Statistics questions - help!!?
955889
Alright.\n\nE(S1) = E{(L1/2 + L2/2)^2}\n= Var(L1/2 + L2/2) + E^2(L1/2+L2/2) (Uses the fact that Var(X) = E(X^2) - E^2(X))\n= 0.25(sigma^2) + 0.25(sigma^2) + (L/2 + L/2)^2 (Since L1 and L2 are independent)\n= 0.5(sigma^2) + L^2.\n\nSo the bias of S1 is 0.5(sigma^2)\n\nE(S2) = E{0.5(L1^2 + L2^2)}\n= 0.5E(L1^2) + 0.5E(L2^2)\n= 0.5(Var(L1) + E^2(L1)) + 0.5(Var(L2) + E^2(L2))\n= 0.5(sigma^2) + 0.5L^2 + 0.5(sigma^2) + 0.5L^2\n= sigma^2 + L^2\n\nSo the bias of S2 is sigma^2.\n\nSo that means that S1 has the lower bias.\n\nPerhaps S3 = L1L2 is unbiased?\n\nE(L1L2) = E(L1)E(L2) (Since both are independent)\n= L*L\n= L^2\n\nBias is zero.  There you go!
412043
I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.
308472
Find yourself a z-statistic table and look up critical value in the table.  Check your stats book on how to interpret the table.\nhttp://www.math2.org/math/stat/distributions/z-dist.htm
783821
5 out of the 8 items in Box A are non-defective.\n3 out of the 5 items in Box B are non-defective.\n\nThe probability that an item from Box A and an item from Box B are drawn and are both non-defective is:\n\n5/8 * 3/5 = 3/8\n\nThe probability that an item from Box A and an item drawn from Box B and both defective is:\n\n3/8 * 2/5 = 3/20.\n\nThis means that the probability that one of the two items drawn is defective is:\n\n1 - p(both are not defective) - p(both are defective)\n=1 - 3/8 - 3/20\n= 19/40\n\nWe can see this by showing all possible outcomes. The items in Box A are shown on the top edge, and the items in Box B are shown on the right edge\n\nN  N  N  N  N  D  D D\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nB  B  B  B  B  C  C  C D\nB  B  B  B  B  C  C  C D\n\nThe events marked "A" represent 2 non-defective items. The events marked "B" represent 1 defective and 1 non-defective item. The events marked "C" represent 2 defective items.\n\nOut of 19 events where there is 1 defective and 1 non-defective item, the defective item came from box A 9 times.\n\nSo if 1 item is defective and 1 is not, the probability is 9/19 that it came from box A
292525
several classes (100 students)of 6th graders went on a field trip to the zoo. 1/5 of them went to c the monkeys. 1/2 of them visited the penguins. 1/4 went 2 c the bats. and the rest went to the snack bar. how many students went to the snack bar? if u hav a heart. u will help \n\nwell...i got 5 students went to the snack bar...\n1/5 of 100=20\n1/2 of 100=50\n1/4 of 100=25\nso \n  20+50+25=95\n  100-95=5 students\n~Brendon
586725
The answer would be C.\n\nIn a two tailed test, you're looking to see if the mean is either significantly greater or less than k which is why it is called two-tailed.  The null hypothesis would be that mu = k and the alternate, not equal.  Thus, c
956188
What you're looking for is the standard deviation.  If the data values are X and the mean is XBAR, then\n\nS.D. = sqrt{[SUM(X - XBAR)^2] / n}\n\nn is the number of data items, and SUM means to add up the squares of the differences between the data values and the mean.\n\nThe lower the standard deviation, the more tightly packed the data.  If the data is normally distributed, about 68% of the data will lie within one standard deviation of the mean, and 95% will be within two standard deviations.\n\n[Edit:  In light of the previous answers, I should add that the standard deviation is the square root of the variance.]
780864
arrange the numbers in ascending sequence then the first 25% make the first quartile and so on\n\nso...\n\n1,1,1,2,2, 3,4,5,9,9, 10,12,12,15,20 21,21,21,23,28\n\nthe first quartile is 1-2,\nsecond 3-9\nthird10-20\nfourth 21-28\n\nIf you want to find the 95th percentile it's 95% of the way up.\nIn the sequence above the 23rd value (21) is 92% and the 24th value (23) is 96%.  So the 95% percentile lies 75% of the way between these values.  (23-21) * 0.75 = 1.5 so the 95th percentile is 22.5\n\nThe median is the 50th percentile (9.5)\n\nIf the median > mean then the distribution is positively skewed\nIf the median < mean then the distribution is negatively skewed\nIf the median = mean then the distribution is normally distributed
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
919178
.................... / ....... &#92;\n................ 0.85 ..... 0.15 (Pass test 1)\n.............. / ....... &#92;\n.......... 0.85 ..... 0.15 (Pass test 2)\n........ / ....... &#92;\n... 0.85 ..... 0.15 (Pass test 3)\n(Fail all 3)\n\n1) How many outcomes does your tree show?\n\nThere are four outcomes:\nP(Pass Test 1) = 0.15 = 15%\nP(Pass Test 2) = 0.85 * 0.15 = 12.75%\nP(Pass Test 3) = 0.85 * 0.85 * 0.15 = 10.8375%\nP(Fail all 3) = 0.85 * 0.85 * 0.85 = 61.4125%\n\n2) What is the probability she will eventually pass an exam.  \n\nThis is the sum of all outcomes where she passes:\nP(Pass Test 1) + P(Pass Test 2) + P(Pass Test 3)\n= 38.5875%\n\nAlternatively you could take the 1 - P(Fails all 3) and get the same answer.\n\n3) What is the probability she will take the exam three times?\n\nIt didn't say whether she passes or fails, so the chance she has to take the test a third time is 0.85 * 0.85 = 72.25%\n\n4) What is the expected number of times she will take the exam?\n15% of the time, she'll take it once\n12.75% of the time, she'll take it twice\n72.25% of the time, she'll take it three times.\n\nSo the expected number of times is:\n0.15 * 1 + 0.1275 * 2 + 0.7225 * 3 = 2.5725\n\nSo she's expected to take it 2.5725 times.\n\n5) What is the expected number of times she will fail the exam?\n15% of the time she will pass test 1 (failures = 0)\n12.75% of the time she will pass test 2 (failures = 1)\n10.8375% of the time she will pass test 3 (failures = 2)\n61.4125% of the time she will fail all 3 (failures = 3)\n\n0.15 x 0 + 0.1275 x 1 + 0.108375 x 2 + 0.614125 x 3 = 2.186625\n\nSo she's expected to fail it 2.186625 times.
387360
The problem is incompletely or improperly described.\n\nEach member of the population has some characteristic value.  The mean is the average of this value taken over the entire population, which, assuming each member is an equal contributor, is M = sum(values)/N, where N is the population.  If each member has the same value, say 515, then mean would be 515 and the standard deviation would be zero.  If each even-numbered member has value 1030, and each odd-number member has value 0, the mean is 515 and the standard deviation is (approximately) 515.\n\nSo, you have provided enough information to solve the problem satisfactorily.
912853
1.) Ms. Jordan 275 / 2500 = 0.11 \n     0.11  x  100 = 11%\n\n2.) 900 students x 1.07 = 963 students .\n\n3.) Virus Scan  300 sec / 0.40 = 750 sec for 100%\n\n4.) Gasoline... 8882 / 0.85 = 10,449 BBLs per day.\n\n5.) Jerry's new salary is\n     18,500 x 1.105 = $ 20,442.50\n\n6.) Chen's Investment wound up being \n     6400 x 0.918 = $ 5875.20\n\nHave a nice day,\nZahbudar
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.
1025917
i should know this, I did a semester of statistics....\n\n hmmmm.....\n\n I guess I should have went to class more...... I hope I passed the exam.....\n\n good luck....\n\n..
677379
If the population has a mean of 58 and a sd of 8, then the sample means of size 100 will have a mean of 58 and a sd of 8/sqrt(100) = 0.8\n\na) Within 3 of 58 means that the sample mean will be between 58-3 = 55 and 58+3 = 61\n\nz = (55-58)/0.8 = -3.75\nz = (61-58)/0.8 = 3.75\n\nFrom the table, the area between the two is almost 1.\n\nb) The mean weight is less than 2.5 of 58 or more means less than 58-2.5 = 55.5\n\nz = (55.5-58)/0.8 = -3.13\n\nFrom the table, the are greater than that is almost 0.\n\nc) z = (57.5-58)/0.8 = -0.63\nz = (58.7-58)/0.8 = 0.88\n\nFrom the table, the area between is 0.811 - 0.264 = 0.547.\n\nd) 56 +- (1.96)(0.8)\n56 +- 1.568\n(54.432,57.568)
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
1035104
x = total hits\ndoubles = 1/7 of x, or x/7\nhome runs = 12.5% of x or 0.125x\nsingles = 82\n\nx = x/7 + 0.125x + 82\nmultiply both sides by 7\n7x = x + 0.875x + 574\nsubtract x from both sides\n6x = 0.875x + 574\nsubtract 0.875x from both sides\n5.125x = 574\ndivide both sides by 5.125\nx = 112\n\nthe kid had 112 hits