Homework 3 solution (100points) Due in class, 9/ (10) 1.19 (page 31)

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1 Homework 3 solution (00points) Due in class, 9/4. (0).9 (page 3) (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 eample, the total area under this density curve is (6 4) =. height = /(6-4) = / 4 6 (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. Similarly, the proportion of values that eceed 4.5 would be (6 4.5) = (c) The median of this distribution is 5 because eactly 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 for which the proportion of the data less than 0 equals.0. That is, the area under the density to the left of 0 must equal.0. Therefore, the area =.0 = ( 0 4), and so Thus, (0).0 (page 3)

2 (a) The density function is f ( ) /[5 ( 5)] / 0 over the interval [-5, 5] and f ( ) 0 elsewhere. The proportion of values that are negative is eactly.5 since the value 0 sits precisely in the middle of the interval [-5, 5]. (b) The proportion of values between - and is [ ( )] =.4. The proportion of the values falling between - and 3 is [3 ( )] = (c) The proportion of the values that lie between k and k 4 is [( k 4) k ] = (8). (page 3) (a)the density function is f ( ) /(0 7.5) /. 5 over the interval [7.5, 0] and f ( ) 0 elsewhere. The proportion of depths less than k is given by the epression ( k 7.5) for k 7. 5 and 0 elsewhere. For k 0, this proportion is (0 7.5) =.0. For k 5, it is (5 7.5) = (b) The proportion of values that are at least k is k). The proportion of values that strictly.5 (0 eceed k is also (0 k) because f () is a continuous density. For k 0, this proportion is.5 0) =.80; for k 5, it is (0 5) =.40. (0.5.5 (c) It helps to draw the picture of the density: central 90% So, the area to the right of should be.95; i.e., ( 0 ) =.95. Similarly, the area to the right of is.05, and so ( 0 ) =.05. Solving these equations gives 8. 5 and

3 4. (0). (page 3) (a) The density curve forms an isosceles triangle over the interval [0, 0]. For this reason, such densities are often called triangular densities. The total area under this density curve is simply the area of the triangle, which is (base)(height) = (0)(.) =. The height of the triangle is the value of f() at = 5; i.e., f(5) = (5) =.. height = (b) Proportion ( 3) = (3 0) f (3) = (3)(.) =.8. Proportion ( 7) = (0 7) f (7) = (3)(.) =.8. Proportion ( 4) = Proportion ( < 4) = (4 0) f (4) = - (4)(.6) =.68. Proportion (4 < < 7) = - [Proportion ( 4) + Proportion ( 7)] = - [.3+.8] =.50. (c) This particular value 0 will have 0% of the waiting times above it, so Proportion ( < 0 ) =.0. That is, (0-0 )f( 0 ) =.0. Since f() = in this region, we have.0 = (0-0 )( ). Multiplying the factors and collecting terms yield a quadratic equation = 0, whose solutions are and.36. Of these values, only 0 = lies within the interval [0,0]. Alternate solution: Notice that f() is symmetric around = 5, so that the triangle that captures the leftmost 0% of the area will have a base of the same length as the triangle that captures the rightmost 0%. The left-hand triangular region is much easier to work with: area =.0 = ( )f( ) = ( )(.04 ), so = 5 and = + 5. Therefore, 0 = 0 - =

4 5. (8).3 (page 3) (a) = (b) 0, (.00004) e e d = (.00004) = 0 - (-e (0,000) ) = e -.8 = ,000 = e ,000 Note: for any eponential density curve, the area to the right of some fied constant always equals e -c, as our integration above shows. That is, Proportion ( >c) = c e d = e -c. We will use this fact in the remainder of the chapter instead of repeating the same type of integration as in part (a) Proportion ( 30,000) = - Proportion ( > 30,000) = - e -c = - e (30,000) = - e -. =.699.

5 Proportion (0,000 30,000) = Proportion ( > 30,000) - Proportion( 0,000) = (-.449) =.48. (c) For the best %, the lifetimes must be at least 0, where Proportion( 0 ) =.0, which e 0 becomes =.0. Taking natural logarithms of both sides, - 0 = ln(.0), so 0 = -ln(.0)/ = / = 5,9.5. For the worst %, we have Proportion( 0 ) =.0, which is 0 equivalent to saying that Proportion( 0 ) =.99, so e =.99. Taking logarithms, - 0 = ln(.99), so 0 = -ln(.99)/ = (8).5 (page 54) (a) Let = number of goblets with flaws in a bo of 6. Then has a binomial distribution with n = 6, 5 =.. Using the formula for the binomial function, Proportion( = ) = 6! )!5! (. ) (. = 6(.)(.5773) =.3799, or, about.38. (b) Proportion( ) = - Proportion( ) = - [Proportion( =0) + Proportion( =)] = -! 0 6 [ 6 (. ) (. )! (. ) (. ) ] = - [ ] = !6!!5! (b) Proportion( 3) = Proportion( = ) + Proportion( = ) + Proportion( = 3) = 6 ) 5 4 (. ) (. + 6 (. ) (. )! (. ) (. ) = =.533.!!5!!!4! 3!3! 7. (0).54 (page 54) (a) Let = number of bits erroneously transmitted. Then, is binomial with n = 0, =.0, so Proportion( ) = =.677 (from Table II). (b) Proportion( 5) = =.043. (c) 'More than half' means or more, so Proportion( ) = =.000.

6 8. (0).56 (page 55) (a) Let denote the number of drivers. Then, Proportion( 0) = =.0 (Table III, = 0). (b) Proportion( > 0) = =.48. (This answer of.48 is due to the limitation of the table. The actual answer should be.4409.) (c) Proportion(0 0) = =.556. Proportion(0 < < 0) = = (8). (page 78) Let X = the number of drivers who travel between a particular origin and destination during a designated time period. X has a Poisson distribution with 0. (a) = = 0 Find P 5 5 P5 5 Using Table III with = 0, we obtain: P (b) Find P P P But, since X is an integer valued random variable, only the integers between 6 and 4 P 6 4 using Table III with = 0, we obtain: satisfy this requirement. So, we find.688 P

7 0. (8)Two fair si-sided dice are tossed independently. a. Let M be the maimum of the two tosses (so M(,5)=5, M(3,3)=3, etc), find the pmf, mean and variance of M. PMF for M M Pr(M) /36 3/36 5/36 7/36 9/6 /36 mpr( m) =4.47 M M ( m M) Pr( m) =.404 M b. Let T be the sum of the two tosses (so T(,5)=6, T(3,3)=6, etc), find the pmf, mean and variance of M. PMF for T T Pr(T) /36 /36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 /36 /36 tpr() t =7 T T ( t T) Pr( t) = T An easier way is to define the number of each toss X and X respectively, then T X X EX EX 3.5 Var( X ) Var( X ).97 X and X are independent Therefore E( T) E( X) E( X ) 7, and Var( T) Var( X) Var( X ) 5.83