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Unformatted text preview: 7.94 = 24 minutes, = 7 minutes, and n = 100 7 x = = 24 minutes and x = = = .7 minutes n 100 The sampling distribution of x is approximately normal because n >30 7.96 = 24 minutes, = 7 minutes, and n = 100 7 x = = 24 minutes and x = n = 100 = .7 minutes (x  ) (22  24) = = 2.86 x .7 a. F or x = 22 : z= P (x < 22) = P (z < 2.86) = .0021 b. F or F or x = 23 : x = 26 : z= (x  ) (23  24) = = 1.43 x .7 (x  ) (26  24) = = 2.86 x .7 z= P (23 < x < 26) = P (1.43 < z < 2.86) = P (z < 2.86)P (z < 1.43) = .9979.0764 = .9215 c. P (x within 1 minute of ) = P (23 x 25) F or F or x = 23 : x = 25 : z= z= (x  ) (23  24) = = 1.43 x .7 (x  ) (25  24) = = 1.43 x .7 P (23 x 25) = P (1.43 z 1.43) = P (z 1.43)P (z 1.43) = .9236.0764 = .8472 d. P (x greater than by 2 minutes or more) = P (x 26) F or x = 26 : z= (x  ) (26  24) = = 2.86 x .7 1 P (x 26) = P (z 2.86) = 1  P (z 2.86) = 1  .99979 = .0021 7.99 p = .88, q = 1  p = 1  .88 = .12, and n = 80 r r pq (.88)(.12) p = p = .88, and p = = = .03633180 n 80 np = (80) (.88) = 70.4 > 5, nq = (80)(.12) = 9.6 > 5 Since np and nq are both greater thatn 5, the sampling distribution of p is approximately normal. b 7.109 = 160 pounds, = 25 pounds, and n = 35 25 x = = 160 pounds and x = = = 4.22577127 pounds n 35 Since n > 30, x is approximately normally distributed. P (sum of 35 weights exceeds 6000 pounds) = P (mean weight exceeds 6000/35) = P (x > 171.43) (x  ) (171.43  160) = = 2.70 x 4.22577127 F or x = 171.43 : z= P (x > 171.43) = P (z > 2.70) = 1  P (z 2.70) = 1  .9965 = .0035 2 ...
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This note was uploaded on 03/10/2008 for the course ECON 41 taught by Professor Guggenberger during the Spring '07 term at UCLA.
 Spring '07
 Guggenberger

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