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- QUESTION 11 The probability density function, f(x), for any continuous random variable X, represents: a. all possible values that X will assume within some interval a ≤ x ≤ b. b. the probability that X takes on a specific value x. c. the height of the density function at x. d. None of these choices.QUESTION 10 Suppose f(x) = 1/4 over the range a ≤ x ≤ b, and suppose P(X > 4) = 1/2. What are the values for a and b? a. 2 and 6 b. Cannot answer with the information given. c. 0 and 4 d. Can be any range of x values whose length (b − a) equals 4. QUESTION 11 The probability density function, f(x), for any continuous random variable X, represents: a. all possible values that X will assume within some interval a ≤ x ≤ b. b. the probability that X takes on a specific value x. c. the height of the density function at x. d. None of these choices. QUESTION 12 Which of the following is true about f(x) when X has a uniform distribution over the interval [a, b]? a. The values of f(x) are different for various values of the random variable X. b. f(x) equals one for each possible value of X. c. f(x) equals one divided by the length of the interval from a to b.…Question 1.2 Consider the function f (x) = (1/24(x^2 +1) 1 < or = x < or = 4) = (0 otherwise) Calculate P (x = 3) Calculate P (2 < or = x < or = 3) Question 1.3 Consider the function f (x) = (k - x/4 1 < or = x < or = 3) = (0 otherwise) which is being used as a probability density function for a continuous random variable x? a. Find the value of K b. Find P (x < or = 2.5)
- Suppose a random variable X has a probability density function as shown in img1.jpg c. Find P(X <= 0.4 | X <= 0.8) Note: I already solve part A and the value of k is 6.The length of time for students to complete a 1-hour timed standardized mathematics test has a probability density function f(t) = 1.5t2 + t, 0 ≤ t ≤ 1In the question given as follows, (a) show that the nonnegative function is a probability density function, and (b) find P(0 x 6). see the equation as attached here
- Question 1 : Suppose that the probability density function (p.d.f.) of the life (in weeks) of a certain part is f(x) = 3 x 2 (400)3 , 0 ≤ x < 400. (a) Compute the probability the a certain part will fail in less than 200 weeks. (b) Compute the mean lifetime of a part and the standard deviation of the lifetime of a part. (c) To decrease the probability in part (a), four independent parts are placed in parallel. So all must fail, if the system fails. Let Y = max{X1, X2, X3, X4} denote the lifetime of such a system, where Xi denotes the lifetime of the ith component. Show that fY (y) = 12 y 11 (400)12 , y > 0. Hint : First construct FY (y) = P(Y ≤ y), by noticing that {Y ≤ y} = {X1 ≤ y} ∩ {X2 ≤ y} ∩ {X3 ≤ y} ∩ {X4 ≤ y}. (d) Determine P(Y ≤ 200) and compare it to the answer in part (a)Suppose X and Y have joint probability density function ?(x, y) = { 6x, 0 < x < 1, 0 < y < 1 − x 0, elsewhere. a. Determine whether the two random variables are dependent or independent. b. If the value of Y is 0.5, what is the probability that X will have a value greater than 0.3? Please answer this question step by step and justify your solutionA lamp has a lightbulb with an average lifetime of 4 hours. The probability density function for the lifetime of a bulb is f(t)=14e−t/4,t≥0.What is the probability that the bulb will fail within 5 hours?