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Normal Distribution!!!急
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發問:
The number of arrivals per minute at a drive-in fast food outlet is modelled by Poisson distribution with mean λ. During a Saturday evening, λ = 0.78.i). Using a suitable approximating distribution, calculate the probability that there are more than 40 arrivals between 7 pm and 8 pm.Due to... 顯示更多 The number of arrivals per minute at a drive-in fast food outlet is modelled by Poisson distribution with mean λ. During a Saturday evening, λ = 0.78. i). Using a suitable approximating distribution, calculate the probability that there are more than 40 arrivals between 7 pm and 8 pm. Due to capacity constrains, the management of the fast food outlet want the probability of more than 40 arricals in an hour to be 0.02. They ask a statistician to determine the value to which λ should be reduced. ii), Show that λ must satisfy the equation 40.5 - 60λ / √60λ = 2.054 我想問 2.054 點計出黎??? iii). Use the substitution u= √60λ to formulate a quadratic equation in u. Solve this equation to show that the value of λ is just less than 05. 吾係好明佢想我點.. 更新: THX呀..可5可以教多我呢條數呀?? Every day Morse attempts the crossword puzzle in his newspaper. The time taken, X minutes, to complete the crossword may be modelled by a X~N (22 , 4.5^2) i). What length of time would be enough for Morse to finish the crossword on 95% of days? 更新 2: Each day Morse takes a train to work. The journey takes 25 minutes. He starts his crossword at the beginning of his journey. iii) Find the peobability that he completes the puzzle by the end of his journey at least twice in a five-day week. 更新 3: iv) Morse chnages his newspaper and finds that sorry9% of occasions he completes the crossword during his morning train journey. Assuming that the time taken, Y minutes, to complete the crossword has the distribution N(18, σ^2), find the value of σ.
最佳解答:
圖片參考：http://www.upload-jpg.com/images.php/d0947320/11.jpg 圖片參考：http://www.upload-jpg.com/images.php/5b98a311/12.jpg 仲有唔明可以再問 ~ 2010-10-29 09:31:48 補充： (i) P(Z>1.645)=0.05 Therefore, X-22/4.5 = 1.645 X=29.4025 (ii) P(0 < X < 25) =P(0 < Z < 25-22/4.5) =P(0 < Z < 0.6666) Using Linear Interpolation, (x - 0.2514)/(x - .2546) = (0.6666 - 0.67)/(0.6666 - 0.66) x=0.2525 So, P(0 < Z < 0.6666) = 0.5 - 0.2525 = 0.2475 (continue) 2010-10-29 09:34:13 補充： P(complete in 25 mins at least twice in five days) =1 - P(complete in 25 mins in 0 day) - P(complete in 25 mins in 1 day) =1 - (1-0.2475)^5 - 5C1(0.2475)(1-0.2475)^4 =0.3619 2010-10-29 09:46:11 補充： (iii) 9% of occasions, he completes the puzzle. P(0 < Y < 25) = 0.09 P(0 < Z < 25-18/σ) = 0.09 P(0 < Z < 7/σ)= 0.09 Using Linear Interpolation, (x - 0.23)/(x - 0.22) = (0.41 - 0.409)/(0.41 - 0.4129) x = 0.2274 Therefore, 7/σ = 0.2274 σ = 30.7779 2010-10-29 09:48:25 補充： last question is 9%?, you typed some meaningless thing there. 2010-10-29 11:14:44 補充： Sorry, i made some mistakes. (ii)P(0 < X < 25) =P(-22/4.5 < Z < 25-22/4.5) =P(-4.8888 < Z < 0.6666) =0.5+(0.5-0.2525) =0.7475 P(complete in 25 mins at least twice in five days) =1 - P(complete in 25 mins in 0 day) - P(complete in 25 mins in 1 day) =1 - (1-0.7475)^5 - 5C1(0.7475)(1-0.7475)^4 =0.9938 2010-10-29 11:22:32 補充： (iii)P(Y < 25) = 0.09 P(Z < 25-18/σ) = 0.09 P(Z < 7/σ)= 0.09 7/σ = 1.34 σ=5.22388
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