what percentage of the population do you expect to find between the mean plus and minus one standard deviation
142505
That would be 99%. Final answer
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!
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.
714755
A.  Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 sd's of the mean, then that would mean that there are a total of 4 sd's between 5 and 55.  So we can approximate this standard deviation by saying\n\n(55-5)/4 = 12.5\n\nB.  Since you are taking a sample of 400, then the standard error of the mean is 12.5/sqrt(400) = 0.625.  Since the sample size is large, then the sample mean is approximately normal.  To be within 2 units, then you would have to be within mu-2 and mu+2.  So we can standardize.\n\nz = (mu-2-mu)/0.625 = -3.2\nz = mu+2-mu)/0.625 = 3.2\n\nThe area between these two values would then be approximately 0.9986.\n\nC. In this case, our hypothesis test would look like this.\n\nH0: mu = 35\nHA: mu < 35\n\nOur test statistic would be (31.75-35)/0.625 = -5.2\n\nWe would then reject the null hypothesis.  So we would conclude that the scientists are correct.
883654
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.
331431
if the data is a number, for instance:\n25 people out of 360 people\nyou would first divide 25 by 360\nthen multiply that by 100\nthat answer is the percent\n\n\nhope that helps! email me at blackbeltgal162@aol.com if you need any more help
988027
since '800-1100' score is within 1 standard deviation (130) of the mean (930)... and assuming that the population is normally distributed, we can say that\n68.27% of the patients scored 800-1100...\n\n(that is if the 68-95-99.7 rule is universal among normally distributed populations)
1008113
The Russian mathematician P. L. Chebyshev (1821- 1894) discovered that the fraction of observations falling between two distinct values, whose differences from the mean have the same absolute value, is related to the variance of the population. Chebyshev's Theorem gives a conservative estimate to the above percentage.\n\nFor any population or sample, at least (1 - (1 / k)2) of the observations in the data set fall within k standard deviations of the mean, where k ³ 1.\n\nUsing the concept of z scores, we can restate Chebyshev's Theorem to say that for any population or sample, the proportion of all observations, whose z score has an absolute value less than or equal to k, is no less than \n(1 - (1 / k2 )). For k = 1, this theorem states that the fraction of all observations having a z score between -1 and 1 is (1 - (1 / 1))2 = 0; of course, this is not a very helpful statement. But for k ³ 1, Chebyshev's Theorem provides a lower bound to the proportion of measurements that are within a certain number of standard deviations from the mean. This lower bound estimate can be very helpful when the distribution of a particular population is unknown or mathematically intractable.
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)
647207
50%.\nSorry, this answer may hurt you, but it is the thruth.
189955
1/7 = 0.1428 = 14.28%\n(1 in seven has to be true)
189610
Look at the table of values for the normal distribution.  95% of the values are included within +/- 1.9 SD of the mean.  So:  1.9 x 6.2 = 11.78 and the interval is\n\n     26 - 11.78 <= Mean <= 26 + 11.78, or\n           14.22<= Mean >= 37.78
920953
1) X/25\n2) X/25\n\nExplanation:\n1) If you divide a set of values by a constant, then both the mean and the standard deviation are divided by that constant.\nSince your question is about the variance, which is the SQUARE of the standard deviation, the variance is divided by the SQUARE of the constant (5, in this case).\n\n2) When you add a constant to a set of values, the mean is increased by the amount of that constant, but the standard deviation is unchanged. This is true because the group of values has all moved up by a constant amount, but their relation to each other is unchanged, and their variation from each other is unchanged. Since the standard deviation is unchanged, the variance (the square of the standard deviation) is also unchanged.
529581
Approximately 95% of the data is within 2 standard deviations of the mean.\nCheck out this website\nhttp://en.wikipedia.org/wiki/Confidence_interval\nand look under the section\n"Confidence Intervals in Measurement"
937201
First find the mean (average).  Then subtract each data number from the mean, and square the difference.  Add up all these squared differences and find the average of that sum.  Then take the square root of that average.  That's the standard deviation.
473228
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
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!