Question

3 The probability density function of a random variable on the interval [9, 16] is f(x) = x. Find the following values. a. Find the expected value The expected value is (Round to two decimal places as needed.) b. Find the variance. The variance is (Round to two decimal places as needed.) c. Find the standard deviation The standard deviation is (Round to two decimal places as needed.) d. Find the probability that the random variable has a value greater than the mean. The probability is (Round to four decimal places as needed.) e. Find the probability that the value of the random variable is within one standard deviation of the mean. The probability is (Round to four decimal places as needed.)

Answer 1

f(x) = 3 x when aux = 16 14 Otherwise (0) 16 MEE() =( X) = xf(x) da = 23 74 x dx = 3 x't? 219 de 3 +

304/716 22/ 3, +1 74 56 -12.66 49 12.66 (6) Vall (X) = E(X*) - €() $(x²) = 16 a 1 2 3 5x dx = 2 2 da 3 (2+12 dx x 74 74 3 169 5/2 da 74

(272) 74x7 X 14197 3 X 2 74x7 3 164.4440 vall X) 164.4440 - 12. 66492 = 164.44 40- 160.3997 = 4.0443 a 4.04 of x (C) The deviation stand ared 6-SD x = var (x) 4.04 = 2.01

The Psu baby that the random nuotiable has a value greater than the mean =P(x > mean) = P(x > 12.66) 16 fra da 12.66 y se da 12.66 14 3 [a2+]12.66 14 1+1 = (u ?- 12.6672) 0.5123 (e) The random deneation probability that the value of the variable is within one standard of the mean = P(4-6 L X L M+6) . - P(12066-2001<x<1206+20

10.65 < x < 14.67) 14.6 Hdx 10.65 14.67 10.65 4.67 43 22 dx 10.65 14.67 Jo.65 312 61 2 (14.67812 _ 10.6512) - 10.65 = 2 * ( 56.1882 – 34.7557) = 3 x 21.4326 = 0.5793