How do you solve this problem & also how do you solve this other question: With a normal distribution Mean=100 SD=30 what % falls between scores of 90 7 110?I used the z=x-m/sd, but I don't know if this is right.
127496
For the first one, I believe that it would be as follows:\nTo exclude the extreme 10%, you remove the 5% at the top (95th percentile to 100th) and the bottom 5% (0 to 5th). I can't remember the formula right off hand, but it will be something like 85 & 115 or right around there.\n\n\nThe second looks like you did it right.\n\nHope this helps. Good luck!
609751
confidence intervals are about what % confident you are that a given person's score falls between x score and x score on any given day.  formulas: \n\n                                   68% CI=X+/-1(Standard error of measurement)\n                                   95%CI=X+/-1.96(SEM)\n                                   98%CI=X+/-2.58(SEM)\n\nto get the SEM:  SD(standard deviation)times the square root of 1 minus the reliability coefficient(rtt)\n\nI might be wrong about the margin of error, but I think it's the SEM(the index of the amount of error that can be expected in obtained test scores due to the unreliability of the test) \n\nhope this helps--statistics suck!
387865
The first sample proportion is p1 = 30/100. Its variance is (30/100)*(1-30/100)/100.\n\nThe second sample proportion is p2=60/80. Its variance is (60/80)*(1-60/80)/80.\n\nWhen you look at the difference, p1-p2, and assume that the two proportions are independent then the standard deviation is\nsqrt(var1+var2)=\nsqrt(  (30/100)(1-30/100)/100 + \n          (60/80)(1-60/80)/80)\nwhich is approximately 0.06666146.\n\nHere you can use the normal approximation for both proportions, and the difference is \n30/100-60/80 = -.45.\nIf you use a two sided test then the z-score is -.45/0.06666146 = -6.75, and the p-value is 1.473087e-11. It is highly likely that the difference is zero! Please note that this last part is not quite right. If you want to do a formal hypothesis test then you need to do a pooled estimate.
422517
Your data set would look like this:\n\n10 10 10 10 10 10 10 10 10 10 20 20 20 20 20 20 20 20 235 235\n\nThe mean is 730/20 = 36.5.\nThe median is (10+20)/2 = 15.\nThe mode is 10.\n\nSo certainly what was stated by the dean isn't a lie.\n\nNow, if I were to calculate, say, the average number of classmates for each person in the statistics classes, there would be 100 9's, 160 19's, and 470 234's.  So the average number of classmates for people taking these classes is approximately 113,920/730 = 156.05.  The median is 234 and the mode is also 234.  So the newspaper isn't really lying either.\n\nI guess you can say that they are looking at this from two different perspecitves, so in that sense, there are two truths.
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.
396942
If the mean is 52,000 and the Std Dev is 31,515 the break even point ($0) is 1.65 Std Dev  (52000/31515) to the left of the Mean.  The answer is .50 (the probablility the result is greater than 52K) plus the probability that the result is between $0 and $52K.  Look up the value of 1.65 Std Dev in your table (or on your calculator) and add .50 to it.  \n\nThat's your result.
544595
The two most common methods of normalizing data are:\n\n1. Subtract the mean, then divide by the standard deviation (or variance)\n\nor...\n\n2. Subtract the minimum, then divide by the range.\n\nThe first method above yields a distribution with a mean of zero and a standard deviation (or variance) of 1.\n\nThe second method yields a distribution which ranges from 0 to 1.\n\nOf course, other methods are possible, such as subtracting the median then dividing by the inter-quartile range.
25210
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.
544464
This is obviously a statistics problem.  Treat each mother as a sample and the data entry the age at which they had their first child.  If all you need is the "mean" point, it is a simple matter of knowing what a "normal distribution" is.  Half of the mothers will be above the mean date and half below (both 50%).  In a "formula" it would be (total of all the ages divided by the number of mothers)  More useful for the analyst is the shape of the distribution - the bell curve and the amount of data in the "tails".  To do this you need to perform a "t test" of the data set.  Since I assume that what you mean is the whole population of women, the distribution can be assumed to be "normal" and the "degrees of freedom" is infinite.  Look up "t test" in your statistics book, also known as "student's t".  The answer is there, in the "degrees of freedom" expressed in "standard deviations" from the mean.  Thus you can talk about "confidence intervals" from this.  The Wikipedia article is surprisingly hopeless for the non-mathematician.  My education is statistics came from "Biostatistical Analysis" by Zar, still in print and one of the only books from my former life I still keep on my shelf.
254561
It is actually binomial with parameter p=0.5 and n=400. Since 400 is quite a large number, u can approximate this by the Normal distribution with the same mean and variance, which are, np=400 and np(1-p)=100.\n\nX ~ N(200,100)\n\nu wanna test that if 216 is within 95% interval.\n\nVariance=100\nSDev=10\n\n216-200=16,\n16/stdev=1.6\n\nso, 216 is 1.6 stdev from the mean 200. which is within +/-1.96(95% confidence interval. hence, the null hy is accepted, with 95% significance level.\n\n\n\nbtw, drbinsomething's solution is equivalent to mine. he directly test P as a random variable while I tested mean nP as the random variable. He didn't mention it, but he took P to be asymptotically Normal as I did. '1.96' only applies when the distribution is Gaussian/Normal.
463322
Is that really the question?  Prob of a normal variable being greater than the mean?  It's just 0.5.  What's the real question, and what has Z to do with it?
44832
There will be a table in your statistics textbook that gives probabilities for a normal (Gaussian) distribution.  Usually it is for a "standard normal" where the mean is 0, and the standard deviation is 1.  The table will show upper tail probabilities for values of x ranging from 0. to 3. (or higher), where the it gives the probability that random sample xi > x.  For general samples with a mean different than 0 and a standard deviation different than 1, you have to apply a conversion to use this table, as follows:\n\nLet m = the mean of a normal distribution\nq = the standard deviation (usually called by the greek letter sigma), yi = boundary of upper tail distribution.\nIn this case, m = 92500, q = 8400, yi = 95000.  The value of x to look up in the table is \n(1) x = (yi-m)/q.  \nThis will tell you the probability that ONE random element, xi > x.\n\nHowever, if you take a sample of N elements, if the standard deviation of the entire sample is q, the standard deviation on the MEAN of the sample is \n(2) q'=q/sqrt(N).\nSo for the sample of 64 families, the standard deviation on the mean is 8400/sqrt(64) = 1050.  This is the value to use for q in equation (1).  Equation (1) then gives x = (95000-92500)/1050 = 2.38.\n\n  So the probability is the same as the probability that x > 2.38 for a standard normal distribution.  From the table (or from an online calculator like the one at\nhttp://bayes.math.montana.edu/cgi-bin/Rweb/buildModules.cgi\nI find that the probability = 0.008656319 , less than 1%
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.
75812
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
329154
If the population has a mean of 58 and a sd of 8, then the sample means of size 100 will have a mean of 58 and a sd of 8/sqrt(100) = 0.8\n\na) Within 3 of 58 means that the sample mean will be between 58-3 = 55 and 58+3 = 61\n\nz = (55-58)/0.8 = -3.75\nz = (61-58)/0.8 = 3.75\n\nFrom the table, the area between the two is almost 1.\n\nb) The mean weight is less than 2.5 of 58 or more means less than 58-2.5 = 55.5\n\nz = (55.5-58)/0.8 = -3.13\n\nFrom the table, the are greater than that is almost 0.\n\nc) z = (57.5-58)/0.8 = -0.63\nz = (58.7-58)/0.8 = 0.88\n\nFrom the table, the area between is 0.811 - 0.264 = 0.547.\n\nd) 56 +- (1.96)(0.8)\n56 +- 1.568\n(54.432,57.568)
547564
Since n = 200, the central limit theorem certainly applies here.  The central limit theorem states that the sample mean has a normal distribution where the parameters are:\nmean = mean of the distribution where the samples were taken (here the uniform distribution)\nstandard deviation = (standard deviation from the distribution where the samples were taken)/sqrt(n).\n\nThe mean of the uniform distribution is (b+a)/2 = (48+24)/2 = 36.    The standard deviation of the uniform distribution is ((b-a)^2)/12 = ((48-24)^2)/12 = (24^2)/12 = 576/12 = 48.\n\nTherefore, the sample mean has a normal distribution with mean = 36 and with standard deviation 48/sqrt(200) = 3.394.\n\nYou'll now need either a normal probability table or a calculator (like a TI-83 or TI-84) to find this probability.  Using a TI-83/84, you would do the probability as follows:\n\nP( sample mean < 35) = normalcdf(-E99, 35, 36, 3.394) which, according to my calculations, equals approximately 0.3841.
572784
P(X<6) = Probability that X is less than 6\nTransform the X value of 6 to Z\n\nZ= 6 - 8 / 2 = -2/2 = -1\n\nLook up the Z value of -1 in the standard normal table and we get a probability of 0.1587 or 15.87%.\n\nI think this is right.