A Competency test has scores with a mean of 69 and a standard deviation of 4.?
605000
The mean describes the average score, the SD describes the shape and range of the normal distribution, or bell curve. \nWith this information, we can say the following:\nAbout 68% of test-takers got between 65% and 73%- a very tight grouping (the area within 1 SD).  Approximately 95% of people got between 61% and 77% (2 SD), and about 98-99% got between 57% and 81% (a range of 3 standard deviations).
773196
Mean: 113\nMedian: 108-116\nMode: 108-116\n\nThe mean was the hard one, I actually took the middle score from each range (eg 90-98=94) and multipled it by the number of times it occured (eg 94x6=564) and added all 5 (564+2266+4816+3388+1170= 12204). Then divided it by the frequency (6+22+43+28+9=108). Total (12204/108= 113). It's impossible to find a mean of a set of ranges, you need exact figures.\n\nThe mode is the most commonly occuring range, the median is the central range. These figures can't be accurate without the full actual IQ scores. But based on the figures provided, the mean, median and mode are as I stated above. If you provide the exact figures I'll work out the exact mean, mode and median.
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.
549424
An IQ of 120 is actually above average.  98th percentile is really good.  What it means is that you scored better than 98% of the people who took the test.  The standard deviations you don't really need to worry about.\n\nBottom line:  You're a smart cookie!  Congrats on your test scores!
674883
the average, IE add up all parts and then divide by the number of parts.  For example if you'd taken 3 tests and gotten a 70 on the first, an 80 on the second and a 90 on the third you add the three to get 240, and then divide by 3 and your mean grade is 80.
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)
642186
69..\n\nand try to paraphrase your friggin homework...
64438
Let me see if I can answer your question in English. :-)\n\nI'm sure you have an equation in your notes or book that you can just start plugging stuff into.  It probably looks like this:\n\ns = square root of (X - X bar)^2/(n-1)\n\nBut if you want to know why it works, keep reading.\n\nYou have a list of numbers.  First, you need to find the average, or mean, of those numbers, by adding them up and dividing by how many there were.\n\nNow, if you subtract each number from the mean, you get a new list of numbers.  These are the deviation scores, and tell you how far above or below average each number was.  If you add them up, you will get zero, because some of them are positive and some negative.  That's not very interesting, so we square them to make them all positive.  Then we add up all those squared deviation scores.\n\nYou probably got a pretty large number, and not the one we want, yet.  We want the Standard Deviation, which is sort of like the average deviation.  So just like you divide by the number of scores to get the average, you need to divide the sum of the squared deviation scores by something.  Sometimes we divide it by the number of scores, like you would think, but for complicated reasons it works better if you divide by one less than that.  So if you started with 5 numbers, divide by 4 now.  (Be sure to check your notes or your book to see what your teacher said you should divide by.)\n\nWe're still one step away.  The number you should have now is called the Variance.  It is also a measure of how spread out your numbers were, but it's bigger than we want.  Remember how we squared the deviation scores earlier?  Now we need to undo that by taking the square root.  What you get is the Standard Deviation you wanted.
127671
if the test administrators know what they are doing, they will give a test that is customized to the age of the person taking it. that means that the iq results should be based on how they compare with others of that age. so if a six-year-old takes the test for six-year-olds, a score of 100 means they are average for six-year-olds. a score of 68 would mean that they are below average intellectual development for that age. a score of 120 would mean they are as intelligent as an eight- or nine-year-old, etc.
107133
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.
956188
What you're looking for is the standard deviation.  If the data values are X and the mean is XBAR, then\n\nS.D. = sqrt{[SUM(X - XBAR)^2] / n}\n\nn is the number of data items, and SUM means to add up the squares of the differences between the data values and the mean.\n\nThe lower the standard deviation, the more tightly packed the data.  If the data is normally distributed, about 68% of the data will lie within one standard deviation of the mean, and 95% will be within two standard deviations.\n\n[Edit:  In light of the previous answers, I should add that the standard deviation is the square root of the variance.]
862471
Your going to compare your mean with the population mean and see if your class really better than the average of the population.\nHo: u = xbar\nHa: u < xbar\n\nalpha = 0.10\none tailed test (of course)\ntherefore: z(table) = -1.28 (i'm not sure)\n\nuse z test since you know the population standard deviation.\n\nz = (x - u)/o/sqrt of n\n\nthat's it.\n\ngood luck
824278
many tests, like physical, mental etc..
502350
If your current grade is 75%, that includes your 72.2% average (add your 5 tests, 361/5) from tests and your good homeworks, etc.\n\nYou need a 75% to meet your needs. So you need to study to improve your odds.\n\nNow, if you wanted to get technical, we could figure out the standard deviation... and then see how many standard deviations 75% is above the average; but, even if I knew that, I'm not very good at figuring out the percentage distribution above average that would be.  But, let's give it a shot.\n\nvariance = sum(x-average)^2/n \n         = (sum(x^2) - 2n*average^2 + average^2)/n\n         = (sum(x^2) + (-2n + n)*average^2) / n\n         = (27517 + (-5)*72.2^2) / 5\n         = 1452.8 / 5\n         = 290.56\nstd. dev. = sqrt(variance) = 17.04582\n\n66% of your tests might fall in 72.2 +/- 17 (55..89)\n95% of your tests might fall in 72.2 +/- 17*2 (38..102)\nthat's just the oddity of std. dev.'s, so pardon :)\n\nso, 2.8 (75-72.2) is 0.1642631 (2.8/17.04582) std. dev. above which is probably just past the average, so it's well below the 83% (1 std dev above)and slightly above 50% (0 std dev).  Even if it was 0.1642.. of (83%-50%) then that's about 56.6%  So study hard, and you're get above your average quite a bit.  If you don't study, you'll fall below.  Good luck.
793773
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...
924135
google ..... maths psychometric test book
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Ho: mean=40 hours/week\nHa: mean > 40 hours/week\nalpha: 95%\nn = 30\ndf = 30-1\n\nOne sample t-test:\n\nt=(mean-assumed mean)/(sample standard deviation/square root of n)\n\nt=(40-45)/(7.5/sqrt 30)\nt=-5/1.369\nt=-3.65\n\nFrom the tables with alpha = 0.05 and df = 29 the t value is 2.045.\n\nSince our calculated t-value is outside of the t value from the table, there is little statistical significance to conclude that the average work week has increased.