Given the following Golf Scores: 102, 88, 100, 100, 99, 67, 100, 100, 98, 77, 99, 88 \n\nWhat Percentage of the scores are within three standard deviations of the mean? Compare your answer to that predicted by Chebyshev's Theorem? \n\nWith the mean at 93.16667 and the Standard Deviation at 10.63
204478
93.16667 =- 3*10.63 \n61.28 to 105.04, so all of the values fit in this range.\n\nThe previous answerers have been wrong about Chebyshev Inequality.  According to that, there should be 1-(1/k)^2 values within k standard deviations of the mean.  So, according to the theorem, there should be 1-(1/3)^2 = 8/9 = 88.89% within 3 SD's of the mean.
478501
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.
541212
z score= (n-mean)/standard deviation\nstandard deviation = square root of variance\n\nz score = (50-24)/ 5\n            = 26/5\n            = 5.2\n\nThe z score of the data is 5.2, meaning n is 5.2 standard deviations from the mean.
301374
Dear Cyber Friend:\nIf the problem states that the data (women heights in this particular case!) is normally distributed, it means the distance from the mean (provided as 63.6) is the same from any point indicated by the standard deviation (provided as 2.5). All you have to do is picture the distribution starting from the first quartile (they are asking you to calculate the third one though). Thus, Q1=mean minus 2 times standard deviation = 58.6, Q2=mean minus 1 time standard deviation = 61.1, Q3=mean plus 1 time standard deviation = 66.1 (your answer to the problem), Q4= mean plus 2 times standard deviation = 68.6 (just in case!)\n\nGot the idea? I hope this helps! Good Luck!
173231
You have to standardize it:\nLow point: (40-50)/10=-1 \nHigh point: (60-50)/10=1\nNumber you want: (x-50)/10=Z\nSo you are looking for the area under the normal distribution curve from -1 to 1\nThis is "one standard deviation" which is common to use, so any statistician will tell you "Oh, that's .3413" but I'm not a statistician so...\nThis curve is symmetric so you want twice the area from 0 to 1. In the Z-table (found in any statistics book or online) you look up z=1.00.  This gives a value of .3413.\nDouble that and you get .6826 or 68.26% probability. So for a thousand students, 1000*0.6826=683 will get between 40 and 60.\nThe other range is found the same way:\n(30-50)/10=2 (2 standard deviations)\nfrom Z table: 0.4772\nDouble that and you get 0.9544 or 95.44% so 954 students would get between 30 and 70.
422517
Your data set would look like this:\n\n10 10 10 10 10 10 10 10 10 10 20 20 20 20 20 20 20 20 235 235\n\nThe mean is 730/20 = 36.5.\nThe median is (10+20)/2 = 15.\nThe mode is 10.\n\nSo certainly what was stated by the dean isn't a lie.\n\nNow, if I were to calculate, say, the average number of classmates for each person in the statistics classes, there would be 100 9's, 160 19's, and 470 234's.  So the average number of classmates for people taking these classes is approximately 113,920/730 = 156.05.  The median is 234 and the mode is also 234.  So the newspaper isn't really lying either.\n\nI guess you can say that they are looking at this from two different perspecitves, so in that sense, there are two truths.
344393
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....
275518
The probabilities that we know are\nP(M) = .486\nP(F) = .514\nP(G and M) = .174\nP(G and F) = .045\nNow, \na) P(G) = P(G and M) + P(G and F)\n= .174 + .045\n= .219\nb) .174/.486 = .3580\nc) .045/.514 = .0875 Oops, this is golf, given female.\nFor c) .0875/.219 = .3995\nPlease note that blahb31 is wrong for b and c\n.174 stands for the P(plays golf AND is male)\nThus, P(G|M) = P(G and M)/P(M) = .174/.486\nFor c)\nP(F|G) = P(G|F)P(F)/P(G)\nP(G|F) = P(G and F)/P(F) = .045/.514 = .0875\nP(G) = .219 (from part a)\nThus, P(F|G) = .0875/.219 = .3995
267158
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.
506603
If 1 man is chosen, then the z score would be\n\nz = (180-170)/30 = 0.33\n\nThe area to the right of this value is 0.3707.\n\nIf 100 men were chosen, then the sample mean has a normal distribution with a mean of 170 and a standard deviation of 30/sqrt(100) = 3.\n\nz = (180-170)/3 = 3.33\n\nThe area to the right of this value is 0.0005.\n\nThe z score for the other part is\n\nz = (160-170)/3 = -3.33\n\nThe area to the left of this value is 0.0005.
542350
Take the three and test them on a launch monitor with a certified club fitter or PGA pro.  They will be able to tell which one is best for your swing.  They are all good drivers.  The major differences are with trajectory and forgiveness.  A fitter or pro will look at your skill level (be honest about your handicap), swing speed, ball speed, and launch angle......and will be able to advise the best one for you.\n\nOne more thing to try....go on eBay and see if you can find one of the newer TM's for a better price.  I have a TM R7-425 with a 75g V2 shaft and I love it !!!!!
411814
10% > is z score of 1.281551\n7% < is z score of -1.475791\n\nz = (x - μ) / σ\n\n1.281551 = (16.3 - μ) / σ\n\n-1.475791 = 15.5 - μ) / σ\n\nMulitply by σ:\n\n1.281551σ = 16.3 - μ\n-1.475791σ = 15.5 - μ\n\nSubtract one from the other:\n1.281551σ + 1.475791σ = .8\n\n2.757342σ = .8\nσ = 0.29013448\n\nPlug that back into the equation:\n1.281551 (0.29013448) = 16.3 - μ\nμ = 15.9281779\n\nstandard deviation is ~.29 oz. and mean is ~15.93 oz.\n\nIf you go back and do that common sense check, you will see this fits in the bell curve between the percentages stated.  Only 10% is more than 16.3 oz and only 7% is less than 15.5 oz.  Since this is a normal distribution, the number has to be between these two, and should be closer to the 16.3 than the 15.5, which it is.  Clearly, the first answer submitted was wrong.  Brian made both z scores negative...
536306
in Statistics in theory there should zero measurements falling beyond the three standard deviations from the mean on either side. though in practical some values do fall in this range. I don't remember the exact statistical names of these values but i think they are referred to as "outliers".
446842
The mode is the data point that appears the most often. From the problem statement, there's a single mode, 5. So, since there are 3 cards, then at least 2 of the cards must be 5. (If all 3 cards are different, then there would be 3 modes.)\n\nSo the cards are 5, 5, and x. (x could be 5 without changing the mode). Now, look at the mean. To find the mean, add up all the values, and divide by the number of values. So add 5 + 5 + x to get x + 10, and divide that by 3. That expression must equal 8.\n\nSo, you now have an equation to solve: (x + 10) / 3 = 8. Solve it as you normally would:\n\nmultiply both sides by 3 to get x + 10 = 24\nsubtract 10 from both sides to get x = 14.\n\nSo the 3 cards are 5, 5, and 14. The mode is 5, and the mean is (5 + 5 + 14) / 3 = 24 / 3 = 8.
436889
when the probability is 0.2:\n\nP((β-12%)*Sqrt(n)/σ>=1.301)=20%\n1.301 is the T value under degree of freedom of 44 and α =10%\n\nI did not pay attention in leaasons of statistics, I dont know if it's correct.
45555
3.56 is the correct answer.\n\n1. Find the mean of your data: 27.076\n\n2. Subtract the mean from each piece of data to find the deviations\n\n27.076 - 24.18 = 2.9\n27.076 - 31.32 = -4.24\n27.076 - 23.59 = 3.49\n27.076 - 31.54 = -4.46\n27.076 - 31.69 = -4.61\n27.076 - 24.88 = 2.2\n27.076 - 26.14 = 0.94\n27.076 - 29.2 = -2.12\n27.076 - 22.35 = 4.73\n27.076 - 25.87 = 1.21\n\n3. Square each of the deviations\n\n2.9^2 = 8.41\n-4.24^2 = 17.98\n3.49^2 = 12.18\n-4.46^2 = 19.89\n-4.61^2 = 21.25\n2.2^2 = 4.84\n0.94^2 = 0.88\n-2.12^2 = 4.49\n4.73^2 = 22.37\n1.21^2 = 1.46\n\n4. Find the mean of the squared deviations (this is the variance)\n\n11.38\n\n5. Find the square root of the mean of the squared deviations (this is the standard deviation)\n\nSQRT(11.38) = 3.37\n\nFor some reason Excel giving 3.56 for the result. Steps showing uses a lot of rounding offs.\n\nHope this helps.
471789
In a normal distribution:\n68% of the points are within 1 standard deviation\n95% of the points are within 2 standard deviations\n\nYou'll have to look at your normal distribution tables.  You want to find the point where the area is 90%.  This happens around z = 1.645 standard deviations.\n\nSo they should advertise the life of the calculators as 2,150 +/- 450 * 1.645\n\nThat works out to a range of between 1,410 and 2,890 hours, or 2,150 +/- 740 hours.