Sec_4_solutions - CEE 304 Section 4 Solutions 1....

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CEE 304 – Section 4 Solutions 1. Transforming Variables Example : ( ) x X f x e - = for x > 0, and 0 otherwise Develop the density function for Y = X ½ Ans . Y = g(X) = X ½ We have a one-to-one relationship between X and Y, so we can use the following formula to find the pdf of Y given the pdf of X: ( ) ( ) Y X dx f y f x dy = This is equivalent to: 1 1 ( ) ( ) ( ( )) Y X dg y f y f g y dy - - = 2 ( ) (2 ) y Y f y e y - = for y > 0, and 0 otherwise 2. An engineering student got a summer job setting off fireworks on the 4 th of July and other summer events. For a particular display, the bomb is shot from the tube so that after t seconds it is at height: H = (100 m/sec) t – 0.5 (10 m/sec 2 ) t 2 = 500 – 5(10 – t) 2 meters Unfortunately the timers on the bombs are not accurate. The student estimates that the timer will yield a time T to detonation after launch that is normally distributed with mean 10 seconds and standard deviation 1.2 seconds. T ~ N( μ = 10 sec, σ 2 = (1.2 sec) 2 ) a.) What is the probability that the bomb explodes more than 495 meters above the ground? ] 495 ) 10 ( 5 500 [ ] 495 [ 2 - - = t P H P = 0.5934
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b.) Above what height can you be 99% confident the bomb will explode? 99 . 0 ] [ = h H P 99 . 0 ] ) 10 ( 5 500 [ 2 = - - h t P Ans . h = 452.26 m c.) What is the pdf for the height at which the bomb explodes? [We can easily derive the cdf for the height of explosion F H (h), but deriving the pdf f H (h) is made difficult by the need to evaluate d Φ (h)/dh. When originally asked on an exam, this question asked to outline the steps required to to obtain the pdf – it did not ask to do the actual derivation. Here we will derive the cdf and then note how to obtain the pdf.]
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Sec_4_solutions - CEE 304 Section 4 Solutions 1....

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