The value of k is 1/3. The 40th percentile m 2.2. The expected amount of raw material that will be left at the end of the week is 2.1 thousand pounds. value of E(x) is 1 , value of var(x) is 8/3.
To determine the value of k for a valid pdf, we need to ensure that the pdf integrates to 1 over the range of possible values for X. The range for X is 1 ≤ x ≤ 4, so we can integrate the pdf over this range to find the value of k: ∫1^4 f(x) dx = ∫1^4 k−k/x^2 dx = k(x - 1/x) |1^4 = k(4 - 1) - k(1 - 1/1) = 3k To make this integral equal to 1, we set 3k = 1 and solve for k to find that k = 1/3. To compute E(X) and Var(X), we can use the formulas for the expected value and variance of a continuous random variable: E(X) = ∫1^4 xf(x) dx = ∫1^4 x(1/3 - 1/(3x^2)) dx = (1/3)x - (1/3)(1/x) |1^4 = (1/3)(4 - 1) - (1/3)(1 - 1/1) = 1 Var(X) = ∫1^4 (x - E(X))^2 f(x) dx = ∫1^4 (x - 1)^2 (1/3 - 1/(3x^2)) dx = (1/3)(x - 1)^2 - (1/3)((x - 1)^2)/x^2 |1^4 = (1/3)(4 - 1)^2 - (1/3)((4 - 1)^2)/4^2 - (1/3)(1 - 1)^2 + (1/3)((1 - 1)^2)/1^2 = (1/3)(9 - 1) - (1/3)(9 - 1)/16 = 8/3 To compute the 40th percentile m, we need to find the value of x such that the probability that X is less than or equal to x is equal to 0.4. This can be expressed as: ∫1^m f(x) dx = 0.4 Substituting in the value of k that we found earlier and solving for m, we find that:∫1^m (1/3 - 1/(3x^2)) dx = 0.4 (1/3)x - (1/3)(1/x) |1^m = 0.4 (1/3)(m - 1) - (1/3)(1 - 1/1) = 0.4 (1/3)(m - 1) = 0.4 m - 1 = 1.2 m = 2.2 Finally, to answer the question about the amount of raw material expected to be left at the end of the week, we can use the expected value that we computed earlier. Since the demand for the raw material is a random variable with expected value 1, we can expect that on average, 1 thousand pounds of the raw material will be used each week. Since there are 2.5 thousand pounds in stock at the beginning of the week, we can expect there to be 2.5 - 1 = 1.5 thousand pounds left at the end of the week.
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