a. there is no dispersion in data\nb. the mean is a good measure of the average\nc. the data is homogeneous\nd. all of these choices is correct \ne. there is no dispersion in the data and the data are homogeneous
208220
d.\n\n\nps i'm not a socioogy major, i'm a  biostats grad student
269427
There are actually many different types of "mean." The most common "mean"  is the arithmetic mean, also called the average in everyday language.\n\nCalculating the arithmetic mean is simple: Add all the values in the sample together and divide by the total number of values in the sample.\n\nThe arithmetic mean is often used to describe data. The problem is that it doesn't always necessarily give the correct picture.  Sometimes it is better to use the median, which is the number found exactly in the middle of the set of values, or the mode, which is the number most frequently reported. It depends on the data involved and the question you are asking about the data.\n\nWhat this all comes down to is: the mean is a measure of central tendency (as are the median and the mode).\n\nThe other two most common means are the geometric mean and the harmonic mean. If you are interested in leanring more, visit Wikipedia.
533526
The answer is C.\n\nHowever, do you know how to get the answer?\n\nThe mean is the average of all of the numbers. Add the four numbers you gave me and divide by 4. That is how you get your number. This is what teachers use to find your average for your school grade.\n\nSorry, im studying to be a teacher, I wanted to make sure you understood.
370695
The arithmetic mean is the average or the sum of the data divided by the sample size.  It is an appropriate measure of central tendancy for data taken for a normal population.  505 of the data is on one side of the mean and 50% on the other (Bell shaped curve).  In this case, the mean equals the median.  \n\nThe median is the exact middle of the data by count.  Splits the data by count not value.  (If you had 9 data points the median would be the 5th number when order from smallest to highest.)  It is a good measure of central tendency for skewed data or data with long tails.  Nonnormal.  For highly skewed data, the mean and median could be far apart.\n\nMode is the most frequent value in the data set.  It is good for central tendency for data with multiple modes (bimodal, trimodal, etc).  It is also good for nominal data (nonnumeric data).
361964
(A) s = 0.2+0.4+0.4 = 1; Pm=s/3=0.33333 – mean value for P; \nsigma=sqrt{(0.3333-0.2)^2 + (0.3333-0.4)^2 + (0.3333-0.4)^2}/3; whereas sigma is standard deviation.\n(B) s = 0.2+0.2+0.5+0.2 = 1.1; Pm=s/4=0.275;\nsigma=sqrt{(0.275-0.2)^2 + (0.275-0.2)^2 + (0.275-0.5)^2 + (0.275-0.2)^2}/4 \ndo I answer your question correctly?
155889
Student's t-test can be used to determines the probability that 2 random variables have the same mean. This can then be used to test whether two sets of data come from the same source.
92715
IMO, fb2kool is on to something.  Luck is used to describe a short series of events that appear to deviate from the average (when they simply contribute to the average).  The strange thing about probability is that the "average" never seems to happen.
507811
a) Can't sell more than 4, so 1*p(selling 1) + 2*p(selling 2) + 3p(selling 3) + 4p(selling4).\n\nb) Find the number n so that p(x>n) is less than 0.05 for a Poisson "vif" mean 3.
223100
The controlled parameter is the one thing that does not change in the experiment. So as others have mentioned, the answer is B.
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.
181122
For experimental data it is 100 * (average of the set - measured member of the set) / (average of the set).\n\nIf you are measuring something, how have you established an "Expected Value"? Why are you conducting an experiment/measurement if you already know what to expect?
141755
There is a difference between a sample and population mean. Because you used only 16 books, the data could be significantly skewed. Therefore, there is not sufficient evidence to claim that the population shows similar charcteristics.
277898
The answer would be C.\n\nIn a two tailed test, you're looking to see if the mean is either significantly greater or less than k which is why it is called two-tailed.  The null hypothesis would be that mu = k and the alternate, not equal.  Thus, c
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)
434906
OK, here goes:\n\nThe data could be\n7 9 10 12 12 13 14\n\nThe median (center number) is 12\nThe mode (most repeated number) is 12\nThe mean (average) is 11.\n\nThe point is that you need to look at all three measures to get a good idea of the data.\n\nHere's another set:\n1 3 10 12 12 19 20\nThis has the same measurements.\n\nIt is only when we look at things like the variance and std deviation that we can really see the difference between those two data sets.\n\nGood luck!
183429
Measures of central tendency are measures of the location of the middle or the center of a distribution. The definition of "middle" or "center" is purposely left somewhat vague so that the term "central tendency" can refer to a wide variety of measures. The mean is the most commonly used measure of central tendency.\n\nWithout knowing something about how data is dispersed, measures of central tendency may be misleading. For example, a residential street with 20 homes on it having a mean value of $200,000 with little variation from the mean would be very different from a street with the same mean home value but with 3 homes having a value of $1 million and the other 17 clustered around $60,000. Measures of dispersion provide a more complete picture. Dispersion measures include the range, average deviation, variance, and standard deviation.
460908
Wal can't be right. The standard deviation of the means of samples is always lower than the standard deviation of the population. The formula is\n\ns / sqrt(n)\n\nwhere s is the standard deviation of the population.\n\nfor a), your answer is 15 / sqrt(10)