Px521 normdist5250102 normdist4550101 if it says area

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P(X>52)=1-Normdist(52,50,10,2) P(45≤X≤57)=Normdist(57,50,10,1)-Normdist(45,50,10,1) If it says area to the right of the curve that means 1-Normdist, if it says area to the left its just Normdist. Ex] if μ=35 and σ=5 P(X<k)=.75→ k=Norminv(.75,35,5) P(X>k)=.6 → k=Normin(.4,35,5) Tdist(x,degrees of freedom, tails) Tinv(probability, degrees of freedom) Ex] P(t 20 >x)=20% → x=tinv(.4,20) P(-k<T<k)=0.90 → =Tinv(0.1,15) P(T>2)= → =Tdist(2,15,1) P(T<-1.75)= → = =Tdist(1.75,15,1) Population Mean~tmultiple, Population Proportion~zmultiple x +- tmultiple * or p +- zmultiple * (sqrt everything) tmultiple= 90% ci→ tinv(10%,n-1)
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95% ci → tinv(5%, n-1) 99% ci → tinv(1%, n-1) zmultiple= 90% ci → norminv(.95 95% ci → norminv(.975) 99% ci → norminv(.995) If no point estimate use mean for the point estimate. *When doing population proportions the sample proportion has to be in decimal form (10/100=.1) Based on your confidence interval, can you conclude that the majority of the customers wait less than 1 minute? Answer: This confidence interval tells that the proportion of the people waiting more than 1 minute are less than 50%, in other words we can conclude that the majority of the customers wait less than 1 minute. 1. What is the probability that 20 hours will pass before a failure is observed? Let Y denote the time between two failures. P(Y>20)=1-EXPONDIST(20,0.1,1)= 0.1353. 2. What is the probability that there will be exactly two machine failures during the next hour? Let X denote the number of failures during the next hour. P(X=2)=POISSON(2,0.1,0)= 0.0045. Let X denote the number of failures during the next hour. P(X=2)=POISSON(2,0.1,0)= 0.0045. 3. What is the expected number of failures in an hour? E(X)=λ= 0.1 (Earlier in the problem: the life of a certain type of device has an advertised failure rate of 0.1 per hour)
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  • Spring '11
  • kjghkjh
  • Statistics, ex, inoculated- BINOMDIST problem

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