of random variable X is given below. What is E[X]?
X | 1 | 2 |
P(x) | 0.6 | 0.4 |
The probability distribution of random variable X is given below. Let Y = 3X2 − 1 be a new random variable. What is σy?
X | 1 | 2 |
P(x) | 0.2 | 0.8 |
In basketball, free throws are unopposed attempts to score points from a fixed distance away from the basket and are awarded to a player when the opposite team breaks particular rules. “Converting” a freethrow means that the ball goes through the basket and a point is successfully scored. A player shoots either two free-throws consecutively, or shoots what’s called a “one-and-one,” in which a second free-throw attempt is allowed only if the first attempt was successful. Assume that a player’s free-throw attempts are independent. Lakshmi converts 75% of her free-throws. She is awarded two consequtive free-throws. The random variable X is the number of points resulting from Lakshmi’s two free-throw attempts. What is E[X]?
Linh converts 65% of their free throws. They get to shoot a “one and one.” The random variable W is the number of points resulting from Linh’s “one-and-one” attempt. What is E[W]?
The Cubs team will play 5 baseball games against the Cardinals team. For any one game it is estimated that the probability of a Cubs win is 0.55, and the outcomes of the 5 games are independent. Let X be the number of games that the Cubs win in the series. What is σX?
The Cubs team will play 5 baseball games against the Cardinals team. For any one game it is estimated that the probability of a Cubs win is 0.4, and the outcomes of the 5 games are independent. Let X be the number of games that the Cubs win in the series. What is the probability that the Cubs win a majority of the games in the series?
The probability distribution of random variable X is given below. What is E[X]? X 4 2...
х 1 4 5 4. The probability distribution of a random variable X is given below -4 3 P(X=x) 0.1 0.2 0.3 0.2 a) Find E(X) 0.2 b) Find Var(X)
Joe is attempting basketball free throws. The probability of success on each attempt is 0.39 and his consecutive attempts are independent of each other. Answer the following questions. If Joe attempts three free throws, what is the probability he has at least one success? (a) Let X be the number of attempts required for Joe to observe his first successful free throw. What kind of discrete random distribution for X? (Binomial, Poisson, or none of them) (b) If Joe attempts...
Cumulative distribution function The probability distribution of a discrete random variable X is given below: Value x of X P(x-x) 0.24 0.11 -2 0.26 0.11 Let Fx be the cumulative distribution function of X. Compute the following: X 5 ? 18+ (-2) - Px (-4) = 0
Suppose that you’re watching the World Series. At the beginning of the Series, you believe that the Cubs are a better team than the Indians…. (Sorry, I must pause here. As an Indians’ fan, it hurts to write that.) Anyways, you believe that the probability that the Cubs will beat the Indians in a single game is at least 0.50. You decide to reject this hypothesis if the Cubs fail to win more than 2 games in the World Series....
The random variable X has the probability distribution table shown below. x 2 4 6 8 10 P(X = x) 0.2 0.2 a a 0.2 (a) Assuming P(X = 6) = P(X = 8), find each of the missing values. a = (b) Calculate P(X ≥ 6) and P(2 < X < 8). P(X ≥ 6) = P(2 < X < 8) =
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X 11), n= 18, p = 0.6
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X=14), n=19, p=0.6
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X=16), n=17, p=0.6
Given a positive random variable that has an arbitrary distribution with , then the probability could only be one of the choices below. Which one could it be? Given a positive random variable Z that has an arbitrary distribution with E (2) 1, then the probability P (Z 2 8) could only be one of the choices below. Which one could it be? 0.5 0.1 O 0.2 0.875 We were unable to transcribe this imageE (Z) = 1 P(Z2 8)
6. The distribution law of random variable X is given -0.4 -0.2 0 0.1 0.4 0.3 0.2 0.6 Xi Pi Find the variance of random variable X. 7. Let X be a continuous random variable whose probability density function is: f(x)=Ice + ax, ifXE (0,1) if x ¢ (0:1) 0, Find 1) the coefficient a; 2) P(O.5 X<0.7); 3) P(X>3). Part 3. Statistics A sample of measurements is given Y 8 4 2 2 0 8. Compute the coefficient of...