statistic question!!?
967657
When the population standard deviation is unknown, thats when you use the t-distribution, in calculating the confidence interval. \n\nThe t-distribution (also known as Student's t-distribution) measures the significance of a difference of means of small samples when two distributions are though to have the same variance, but possibly different means.\n\nBut as the sample size increases, the t-distribution gets close to the normal distribution.\n\nHope this helps!
877981
For the first part, per hour the rate is 30.  so we have lambda = 30, so we know the mean and variance (look at the answer before mine).  To find the Probability the 240th person enters in the 8th hour, find the probability 240 or more people arrive in an 8 hour day, then times that by the probability less than 240 arrive in 7 hours.\n\nso (1-P(<240 in 8))*P(<240 in 7)\n\nRemember, The poisson process is memoryless.\n\nSteve
49083
This may not be exactly what you are looking for but check it out: http://www.whois.sc/internet-statistics/
605382
just for personal inquiry, where'd you get that statistic?\n(oh and i totally agree, that is kinda sad)
354101
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
885649
The bigger the sample size, the more accurate the estimate.\n\nSay I was estimating the percentage of red m&m's in a pack.\n\nIf I sampled 10 m&m's, I might I might find that the sample contains 1 red one. This could lead me to think that a pack contained 10% red m&m's.\n\nHowever, if I sampled 100 m&m's, I might find that the sample contains 15 red m&m's, leading me to think that a pack contains 15% red m&m's. This answer is impossible using the sample size of 10, as you can't get 1.5 m&m's of a particular colour.\n\nIf I sampled 1000 m&m's I might find that I have 143 red ones, leading me to think that the pack contains 14.3% red m&m's.\n\nSo, as you can see, the bigger the sample size, the closer the estimation comes to the actual answer.
286076
What kind of statistic are you looking for?
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
282573
Statics?  Statistics?
280200
If a random sample size of 25 corresponds to a standard deviation of 30, then you can expect a random sample of size 25N to roughly correspond to a standard deviation of 30 / sqrt(N).\n\nIf 99% of the area under the bell curve is between -Z and +Z standard deviations, then you want 30 Z / sqrt(N) to be less than or equal to 0.5.\n\nMaybe a little more because of the variation of the estimated standard deviation from the true standard deviation.\n\nAlso check that the theory and the measurements both suggest it is appropriate to use the normal or Gaussian distribution.
629177
P(selecting a person at random and he has the symptom and is detected by the procedure)\n\n= P(patient having it) and P(detecting it using the procedure)\n\n= (0.5/100)*(98/100)\n= 0.0049\n\nBut if the person is not randomly selected and you are sure s/he has the problem and its just a matter of detecting it with the procedure, then Probability = (98/100) = 0.98
522796
a) Sensitivity = P(Test + | Has disease), so out of the people who have the disease, 93.4% would test positive for it.\n\nSpecificity = P(Test - | Doesn't have disease), so out of the people who don't have the disease, 98.6% would test negative for it.\n\nb) This is again a Bayes' Rule problem\nP(T+|D) = 0.934\nP(T-|ND) = 0.968\nP(D) = 0.002\n\nP(D|T+) = P(D)P(T+|D)/(P(D)P(T+|D) + P(ND)P(T-|ND))\n= (0.002)(0.934)/(0.002(0.934) + 0.998(0.968))
158411
Start from www.statistics.com or search "statistics books" in Yahoo Answers. Greetings.
735064
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
532490
go to this site, its very good,\n\nit gives clear explanations and statistics
133902
square root of the sample size
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.