Hw3anskey - Peter Zeitz HW3 Answer Key Part I Multiple...

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Peter Zeitz HW3 Answer Key Part I: Multiple Choice 1) d 2) c n = 500 ^ p = 0 : 75 z 1 ° 0 : 05 2 = 1 : 96 ° ° x = p ^ p (1 ° ^ p ) p n h ^ p ° z 1 ° 0 : 05 2 ° ° x ; ^ p + z 1 ° 0 : 05 2 ° ° x i = [0 : 713 ; 0 : 787] 3) c 4) b n = z 2 ° 2 x E 2 = (2 : 58) 2 (18 : 65) 2 4 : 8 2 = 100 : 49 round up - > 101 5) c since E ° ° X ± = E ° 1 n P X i ± = 1 n P E [ X i ] = 1 n = ± 6) d 7) b 8) c 7.94 ± = 24 , ° = 7 , n = 100 = ) ± ° x = ± = 24 : ° ° x = ° p n = 7 p 100 = 0 : 7 The sampling distribution of ° x is approximately normal because n > 30 : 7.96a . For ° x = 22 : z = x ° ± ) ° ° x = 22 ° 24 0 : 7 = ° 2 : 86 : P x < 22) = P ( z < ° 2 : 86) = 0 : 0021 7.96b. For ° x = 23 : z = x ° ± ) ° ° x = 23 ° 24 0 : 7 = ° 1 : 43 : For ° x = 26 : z = x ° ± ) ° ° x = 26 ° 24 0 : 7 = 2 : 86 P ( z < 2 : 86) ° P ( z < ° 1 : 43) = 0 : 9215 7.96c. P x within 1 minute of ± ) = P (23 ± ° x ± 25) For ° x = 23 : z = x ° ± ) ° ° x = 23 ° 24 0 : 7 = ° 1 : 43 : For ° x = 25 : z = x ° ± ) ° ° x = 25 ° 24 0 : 7 = 1 : 43 P ( z < 1 : 43) ° P ( z < ° 1 : 43) = 0 : 8472 7.96d. P x greater than ± by 2 minutes or more ) = P x ² 26) For ° x = 26 : z = x ° ± ) ° ° x = 26 ° 24 0 : 7 = 2 : 86 : P ( z ² 2 : 86) = 1 ° P ( z < 2 : 86) = 0 : 0021 7.99 p = 0 : 88 , q = 0 : 12 ; n = 80 : ± ° p = p = 0 : 88 , ° ° p = p pq n = q (0 : 88)(0 : 12) 80 = 0 : 036 np = (80) (0 : 88) = 70 : 4 > 5 , nq = (80) (0 : 12) = 9 : 6 > 5 Since np and nq are both greater than 5, the sampling distribution of ^ p is approximately normal. 7.109 ± = 160 ; ° = 25 ; n = 35 : ± ° x = ± = 160 , ° ° x = ° p n = 25 p 35 = 4 : 23 Since n > 30 ; ° x is approximately normal P ( sum of 35 weights exceeds 6000 pounds ) = P ² mean weight exceeds 6000 35 ³ = P x > 171 : 43) For ° x = 171 : 43 : z = x ° ± ) ° ° x = 171 : 43 ° 160 4 : 23 = 2 : 70 P ( z ² 2 : 70) = 1 ° P ( z < 2 : 70) = 0 : 0035 8. 99 ° = 578
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