Between the mean and 3.01 standard deviations from the mean\nAnswer choices\n99.87%\n49.86%\n49.87%\n50.13%
214662
49.87\n\nThe entire area is 1. So half the area (from the mean to the highest STD measured) will approach 50%, never more.\n\nThe chart on the link below shows 99.74% for the curve from -3 to +3, so half is 49.87, slightly higher at 3.01, but this is the answer due to rounding.\n\nJOE
203120
your confidence interval of .4873 to .5526 indicates the range of the true mean with 95% confidence.  Since values above and below .50 are in this range, the poll is a tossup.
658433
If you are using a sample set of data you can press stat/edit - then press enter/enter a list of numbers/ press stat again/move the arrow to calc/ select 1-Var stats/ press enter/press 2nd and then 1 (or L1)/ press enter.\n\nx bar = the sample mean\nSx = the sample standard deviation\nVariance = Sx squared
257872
If you can't answer this, you have no idea what the empirical rule is.\n\nAbout 68% of all values fall within 1 standard deviation of the mean for the normal distribution.
102837
Well a) is half a SD away from the mean so look in your standard deviation table for sigma = 0.5\n\nb) is a bit more complicated where you have to find the probability of it being ABOVE 190 (i.e. half a standard dev away... about 70%) and BELOW 230 (1.5 st devs... about 90%) and multiply them together\n\nc) is looking for 1.25 standard devs and taking 1 minus this figure as you want to see BELOW this high number (answer will be around 85% or so)
122927
Use standard deviation formula and put the percent deviation < tolerance limit
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.
495271
first fin the mean with the sample of 280 \n\nmultiply 280 to .184 to get the mean \n\nmultiply 280 to .184 and to .816 (from 1-.184) then take the \n\nsquare root that will be your standard deviation \n\n\ntake the z score you will get   ( 56 - 51.52 )/6.483851\n\n=.6909   find teh valu in the standard normal distribution table \n\nyou will get 0.2549    subtract it from .5  because you need only \n\nthe probability higher than of 56  you will get \n\n24.51%
277087
Since we are not given any raw data, we must assume normality and use the population standard deviation.\n\nThe 90% CI for the population mean is given by:\n(xbar-z(.05)*sigma/sqrt(n), xbar+z(.05)*sigma/sqrt(n))\n\nHere we are told xbar = 12.4, sigma = 6, and n = 54.  Using a table of normal probabilities, we find z(.05) = 1.645.  Simple calculation shows the CI is:\n(11.0569, 13.7431)\n\nFor part (2), we want our margin or error to be 1 with a 90% confidence level.  Our population standard deviation is still 6 and z(.05) is still 1.645.  Thus, we want:\n1.645*6/sqrt(n) = 1\n9.87/sqrt(n) = 1\nsqrt(n) = 9.87\nn = 97.4169\nRound up to get n = 98.\n\nFor your additional comments:\nIncreasing the confidence level also increases the margin of error--not (a).\nThe sample standard deviation could be higher or lower than the population standard deviation, so it could increase or decrease the margin of error, but there is no guarantee--not (b).\nIncreasing the sample size to 100 makes for a better estimate of the mean and would definitely decrease the margin of error.\n\nThus, (c) is the correct choice.
421757
Quickly, based on my own past learning of these things:\nThe mean is used when the distribution is normal, no lumps on either end or any place but in the middle.\nThe standard deviation is how far apart scores are from each other, another way to think of it is how alike they are.  \nThe mean and standard deviation go together in a NORMAL distribution as above... one big bell shaped hump in the middle.\nStandard error of the mean:  \n\nIt may be easier for you to check out this link:\nhttp://research.med.umkc.edu/tlwbiostats/stnderrmean.html
42364
You have all the information you need.\n\nThe Mean is the same as the average. It's just another term for a simple average. So it's $7500\n\nThe standard error of the mean is the standard deviation divided by the square root of the number of samples. So it's just $650 divided by 10, or 65.
83169
Figure out how many standard deviations from the mean those values are:  (13500-10000)/2000 = 1.75; (8000-10000)/2000 = -1; 11500-10000)/2000 = 0.75.\n\nNow look up the areas under the curve for those z-values and those are the answers.  While you're at it, find the value 0.10 in the table and see how many z-values (standard deviations) that is from the mean.  Multiply by 2000.\n\nNote that you have to add the values in part b - they're on opposite sides of the mean.
181189
X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.
494376
1 = 100%\n2/3 = 0.6666666... = 66.6666.. %\nso,\n1-2/3 = 166.66666...%\n\nHope this helps!
314325
1. 63.75 or 63 3/4\n2. 63\n3. 24.75 or 24 3/4
278775
Do you have area tables for the standard normal distribution? If so, look in the table for 0.2514 and read of the z-value (approx 0.68). That means 25.14% is 0.68 standard deviations from the mean. If 30 is 0.68 standard deviations from the mean, then 30/0.68 = 1 standard deviation. Do the sum then go and get your A+ grade.\n\nIf you haven't studied the standard normal distribution, then your teacher is asking too much too soon.
196025
Range: 19\nStdev: 6.27\n\nRange is just the max value minus the min value.  To learn more about calculating standard deviation, visit http://en.wikipedia.org/wiki/Standard_deviation\n\nSo, I saw someone got a different standard deviation than me, so I checked it again on Excel and I still got 6.27.  Instead of dividing by 7, I divided by 6 (n-1).\n\nIf these data are a sample of a larger population, then you divide by 6 (n-1) to estimate the stdev of the population, so the stdev is 6.27.  If you just want to know the stdev of those numbers, then you divide by 7 (n) so the stdev is 5.81.