Urgent! Please take this challenge! :( statistics counting rate and standard deviation (stn or std)?
118025
No, it's answer a, 100.\n\nRadioactive emissions follow a Poisson distribution.  The mean  of this Poisson distribution is approximately 9934 (the observed count).\n\nFor the Poisson, the variance equals the mean, so our estimate of the variance is also 9934.  The standard deviation is the square root of the variance, so our estimate here would be 99.7 which is approximately 100.
701818
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....
655920
The standard deviation is the square root of the variance. Now, the variance is the "average squared difference (distance) of each observation from the mean".\n \nLets think about it. You want a measure of how dispersed the data is, right? Well, suppose you have heights for your data and your trying to compare two groups. For example, you want to know if the height´s of New Yorkers vary more or less than the heights of Floridians. You are NOT trying to see which are taller.\n \nSo you can calculate the average height of each group and then see haw far away from this average each individual height is and then calculate the average distance from the average height. If everyone had the same height, the average distance would be ZERO, reflecting that the population´s height is very CONCENTRATED around its average.\n \nBut why do you use the squared difference? Because if you were to calculate the average of "x - average(x)" you would always get zero, so by squaring and taking the square, you always get a positive number (the distance is always positive).\n \nGot it?
967643
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.
1034794
I don't think this stat is available. \n\nIS\nIT?
874605
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.
408817
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.
1047890
class C\n{\npublic:\nenum { NUM_VALUES = 10 };\n\nprivate:\ndouble m_values[NUM_VALUES];\n\npublic:\n// Sets the number stored at index i in m_values to v\nvoid SetValue(int i, double v) throw( std::exception )\n{\nif ( i < 0 || i >= NUM_VALUES ) throw std::exception("Index out of range");\nm_values[i] = v;\n}\n\n// Gets the number stored at index i in m_values\ndouble GetValue(int i) const throw( std::exception )\n{\nif ( i < 0 || i >= NUM_VALUES ) throw std::exception("Index out of range");\nreturn m_values[i];\n}\n\n    // Returns the mean of the numbers stored in m_values\n    double GetMean(void) const \n    {\n        double sum = 0;\n        for ( int i = 0; i < NUM_VALUES; i ++ ) sum += m_values[i];\n        return sum / NUM_VALUES;\n    }\n\n    // Returns the sample standard deviation of the numbers stored in m_values\n    double GetStandardDeviation(void) const \n    {\n        double mean = GetMean();\n        double sum = 0;\n        for ( int i = 0; i < NUM_VALUES; i ++ ) sum += (m_values[i]-mean)*(m_values[i]-mean);\n        return sqrt(sum/(NUM_VALUES-1));\n    }\n};
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.
197113
To calculate by hand, you need to know the total number of months (n), which is 24.  Then you need the total number of accidents, which is 1x0+2x1+5x2+7x3+6x4+3x5 or 72.  Then find the mean number of accidents per month, 72/24 = 3.  Tis is called x bar (its symbol is x with a bar over it).\n\nNext you have to subtract each number of accidents from x bar, square this difference, and multiply it by the number of months that had that many accidents.\n\n(0-3)^2 = 9 x 1 = 9\n(1-3)^2 = 4 x 2 = 8\n(2-3)^2 = 1 x 5 = 5\n etc.\n\nAdd up all the answers, and divide it by 24.  This is called the variance.\n\nFinally, take the square root of the variance.  That is the standard deviation.\n\nMany times statisticians deduct one from the total when calculating population standard deviation.  That would be done if not all of the data is in your chart.  Based on the fact that x bar came out exactly 3 using n = 24, I bet that's the number your book intended for you to use.  Hope that helps!
919154
It depends on whether you want the mean number of heads, the total number of heads, etc.\nIf you want the total, multiply the standard deviation by sqrt(n), where you have n tosses.\nIf you want the average, divide by sqrt(n).\n\n(Note, the formula talked about below is for the *margin* of error, not the standard error. And thats only for means.)
461783
This is a Two Tailed test ... in other words, you are trying to determine if you have statistical evidence to suggest that the average change was something other than zero disruptions different between weeks 1 and 2 ... just because you saw a change of 5, doesn't mean anything ... it could just be a random fluke !\n\nHo (Null Hypothesis):  Sample Aveage change in distruptions was zero -- no change, or X-Bar = 0\nHa (alternative):  Sample Average change was not 0, or X-Bar <> 0\n\nAt the 0.05 (5%) level of significance (in total), you would have 0.025 probability in the upper end of the distribution and another 0.025 probability in the other end ... that's a total of 0.05 or 5%\n\nUsing your Z-Score table, you find the Z-Score (the number of Standard Deviations in a Standardized Normal Distribution with mean of 0 and Standard Deviation of 1) associated with 0.025 or 2.5% ... this is:\n\n-1.959961082 Standard Deviations from the mean for the low tail and 1.959961082 standard deviations from the mean for the high tail.\n\nSo, the number of standard deviations from a presumed mean of zero (under the null hypothesis) for the sample at hand is:\n\n( 5 - 0 ) / 2  = 2.5\n\nThat is to say, that if the Null Hypothesis of NO CHANGE in behavior were true, a legitimate value of 5 drawn purely randomly could still happen, but that would be 2.5 Standard Deviations away.\n\nBut, your upper and lower limits were only +/- 1.959961082 !!!\n\nSo 95% of the time, you would expect to see something between +/- 1.959961082 standard deviations from zero (within this range) ... but this time, you saw a value associated with a whopping 2.5 standard deviations from the presumed mean of zero ... (we'll call 1.959961082 = 1.96 here)\n\n<< I ACTUALLY MADE A LITTLE GRAPHIC HERE, BUT IT WON'T DISPLAY PROPERLY >>\n\nSo you would reject Ho.\n\nConclusion:  Based upon the sample taken, beleieved to be truely random and under the assumptions of normality, we have statistical evidience which would tend to suggest that the observed sample difference of 5 is statistically significant at the 95% confidence level, thereby, leading us to conclude, based upon the evidence observed, that the incentive system may have played a roll in the positive inprovement of the respondents.\n\nDoes this help ?   My contact information is in my profile.\n\nRegards,\n\nAlan
103664
Diagnostics and Statistics Manual V4
1156399
This is a good question even for a statistician.\nThe problem is your set is small and I do not know what the deviation between samples is, however the repeatability and reproducibility information can bind the window \nthrough which the third value will fall through or perhaps even remain in.  \n\nLet me know.
149503
Right, so feed the numbers into your calculator and press the SD button.\n\nOr enter it into Excel and use teh standard deviation function.
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!
286076
What kind of statistic are you looking for?