Find the 95% confidence interval!?
550389
95%CI = Mean ± (1.96 * (Sample Standard Deviation) / root(Sample Size))\n\n= 114 ± (1.96 * 13)/ root (50)\n=114 ± 3.6 = 110.4 to 117.6\n\nThe answer is (C)
828638
Assuming the population is normal, to calculate the percentage of subjects obtaining 65 or below, we find the Z score that corresponds to 65:\n\nZ score \n= ( score - mean ) / standard deviation\n= (65 - 80) / 6\n= -2.5\n\nLooking up a Z-score of -2.5 in a 'table of standard normal probabilities' (or using statistical software), gives 0.0062 which is 0.62%. \n\nAs for the confidence interval.. it is 'estimate' +/- 'margin of error'\n80 +/- 1.95 * 6 / sqrt(100)\n80 +/- 1.176\nThe 95% confidence interval (78.824,81.176)\n\n(I.e we can be 95% confident that the true population mean lies between 78.824 and 81.176, and this is based our the sample of 100 subjects which had a mean of 80)
650317
If you have a computer program that can compute binomial probabilities, you can use that.\n\nIf you are doing this by hand, however, I would suggest using the normal approximation to the binomial here.\n\nn = number of trials = 200\np = prob. of success = 1/6 if we consider getting a five to be a success.\n\nThen np = 200(1/6) = 33.33 and n(1-p) = 166.67 are both >= 5 or 10 (the cutoff is different for different classes) then the sample size is large enough to use the normal.  This binomial is approximately normal with mean = np = 33.33 and SD = sqrt(np(1-p)) = sqrt(200(1/6)(5/6)) = 5.27.\n\nNow for the continuity correction.  The rectangle over 40 goes from 39.5 to 40.5.  Since we are finding the probability that the number of 5's is at least 40, then we should start at 39.5 so that all of 40 is included.\n\nz = (39.5-33.33)/5.27 = 1.17\nArea to the right of 1.17 on the standard normal is 0.121.
463856
Gosh this is a bit vague - but I think you're referring to why we use 1/F when getting a lower confidence interval or doing a 2-sided test.\n\nThe definition of an F m,n-dsitribution is:\n\n[Chi-square(m)/ m] ÷ [Chi-square (n) / n]\n\nTherefore if we did 1 / Fm,n this would be:\n\n[Chi-square(n)/ n] ÷ [Chi-square (m) / m]\n\nSince "dividing by a fraction is the same as turning it over and mulitplying by it".\n\nSo suppose we want a lower 5% limit for F4,5, ie:\n\nP(F4,5 < ? ) = 0.05\n\nWe can invert both sides:\n\nP(1/F4,5 > 1/?) = 0.05\n\n(Note the sign changes, eg 2 < 3 but 1/2 > 1/3)\n\nBut using the result from earlier 1/F4,5 = F5,4:\n\nSO we now have:\n\nP(F5,4 > 1/?) = 0.05\n\nSo what we do is look at the upper 5% point in the F5,4 -tables which is 6.256.\n\nSo 1/? = 6.256\n\nHence ? = 1/6.256\n\nSo the shortcut rule is to get the value we do "one over the F distribution with the number swapped round"\n\nHope this helps.
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.
715667
1.2 - .86 = .34\n34%\ni think\n\n\nhaha i only clicked on this question cuz my names amy! lol
623028
Answer is 10.2% of the sample(25) have a mean score within 3 points of the population mean score of 220.\n\nSD=10 --> 34% (normal distribution)\n   3 from mean = 3/10 x 34% = 10.2%.
983117
C 49.6%\n\n(or A with rounding to 0 decimal places/1 significant figure)
218675
I suppose the percentage must be token from 25.\n25 - 100% = 25 - 25 = 0\n25 - 25% = 25 - 25x25/100 = 25 - (6 + 1/4) = 18 + 3/4\n\nWhen I see the percentage just as a number, then\n25 - 100% = 25 - 1 = 24\n25 - 25% = 25 - 1/4 = 24 + 3/4
714513
Although your question first appears overly brief, perhaps it does have enough information to give an "interval about 36.84" that encompasses only 10% of the bell curve's area.\n\nMy estimation is that we should therefore choose a value for alpha 'a' that is HALF the area you want ... because the "shaded area" will extend symmetrically on either side of the mean ... 5% each way.\n\nMean = 'mu' = u = 36.84\nStd Dev = 'theta' = o = 3.34\nProbability area = p = 0.50 - 0.05 = 0.45 (left side of curve)\nCritical value = z = 1.65 [from statistical table at back of textbook]\n\nWe wish to find the value of the "raw score" x that corresponds to the given z; we will simply use symmetry to find x2 from x1.  The solution will be given in interval notation [x1,x2], the "confidence limits."\n\nz = -1.65 = [(x - 36.84)/3.34]  NOTE: z is negative for "left side"\n\nSolving for x, we find x1 = 31.33 (rounded).\n\nRepeating the same calculation with positive z= 1.65, we find x2 = 42.35 (rounded).\n\nThe interval I believe you are seeking is [31.33,42.35].\n\nHope this helps!\n\nThe link below contains several Java-based statistical tools you may find helpful.\n\nHope this helps!
660150
That would not be possible without the sample values or some other information.  You can make some statements about possible sample means.  For example, since the standard error of the sample mean would be 0.29/sqrt(25) = 0.058, then\n\nThe probability that the sample mean will be between 3.02-0.058 and 3.02+0.058 is 68%.\nThe probability that the sample mean will be between 3.02-0.116 and 3.02-0.116 is 95%.\n\nand so on.
757107
10.7%\n\n(1030 - 920) / 1030 = 0.107
714579
P (X>=5) \n= 1 - [  P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)] \n\nRandam sample of 100 days from the past 10 yrs... \nC(360*10, 100)
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
1109248
First make sure you want a prediction interval for the default rate and not a confidence interval.  The confidence interval will be for the unobservable rate (average rate) while the prediction interval will be for the individual occurances of either default or not default per random 1000 observations.\n\nSecond, make sure you want the interval for the rate per 1000 or the interval for the rate with a sample of 1000: these are actually different!\n\n\nIf prediction interval for the rate per 1000 observation is your poison then that can be approximated by:\n\np +- A*S*sqrt(1+1/n)\nwhere n is the number of samples of 1000\np is the average default rate of the samples\nA is the 95%ile of the student -t distribution (yes! 95th for a 90% PI: 5% on either end of the +-)\nS is the standard deviation of the rate (think binomial).\n\nYou can look up many ways of creating the interval for a binomial proportion but this is the basic theme.
701818
Gosh.   I have no idea how to do it in excel.... \n\nbut for a normal distribution with an error function that always returns 1 (no error)  then the percentage of the population between -n standard deviations and +n standard deviations is n/(sqrt(2) \n\nSo, if you want there to be 90% of the population between -n and +n so that 5% is below -n and 5% is above +n then \n\n0.9 = n/sqrt(2)\n0.9 * sqrt(2) = n\n\n1.27279221 = n \n\nso,  now we take 1.27279221 time the standard deviation of 0.87 and then subtract it from 10.42 \n\nUnderweight = 10.42 - (1.27279921 * 0.87)\n\n= 9.31266469\n\nand then you do the same thing for 80% (except you add) to get the weight at 10% over....
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.