mean number of errors per chapter is 0.8. Find the probability that there are less than 2 errors in a chapter?
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In the experiment of testing for the number of errors in a chapter...\n\nLet the event an error appearing in a chapter be a success..\n\nLet "x" represent the number of success.\nThe number of errors in a chapter can be either 0,1,2,...\n==> "x" can carry the values 0,1,2,3,...\n==> "x" is a discrete random variable with range = {0,1,2,..} [Countably infinite]\n\nThe probability distribution of "x" is a poisson distribution. \n[Since there are a large number of trials all of which are identical and independent with only two possible outcomes in each trial, either a success (there being an error) or a failure (there not being an error), with the probability of success in all the trials remaining the same all throught ]\n\nThe probability function of a poisson distribution\n==> P(X = x) = {e^(-m) * m^(x)}/x!\n \nParameter of the distribution\n==> Mean of the distribution\n==> Mean number of errors per chapter =0.8 \n==> m = 0.8\n\nTherefore P(X = x) = {e^(-0.8) * (0.8)^(x)}/x!\n\nProbability that there are less than 2 errors in a chapter \n==> P(x < 2) \n= P(x=0) + P(x=1)\n= {e^(-0.8) * (0.8)^(0)}/0! + {e^(-0.8) * (0.8)^(1)}/1!\n= {e^(-0.8) * 1)/1 + {e^(-0.8) * (0.8)}/1\n= e^(-0.8) + {e^(-0.8) * (0.8)}\n= e^(-0.8) [ 1 + 0.8]\n= e^(-0.8) [1.8]\n\n\nhttp://www.futureaccountant.com/probability/
781591
P(wrong size) = 0.2\nP(defective) = 0.35\nP(not defective and right size) = 0.60\n\nThus P(defective or wrong size) = 1 - P(not defective and right size) (These two events are opposites.  Look at the Venn diagram for help with this, it's DeMorgan's law if you're curious)\n= 1 - 0.60 = 0.40\n\nP(defective or wrong size) = P(defective) + P(wrong size) - P(defective and wrong size)\n\nSo 0.40 = 0.2 + 0.35 - P(defective and wrong size)\n0.40 = 0.55 - P(defective and wrong size)\nP(defective and wrong size) = 0.55 - 0.40 = 0.15
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Pick one knight anywhere.\nIf you pick either of the knights directly adjacent to him, you are done.  This happens 2/24 of the time.\n....OOXXOO...\n\nIf you pick a knight that is separated by exactly one seat (2/24 of the time), then there are 3 out of 23 seats that are adjacent.\n...OXOXO....\nIf you pick a knight that is more than one seat away (20/24), then there are 4 out of 23 seats that are adjacent.\n...OXO...OXO...\n\nSo the probability is:\n1/12 + 1/12 * 3/23 + 10/12 * 4/23\n\nAdding it up:\n23/276 + 3/276 + 40/276, or 66/276 = 33/138\n\nThis is approximately 23.91%
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The binomial distribution is given by:\n\n    N!\n----------- x (p)^n x (1-p)^(N-n)\nn!(N-n!)\n\nWhere N is the number of trials, n is the number of successes and p is the probability of success\n\nIn this question, n, the "number of successes" is the number of failures caused by operator error. p, the "probability of success" is the probability that the failure is caused by operator error (i.e. 30%).\n\nSo, in the questions you need to substitute the following for each variable:\n\ni) N=20, p=0.3\n\nAs this question uses the phrase "at least", you will need to use the equation more than once and add all the results. Repeat the calucation for each n for n=10 to n=20 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=10, 0.03081708\nn=11, 0.01200665\nn=12, 0.00385928\nn=13, 0.00101783\nn=14, 0.00021811\nn=15, 0.00003739\nn=16, 0.00000501\nn=17, 0.0000005\nn=18, 0.00000004\nn=19, 0.00000000\nn=20, 0.00000000\n\nTotal: 0.0479619 or 4.80% chance that at least 10 are caused by operator error\n\nii) N=20, p=0.3\n\nThis question is similar to q1, however you need to repeat the calculation for each n for n=0 to n=4 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=0, 0.00079792\nn=1, 0.00683934\nn=2, 0.02784587\nn=3, 0.07160367\nn=4, 0.13042097\n\nTotal: 0.23750778 or 23.75% chance that no more than 4 are caused by operator error\n\niii) In the random sample, only 5 are errors. Work out the probability of this being the case using N=20, n=5, p=0.3. If it seems reasonable, then the 30% figure is probably right, and the sample just contains a low number of failures due to operation error. If the probability is impossibly small, then the 30% is not correct.\n\nThe probability I calculated for n=5 was 0.178863051. This indicates a 17.89% chance that there would be 5 errors in the sample. This is not so outrageous to suggest that 30% is an incorrect percentage of operator errors. It's just that in this sample, the number of actual errors was less than the number of expected errors (which is 30% x 20, or 6 errors).
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If the claim is correct, then the mean of samples of size 25 has a normal distribution with a mean of 10 (the population mean) ounces and a standard deviation of 0.5/sqrt(25) = 0.1.\n\nz = (10.15-10)/0.1 = 1.5\n\nAccording to the standard normal table, the probability is 0.0669.\n\nThis is due to the fact that the population is normal.\n\nedit: You guys, it is not three sd's above.  You have to think in terms of standard error.  It is 1.5 standad errors above.  Do the math.
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the Chebyshev's inequality goes Pr(|X-mean| > or = k.sd)< or = 1/k^2\n\ni m assuming u meant standard deviation when u said 4.3818, which is 2 sd.\n\nThe prob of points outside 2sd at both sides is not more than 1/k^2.. or 1/2^2=1/4.\n\nSo, the probability of points within 2sd will be at least 1-1/4=3/4.\n\nhow many data points? if u have 100 data points in total, there will be at least 100x3/4=75 data points within the mean+/-4.3818 range.\n\nNote: The exactly probability depends on the type of distribution the data follows, for example, for Normal/Gaussian, 95% lies within +/- 1.96sd. Chebyshev's inequality states the minimum possible for any type of distribution.
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heres 1 from my university. dont know how much itll help.\n\nhttp://www.upscale.utoronto.ca/PVB/Harrison/ErrorAnalysis/index.html
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a) \nThe probability Anna passes is 1/2\nThe probability Boris passes is 4/9\n\nThe probability that they both pass is the product of the two probabilities:\nBoth = 1/2 * 4/9 = 4/18\nReduce this to 2/9:\n\nProbability they both pass = 2/9\n\nb) \nFirst, a common mistake would be to assume that the chance that they both fail is 1 minus the probability they both pass, but it isn't.  Instead, first figure the probability that Danny passes.  This is just the reverse of the problem above.\n\nThe probability Chloe passes = 3/5\nThe probability Danny passes = d\nThe probability they both pass = 7/15\n\n3/5 * d = 7/15\nd = 7/15 / 3/5\nd = 7/15 * 5/3\nd = 35/45\nd = 7/9\n\nSo the probability that Danny passes is 7/9\n\nTo figure the chance they both fail, first figure their individual probabilities of failure.\nProbability that Chloe fails = 1 - 3/5 = 2/5\nProbability that Danny fails = 1 - 7/9 = 2/9\n\nProbability they both fail is the product of 2/5 and 2/9.\n\nChance they both fail is:\n2/5 (Chloe fails) x 2/9 (Danny fails) = 4/45
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1.  Using your Normal Distribution table (sorry, mine is buried under my dirty laundry), find the Std Devs that represents .18 (.50 - .32).  Multiply by 30 (Std Dev).  This number represents the number of pages short of 150 (the mean).  It's probably around 140.\n\n2.  The second group (.18 below 150 pages + .25 above 150).  Repeat the process for 1 above. (about 140 to 165)\n\n3.  The third group is from .25 to .49 above the mean (about 165 to 240)\n\nGood luck.
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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.
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I think that you would need at least 2 values in each cell so that the within groups SD can be estimated, so if each of the two ways has two factors then you would need a sample of at least 8, so that there is two for each cell.  Could be wrong though.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
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Note: they aren't asking just for the chance that at least 2 children have blue eyes.  They are asking for the probability, given that you know at least one already does.\n\nThis is conditional probability.\n\nP(B | A) = P(A and B) / P(A)\n\nSo the probability that you have at least 2, given you have at least 1 is the probability you have at least 2 divided by the probability you have at least 1.\n\nFor that part though, you would use binomal distribution.\n\nFirst, what is the probability that at least one child has blue eyes?\nThis would be 1 minus the probability that nobody has blue eyes.  \nP ( blue >= 1 ) = 1 - (0.75)^5 \nP ( blue >= 1 ) = 0.762695312 \n\nNow the probability that at least *two* children have blue eyes:\nP( blue = 2 ) + P ( blue = 3 ) + P ( blue = 4 ) + P ( blue = 5 ).  Again you can figure the negative statement and subtract from 1.  So this would be equivalent to:\n1 - P ( blue = 0 ) - P ( blue = 1 ).\n\nThis would be:\nP ( blue >= 2 ) = 1 - (0.75)^5 - (0.25)(0.75)^4 * C(5,1)\nThe C(5,1) means the ways to choose 1 person from 5, which is just 5.\nP ( blue >= 2) = 1 - (0.75)^5 - (0.25)(0.75)^4 * 5 \nP ( blue >= 2) = 0.3671875\n\nSo putting it all together:\nP ( blue >= 2 | blue >= 1 ) = P ( blue >= 2 ) / P ( blue >= 1 )\n= 0.3671875 / 0.762695312\n= 0.481434059\n≈ 48.1%
1076864
Put the data of each set into bins according to their leading digit. Then according to Benford's law the distribution should be as follows:\nLeading digit Probability\n1 30.1%\n2 17.6%\n3 12.5%\n4 9.7%\n5 7.9%\n6 6.7%\n7 5.8%\n8 5.1%\n9 4.6%\n\nUsing a chi-squared or Kolmogorov test it is possible to determine which of the four data sets differs significantly from Benford's distribution.
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In 36 possibles outcomes\nprobability of doubles\n1-1\n2-2\n3-3\n4-4\n5-5\n6-6\n6 in 36 or 1 in 6 or 16,666%\nOf the sum being 4\n1 +3\n2+2\n3+1\n3 in 36 or 1 in 12 or 8,333%\nIf it is combined probability you can sum the two\n16,66+8,33=25%
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It depends on whether you want the mean number of heads, the total number of heads, etc.\nIf you want the total, multiply the standard deviation by sqrt(n), where you have n tosses.\nIf you want the average, divide by sqrt(n).\n\n(Note, the formula talked about below is for the *margin* of error, not the standard error. And thats only for means.)
906595
C.\nThis is a binomial probability problem.\nThe binomial formula says \nP(r out of n) = (nCr)(p^r)(q^(n-r)), where n is the number of trials (here 6 rolls), r is the number of successes (here 0, 1, 2, and 3 twos), p is the probability of success (here 1/6), and q is the probability of failure (here 5/6)\nYou need to plug in 0, 1, 2, and 3 for "r" into this formula.  If you do, you get:\nP(0 twos) = .3348979767\nP(1 two) = .401877572\nP(2 twos) = .200938786\nP(4 twos) = .05358367627\nYou add up these answers to get your final answer, about .9913