Enter the correct number to make this a valid probability mass function. 3 10 P(X =...
Enter the correct number to make this a valid probability mass function. X 1 3 10 P(X = x) 0.28 0.1 数字
Explain why the probability mass function P(X = 1000) = 0.1, P(X = 1500) = 0.2, P(X = 2000) = 0.3, P(X = 2500) = 0.3, P(X = 3000) = 0.1 is not practical as a distribution for the number of phone calls to a help-desk call center during a day
Which of the following tables shows a valid probability density function? Select all correct answers. Select all that apply: 0.04 0.89 0.07 P(X = z) P(X = z) 10 3 10 4 0 3-10 1-2 1-515 X 0. 0 0 0 0 X 01234 0123
Which of the following distributions is(are) valid discrete probability distribution(s)? 2. 3. 4. X p(x) X P(X) X p(x) X P(X) 0.3 0 0.3 0 0.2 0 0.1 1 0.4 1 -0.2 1 0.7 1 0.1 2 0.3 2 0.9 2 0.2 N 0.8 O All are valid O 1,3, and 4 only 1 and 4 only 1 only 4 only
(d) Find the probability mass function of X given Y = 3 (ie, p(x|y = 3)) 7. (10 points) Consider two jars, Jar M and Jar W. In Jar M, there are 3 balls numbered 0, 1, 2. In Jar W there are 3 balls numbered 1, 2, 3. A ball is drawn from Jar M, then a ball is drawn from Jar W. Define M as the number on the ball from Jar A and W the number on...
Assume that the probability mass function of X is given by P(X = 1) = P(X = 2) = P(X = 3) = 1/3 A random sample of n = 36 is selected from this population. Find the probability that the sample mean is greater than 2.1 but less than 2.5, assuming that the sample mean would be measured to the nearest tenth.
Fill in the table below such that it is a valid joint probability mass function for two independent discrete random variables X and Y .x Pxxtx.y Pi() 0.25 Px) 0.20 0.40 Check
Question 1. A Discrete Distribution - PME Verify that p(x) is a probability mass function (pmf) and calculate the following for a random variable X with this pmf 1.25 1.5 | 1.7522.45 p(x) 0.25 0.35 0.1 0.150.15 (a) P(X S 2) (b) P(X 1.65) (c) P(X = 1.5) (d) P(X<1.3 or X 221) e) The mean (f) The variance. (g) Sketch the cumulative distribution function (edf). Note that it exhibits jumps and is a right continuous function.
f(x) = *, x = 1, 2, 3, 4 © Verify that f(x) is a valid probability mass function(pmf). © P(X < 2.5) * P(X > 1) © Compute the mean and variance
1. (10 points) The joint probability mass function of X and Y is given by p(1,1)= P(2,1)= 0, P(3,1) = 2 P(1, 2) = 1 p(2, 2) = 2 P(3, 2) = 16 p(1,3) = (2, 3) = 0, P(3, 3) = 5 Find (a) PX\Y(3,1); (b) E[X Y = 2) and (c) Fyx (2/1).