ST 558 Homework 6

# ST 558 Homework 6 - Kelly Harrison ST 558 D Homework 6...

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Kelly Harrison 2/21/10 ST 558 D Homework 6 PROBLEM 1 μ = 30 σ = 6 a) P(x > 17) = P(z > (17-30)/6) = P(z > -2.17) = .9850 b) P(x < 22) = P(z < (22-30)/6) = P(z < -1.33) = .0918 c) P(32 < x <1) = P((32-30)/6 < z < (41-30)/6) = P(.33 < z < 1.83) = .3371 d) 80% z = .84 x = μ + z σ = 30 + .84(6) = 35.04 e) Middle 75% z = ± 1.15 x = 30 ± 1.15(6) = (23.1, 36.9). PROBLEM 2 μ = 18, σ = 2.5 a) P(x < 15) = P(z < (15-18)/2.5) = P(z < -1.2) = .1151 b) P(x < k) = .2236 P(z < z k ) = .2236 z = -.76 x = μ + z σ = 18 + (-.76)(2.5) = 16.1 c) P(x > k) = .1814 P(z > z k ) = .1814 z = .91 x = μ + z σ = 18 + .91(2.5) = 20.275 d) P(17 < x < 21) = P(-.4 < z < 1.2) = .5404.

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PROBLEM 3 A soft drink machine is regulated so that is discharges an average of 200 milliliters per cup. μ = 200 and σ = 15 a) P(x > 224) = P(z > (224-200)/15) = P(z > 1.6) = .0548 b) P(191 < x < 209) = P((191-200)/15 < z < (209-200)/15) = P(-.6 < z < .6) = .4515 c) P(x > 230) = P(z > (230-200)/15) = P(z > 2) = .0228 .0228(1000) = 2.8 Answer: Approximately 2.8 cups. d) Smallest 25% z = -.67 x = μ + z σ = 200 + (-.67)(15) = 189.9. PROBLEM 4 A company pays its employees an average wage of \$15.90 an hour with a standard deviation of \$1.50. μ = 15.90 and σ = 1.50 a) P(13.75 < x < 16.22) = P(13.75-15.90)/1.5 < z < (16.22-15.90)/1.5) = P(-1.43 < z < .21) = .
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