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Solution1_2

# Solution1_2 - STAT 350 Solutions to Homework 1 and Homework...

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STAT 350 Solutions to Homework 1 and Homework 2 19. (a) The density curve forms a rectangle over the interval [4, 6]. For this reason, uniform densities are also called rectangular densities by some authors. Areas under uniform densities are easy to find (i.e., no calculus is needed) since they are just areas of rectangles. For example, the total area under this density curve is ) 4 6 ( 2 1 = 1. height = 1/(6-4) = 1/2 4 6 x (b) The proportion of values between 4.5 and 5.5 is depicted (shaded) in the diagram below. The area of this rectangle is ) 5 . 4 5 . 5 ( 2 1 = .5. Similarly, the proportion of x values that exceed 4.5 would be ) 5 . 4 6 ( 2 1 = .75. 4 6 x 4.5 5.5 (c) The median of this distribution is 5 because exactly half the area under this density sits over the interval [4,5]. (d) Since 'good' processing times are short ones, we need to find the particular value 0 x for which the proportion of the data less than 0 x equals .10. That is, the area under the density to the left of 0 x must equal .10. Therefore, the area =.10 = ) 4 ( 0 2 1 x , and so . 20 . 4 0 x Thus, . 20 . 4 0 x 21. (a) The density function is 5 . 12 / 1 ) 5 . 7 20 /( 1 ) ( x f over the interval [7.5, 20] and 0 ) ( x f elsewhere. The proportion of depths less than k is given by the expression ) 5 . 7 ( 5 . 12 1 k for 5 . 7 k and 0 elsewhere. For 10 k , this proportion is ) 5 . 7 10 ( 5 . 12 1 = .20. For 15 k , it is ) 5 . 7 15 ( 5 . 12 1 = .60. (b) The proportion of x values that are at least k is ) 20 ( 5 . 12 1 k . The proportion of x values that strictly exceed k is also ) 20 ( 5 . 12 1 k because ) ( x f is a continuous density. For 10 k , this proportion is ) 10 20 ( 5 . 12 1 = .80; for 15 k , it is ) 15 20 ( 5 . 12 1 = .40.

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(c) It helps to draw the picture of the density: 7.5 20 x x 1 x 2 central 90% So, the area to the right of 1 x should be .95; i.e., ) 20 ( 1 5 . 12 1 x =.95. Similarly, the area to the right of 2 x is .05, and so ) 20 ( 2 5 . 12 1 x = .05. Solving these equations gives 125 . 8 1 x and . 375 . 19 2 x 23. (a) x = .00004 0 (b) dx e x 0000 , 20 00004 . 00004 .
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