SolutionHW3

SolutionHW3 - Stat 500 Homework 3 Solutions STAT500 HW#3...

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Unformatted text preview: Stat 500 Homework 3 Solutions STAT500 HW#3 Solutions 1) Random sample of 25 generated without replacement from Minitab (Will most likely differ from yours): 2 783 364 84 652 104 562 522 667 240 125 526 584 341 134 155 378 86 428 409 219 789 296 708 21 2) a) P(Y>7) = P(Z> 3 . 1 5 7- ) ≒ P(Z>1.5385) =1-P(z<1.5385 ) ≒ 1-0.93804=0.06196 b) P(Y>5.5) = P(Z> 500 3 . 1 5 5 . 5- )=P(Z>8.60) ~ 0 No, the results of this survey do NOT appear to be consistent with the 1990 census, because if the census results are correct, the probability that the average viewing time would exceed 5.5 hours is very small, close to zero. 3) Individual baggage weight has u = 95 and σ = 35. ⇒ Total weight has mean μ Σ y = nμ = 200 × 95 = 19000 with the standard deviation σ Σ y = σ n = 35 200 × = 494.97. So P ( ∑y > 20,000) = P {z > (20,000 – 19,000)/494.97} = P (z > 2.02) = 1 – 0.9783 = 0.0217. The probability that the total weight of the passengers’ baggage will exceed the 20000-pound limit is 0.0217 or 2.17%. 4) a) μ = 160 and σ = 20 ⇒ P (y<150) = P {z < (150 -160)/20} = P (z < - 0.5) = 0.3085. If a single measurement is taken, the probability that this measurement will fail detect that the patient has high blood pressure is 0.3085....
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This note was uploaded on 10/08/2010 for the course STAT 500 at Penn State.

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SolutionHW3 - Stat 500 Homework 3 Solutions STAT500 HW#3...

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