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Unformatted text preview: P( Z < z ) = 0.04. The area to the left is 0.04 = Φ ( z ). z = – 1.75 . t x = μ + σ ⋅ z . x = 3,000 + 200 ⋅ ( – 1.75 ) = 2,650 hours . b) (3) What is the probability that a bulb would last over 3,150 hours? P( X > 3,150 ) = > 200 000 , 3 150 , 3 Z P = P( Z > 0.75 ) = 1 – Φ ( 0.75 ) = 1 – 0.7734 = 0.2266 . c) (4) What is the probability that the average lifetime of eight randomly and independently selected bulbs is over 3,150 hours? Since the distribution we sample from is normal (Case 2), . n Z X =σ μ . P ( X > 3,150 ) = > 8 200 000 , 3 150 , 3 Z P = P ( Z > 2.12 ) = 1 – Φ ( 2.12 ) = 0.0170 ....
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 Spring '08
 Kim
 Statistics, Normal Distribution, Probability, Standard Deviation, bulbs

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