Suppose X is a continuous random variable with the following probability density func-tion:f (x) = c(2 − x), 0 ≤ x ≤ 2(a) [2 marks] Plot the probability density function, clearly marking on your plot thevalues of f (x) when x = 0 and x = 2 and labelling both axes.(b) [2 marks] Explain and show why the value of c must equal 12 .(c) [2 marks] Find P (1 < X < 1.5).(d) [3 marks] Find P ({1 < X < 1.5}|{X > 0.5}).(e) [2 marks] Find P ({X > 0.5}|{1 < X < 1.5}).(f) [4 marks] Suppose that we have another random variable U that is uniformlydistributed with parameters 2 and 4. That is, the probability density function of Uis given byf (u) = 12, 2 ≤ u ≤ 4.Further, U and X are independent. In addition, someone has told us that therandom variable Y defined byY = X + Uhas expected value equal to 113 and variance equal to 59 . Based on this information,calculate the expected value and variance of X
Question
Suppose X is a continuous random variable with the following probability density func-tion:f (x) = c(2 − x), 0 ≤ x ≤ 2(a) [2 marks] Plot the probability density function, clearly marking on your plot thevalues of f (x) when x = 0 and x = 2 and labelling both axes.(b) [2 marks] Explain and show why the value of c must equal 12 .(c) [2 marks] Find P (1 < X < 1.5).(d) [3 marks] Find P ({1 < X < 1.5}|{X > 0.5}).(e) [2 marks] Find P ({X > 0.5}|{1 < X < 1.5}).(f) [4 marks] Suppose that we have another random variable U that is uniformlydistributed with parameters 2 and 4. That is, the probability density function of Uis given byf (u) = 12, 2 ≤ u ≤ 4.Further, U and X are independent. In addition, someone has told us that therandom variable Y defined byY = X + Uhas expected value equal to 113 and variance equal to 59 . Based on this information,calculate the expected value and variance of X
Solution
(a) To plot the probability density function, you would need to graph the function f(x) = c(2-x) from x = 0 to x = 2. At x = 0, f(x) = c(2-0) = 2c. At x = 2, f(x) = c(2-2) = 0. The y-axis represents the probability density and the x-axis represents the value of the random variable X.
(b) The value of c can be found by integrating the probability density function from 0 to 2 and setting it equal to 1 (since the total probability must be 1).
∫ from 0 to 2 of c(2-x) dx = 1 c * [2x - 0.5x^2] from 0 to 2 = 1 c * [4 - 2] = 1 c = 1/2
(c) To find P(1 < X < 1.5), you would integrate the probability density function from 1 to 1.5:
∫ from 1 to 1.5 of 0.5(2-x) dx = [x - 0.25x^2] from 1 to 1.5 = 0.125
(d) P({1 < X < 1.5}|{X > 0.5}) is the same as P(1 < X < 1.5) since the condition {X > 0.5} is always true when 1 < X < 1.5. So the answer is 0.125.
(e) P({X > 0.5}|{1 < X < 1.5}) is 1 since the condition {1 < X < 1.5} implies that X > 0.5.
(f) The expected value of X is given by the integral of xf(x) from 0 to 2. The variance of X is given by the integral of (x - E[X])^2 * f(x) from 0 to 2. The expected value and variance of U are 3 and 1/3 respectively. The expected value of Y is E[X] + E[U] = 1 + 3 = 4. The variance of Y is Var[X] + Var[U] + 2Cov[X, U]. Since X and U are independent, Cov[X, U] = 0. So Var[Y] = Var[X] + 1/3. Given that E[Y] = 113 and Var[Y] = 59, we can solve for E[X] and Var[X].
Similar Questions
a) [2 marks] Plot the probability density function, clearly marking on your plot thevalues of f (x) when x = 0 and x = 2 and labelling both axes.(b) [2 marks] Explain and show why the value of c must equal 12 .(c) [2 marks] Find P (1 < X < 1.5).(d) [3 marks] Find P ({1 < X < 1.5}|{X > 0.5}).(e) [2 marks] Find P ({X > 0.5}|{1 < X < 1.5}).(f) [4 marks] Suppose that we have another random variable U that is uniformlydistributed with parameters 2 and 4. That is, the probability density function of Uis given byf (u) = 12, 2 ≤ u ≤ 4.Further, U and X are independent. In addition, someone has told us that therandom variable Y defined byY = X + Uhas expected value equal to 113 and variance equal to 59 . Based on this information,calculate the expected value and variance of X.
Calculating Probability DensityIn a uniform PDF, all the possible values have the same probability density. The figure below shows such a uniform PDF, where the possible values are from 0 to 10. For this graph, what is the value of the probability density from X = 0 to X = 10?
Suppose a continuous random variable X has the CDF(a) What is the range of possible outcomes for X?(b) As far as you can tell from the graph, what is P( X ≤ 1.5 )?(c) What feature of the CDF tells you that X is a continuous random variable?
Regarding continuous probability functions, the area under the graph represents the ________.
Suppose that the random variable X is continuous and takes its values uniformly over the interval from 0 to 2. What is the value of the probability P{X ≤ 0.4 or X > 1.2}?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.