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hw2_soln2 - HOMEWORKS 5-8 1.64(a Let x denote the number of...

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HOMEWORKS 5-8 1.64 . (a) Let x denote the number of units that fracture. Then x is binomial with n = 25, π = .20 and we can use Table II. Proportion(x 10) = .011 + .004 + .002 + .000 + ….+ .000 = .017. (b) Proportion(x 5) = .004 + .023 + .071 + .136 + .187 + .196 = .617. (c) Proportion(5 x 10) = .196 + .163 + .111 + .062 + .030 + .011 = .573. (d) Proportion(5< x < 10) = .163 + .111 + .062 + .030 = .366 2.4. The three quantities of interest are: . and , , 1 1 + + n n n x x x Their relationship is as follows: [ ] + + = + + = + + = + = + + = + + = + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n x x n x x n n x x n x n x n n n n n i n i n i i n For the strength observations, 34 . 43 10 = x and 5 . 81 11 = x . Therefore, 81 . 46 1 10 5 . 81 ) 34 . 43 )( 10 ( 11 = + + = x 2.16 (b) ( 29 ( 29 c x nc n x n c x n y n y i i i - = - = - = = 1 1 1 1 Since c x y - = we know that ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 1 1 1 x i i i y s n x x n c x c x n y y s = - - = - - - - = - - = 2.17 . (a) x = i i n x 1 = 14438/5 = 2887.6. The sorted data is: 2781 2856 2888 2900 3013, so the sample median is x ~ = 2888. (b) Subtracting a constant from each observation shifts the data, but does not change its sample variance (Exercise 16). For example, by subtracting 2700 from each observation we get the values 81, 200, 313, 156, and 188, which are smaller (fewer digits) and easier to work with. The sum of squares of this transformed data is 204210 and its sum is 938, so the computational formula for the variance gives s 2 = [204210-(938) 2 /5]/(5-1) = 7060.3.
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