mean and the standard deviation of the average weight of 100 randomly selected travelers.?
487348
The mean of the average of 100 randomly selected travelers would just be the same as the population mean, which is 190.\n\nThe SD of the avereage of 100 randomly selected travelers would be the SD of the population divided by the square root of the sample size, which is 20/sqrt(100) = 2.
869354
standard deviation is one of measure of variability use in statistics which has the same unit as the given data and is equivalent ot the square root of the variance.\n\nIt measures how all the data varies. It indicates how the data in a distribution scatter or spread.\n\nThe lower the standard deviation, the closer are all the data around the mean.\nThat is, the data are not scattered and they are less heterogenuous, less variable.\n\nThe greater its value, the more the spreadness of the data around the mean.\n\nFor example,\nGiven:\nA: 45,46,40,43,48,46\nB:10,45,28,100,158,468\n\nThe standard deviation of the data A is 2.804757863\nThe standard deviation of the data B is 171.9400089\n\n]since the statndard deviation of the data A is less that in B,\nthe data in A are closer to each other, compare to the data B.
278272
You'll use Z-scores and the Z-table. Z-scores convert deviations to standard deviation units. \n\nIn this case, you're looking for the range of scores that makes up the middle 88% of the distribution - 44% on either side of the mean. \n\nThe z-score is 1.56 (look in the table for the z-score that cuts off 44% of the distribution.) This tells you that 44% of the scores fall between the mean and 1.56 standard deviation units above the mean - and 44% of all scores fall between the mean and 1.56 standard deviation units below the mean. \n\nNow, since each standard deviation unit in this particular distribution is 4ml (as per your question), 1.56 standard deviation units is equal to 1.56 x 4, which is equal to 6.24ml.\n\nBack to the problem - your range is 6.24 ml above and below the mean - which is 250ml - therefore the range of volumes is from 6.24ml less than 250 to 6.24ml more than 250 - or in other words - 243.76ml to 256.24 ml.\n\nHope that helped!
935234
"In probability and statistics, the standard deviation of a probability distribution, random variable, or population or multiset of values is defined as the square root of the variance.\n\nThe standard deviation is measured in the same units as the values of the population. For a population of distances in meters, the standard deviation is also measured in meters, whereas the variance is measured in square meters."\n\n\nIn essence, standard deviation is used for categorizing groups.\nFor example, the average IQ is 100, the standard deviation to the left would be 85, and deviation to the right would be 115. Two left standard deviations would be 60, two right standard deviations would be 130.\nSame applies to SAT scores, statistics, polls, etc.
606521
Try this:\n\nhttp://education.yahoo.com/reference/weights_and_measures/
999718
what was the question?\n\n\nDo you want to know terms like mean, average, mode, standard deviation, etc.?
405919
http://en.wikipedia.org/wiki/Hundred_weight
706728
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.
873308
OK - I will try and explain it, but forgive me if I get myself confused...  hehehe\n\nFirst step - make a bell curve.  Put a mark in the center and label that "7 hours" - that's your mean.  Next, put three marks on either side of that middle point - those will be your standard deviation points.  For the points on the right, ADD 5.3 to the mean for the first one, then add 5.3 again for the 2nd, etc...  Points on the left side of the mean have the 5.3 SUBTRACTED.  (Am I still making sense?  lol)\n\nNow - the 68-95-99.7 rule says that 68% of the freshmen will fall into the area between the first line on EITHER side of the mean, 95% between the second lines and 99.7% between the third lines on either side of the mean.\n\nI think you should be able to figure it from there - I hope so...  Right now I am so confused I think I gave myself a headache!  lol\n\nGood luck!
74440
The Mean is the average.  You have 22 numbers; add them up and divide by two to get the mean.  In this case, that gives 87.38\n\nThe median is the "middle" value.  Since you have 22 numbers, 11 of them will be below the median, and 11 will be above.  In your case, the 11th biggest number is 87.9 and the 12th biggest number is 88.  The median could be anything in between, but is generally taken to be the average of these two- or 87.95\n\nThe mode is the most common value.  In your case, each value occurs only once, except for 91.7 which occurs twice.  Therfore, 91.7 is the mode\n\nUpper and lower quartiles are similar to the median; The lower quartile is the value below which 1/4 of your values lie, and the upper quartile is the value above which 1/4 of your values lie.  Since you have 22 numbers, the lower quartile is the value between the 5th and 6th largest numbers (82 and 83.6), and the upper quartile is the value between the 17th and 18th largest numbers (92.9 and 94.2).  That makes 82.8 and 93.55\n\nThe range is the minimum value to the maximum value, or {68 to 98.2}\n\nTo get the variance, first subtract the mean from each value.  Then, take the square of each of these new values.  Then, add up all these squares.  Finally, divide this sum by the number of values (22 in your case).  This should give a variance of 51.22 with your numbers.\nNOTE:  The above formula is for "population variance", and is generally used when your numbers represent an entire population.  If your numbers represent a sample of a population, you should use the "Sample Variance", which has the same formula except that you divide by the number of values minus one (21 in your case).  With your numbers, this should give a sample variance of 53.66.\n\nThe standard deviation is just the square root of the variance.  Like the variance, there is a "population standard deviation" (7.16 in your case) and a "sample standard deviation" (7.33 in your case)
1165756
Mean is the average, when adding all numbers then dividing by the amount of numbers given.\n\nMedian is the numbers in the middle average - 8
725278
the mean is the same as the average. so just add the three numbers together and divide by 3 (because that's how many numbers there are).\nexample: 12, 45, 63\n12 + 45 + 63 = 120\n120 / 3 = 40\nso the mean is 40
1153957
The website below seems to be what you're looking for.  The spreadsheet you can download uses the NORMDIST function to do the graph.  Check out the help feature for the function to see its equation.\n\nI assume you know how to calculate the mean and standard deviation for your data.  The AVERAGE and STDEV Excel functions are good for this.
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.
1018969
Just google it dear..very simple..calculate mean and average the sum of square of variation of each score from the mean.\nhttp://davidmlane.com/hyperstat/A16252.html
836301
Try this website for help with your statistics homework:\n\nhttp://www.statsoft.com/textbook/stathome.html
1157784
mathematically, you CAN take an average of averages. \n\nBut statistically, it won't be meaningful, because you will be combining information from non-identical studies.