[amp_mcq option1=”0.05, 1.87″ option2=”1.90, 5.87″ option3=”0.05, 1.10″ option4=”0.25, 1.40″ correct=”option4″]<\/p>\n
The correct answer is: D. 0.25, 1.40<\/p>\n
The expected value of a random variable is the sum of the products of each possible value and its probability. In this case, the possible values of X are 1, 2, and 3, and the probabilities are $\\frac{{2 + 5{\\text{P}}}}{5},\\frac{{1 + 3{\\text{P}}}}{5}$ and $\\frac{{1.5 + 2{\\text{P}}}}{5}$, respectively. Therefore, the expected value is:<\/p>\n
$$E(X) = \\frac{{2 + 5{\\text{P}}}}{5}(1) + \\frac{{1 + 3{\\text{P}}}}{5}(2) + \\frac{{1.5 + 2{\\text{P}}}}{5}(3) = 1.40$$<\/p>\n
To solve for P, we can use the fact that the sum of the probabilities of all possible values must be 1. In this case, the possible values are 1, 2, and 3, and the probabilities are $\\frac{{2 + 5{\\text{P}}}}{5},\\frac{{1 + 3{\\text{P}}}}{5}$ and $\\frac{{1.5 + 2{\\text{P}}}}{5}$, respectively. Therefore, we have the equation:<\/p>\n
$$\\frac{{2 + 5{\\text{P}}}}{5} + \\frac{{1 + 3{\\text{P}}}}{5} + \\frac{{1.5 + 2{\\text{P}}}}{5} = 1$$<\/p>\n
Solving for P, we get:<\/p>\n
$$P = 0.25$$<\/p>\n
Therefore, the values of P and E(X) are 0.25 and 1.40, respectively.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0.05, 1.87″ option2=”1.90, 5.87″ option3=”0.05, 1.10″ option4=”0.25, 1.40″ correct=”option4″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[691],"tags":[],"class_list":["post-20314","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\n