inter-quartile range for a distribution with a means of 100 SD of 30?
462124
I assume the distribution is Normal, there is not enough information otherwise.\n\nInter-quartile range is the difference between the 75th and the 25th percentiles (aka 3d and 1st quartiles). For standard normal (mean=0, SD=1), there are are 0.67449 and -0.67449. You can get those from tables at the end of any statistics books, or NORMINV function of Excel (give it 0.25 and 0.75 as inputs).\n\nTo covert them to your distribution, multiply by SD and add mean, so you get 120.23 and 79.765, with difference being 40.469, and that is your answer.\n\nIf you're smart, you'll notice that mean cancels out, quartiles of standard normal only differ by a sign, so a faster way to compute it is  2*0.67449*30=40.469
677379
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)
414176
If you want to take help of stats then use the concept of quartiles.\n\nHere you will get three categories but they will not be equal.\nFirst one will contain lower 25% of values,\nSecond one will contain mid 50% of values,\nWhile last one will contain upper 25% of values.\n\n\nBy the way do you have any information about RUCHA VAIDYA smallest yet the brightest star? She has acted in movies like Ek Ajnabee and Main Aisa Hi Hoon. She has also done innumerable ads on TV. Please reply if you have any information about her. Absolutely any Kind of good information about her (even anything personal) is acceptable. Please answer my this question from celebrity or movies section. Please Please Please reply
650317
If you have a computer program that can compute binomial probabilities, you can use that.\n\nIf you are doing this by hand, however, I would suggest using the normal approximation to the binomial here.\n\nn = number of trials = 200\np = prob. of success = 1/6 if we consider getting a five to be a success.\n\nThen np = 200(1/6) = 33.33 and n(1-p) = 166.67 are both >= 5 or 10 (the cutoff is different for different classes) then the sample size is large enough to use the normal.  This binomial is approximately normal with mean = np = 33.33 and SD = sqrt(np(1-p)) = sqrt(200(1/6)(5/6)) = 5.27.\n\nNow for the continuity correction.  The rectangle over 40 goes from 39.5 to 40.5.  Since we are finding the probability that the number of 5's is at least 40, then we should start at 39.5 so that all of 40 is included.\n\nz = (39.5-33.33)/5.27 = 1.17\nArea to the right of 1.17 on the standard normal is 0.121.
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.)
898911
Take the average of the absolute value of each of the numbers minus the mean...\n\n( |2.44-2.23| + |3.54-2.23| + |1.37 - 2.23|  + |1.5 - 2.23| + |2.3-2.23| )/5
935926
It means that there is a large gap in the middle of the data set with one or a small ammount of entries being grossly larger than the majority of the set:\n   the set 1,2,3,4,30 has a mean of 8 and a median of 3\n   the set 20,22,28,32,33,34,64,68,71 has a mean of 49 and \n   a median of 33
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)
761495
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.
567661
the Chebyshev's inequality goes Pr(|X-mean| > or = k.sd)< or = 1/k^2\n\ni m assuming u meant standard deviation when u said 4.3818, which is 2 sd.\n\nThe prob of points outside 2sd at both sides is not more than 1/k^2.. or 1/2^2=1/4.\n\nSo, the probability of points within 2sd will be at least 1-1/4=3/4.\n\nhow many data points? if u have 100 data points in total, there will be at least 100x3/4=75 data points within the mean+/-4.3818 range.\n\nNote: The exactly probability depends on the type of distribution the data follows, for example, for Normal/Gaussian, 95% lies within +/- 1.96sd. Chebyshev's inequality states the minimum possible for any type of distribution.
374562
I think that you would need at least 2 values in each cell so that the within groups SD can be estimated, so if each of the two ways has two factors then you would need a sample of at least 8, so that there is two for each cell.  Could be wrong though.
986393
Take a set of say, 51 data points of varying values.\n\nArrange them from least to greatest.\n\nThe 26th data point is the mode.\n\nThis is different from the median, which is the sum of the data points divided by 51.  The median will not necessarily be equal to any of the data points, while the mode, by definition, will.  If the distribution is normal, aka a bell curve, the median and the mode will be very close.  If the distribution is not a bell curve then the median and the mode may be quite different.
979515
the formula for sd.dev of binomial distribution is "\nsqrt(n*p*q).\ncan't understand the table so use the fomula urself.hope this helps
266644
I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
303069
Do a confidence interval calculation and see if the confidence level for the data you've collected is above that required for 98%.\n\nI can show you how but given your other questions, it's clear you haven't really done any work on this yet and you're just looking to get someone to do your homework for you.\n\nLet me know if you still can't do it after you've tried a while, and I might help you out.
1087248
The previous answer is correct.  Also, if the standard deviation is greater than your mean, then that may indicate your data has a large range, and most likely have an outlier.  For example, if you data is {1,1,1,100}.\nThe mean is 25.75\nThe standard deviation is 49.5.\nThe range of data is 100-1=99, with the outlier being 100.\n\nSometimes, experimentalist would throw out the outlier data and recompute statistics for the new set of data.  So if we throw out the outlier 100 and can rationalize that 100 is a corrupted data point because our previous trials were all 1's, then our new data set is {1,1,1} and\nThe mean is 1\nThe standard deviation is 0.\nThe range of data is 1-1=0.
73222
Your average will be -25\nYour range will be 377\nAverage is taking the sum of the numerical vales and dividing them by the number of data points listed.\n\nie: 95 + 70 - 8 - 282 + 0 = -125/5 = -25\n\nYour range is the difference of the highest number of the group and the lowest number of the group.\n\nie: Highest = 95  Lowest = -282\n95- (-282) = 377\n\nI hope it helps.  bmsc2082yahoo.com and tell me what happens.