How do you calculate percentiles in statistics?
111533
Percentiles are what percent of the sample are above that value and what percent are below that value. So if you have a sample size of 6, if you rank all the scores from lowest to highest.\n\nSay we have the following scores \nAlex 99\nBob 95\nCharlie 92\nDan 85\nEvan 75\nFrank 69\n\nSo Alex is in the 100% percentile because 0 % of the people are better than him.\nBob is in the 80% percentile because 20 % of the people are better than him.\nCharlie is in the 60% percentile because 40 % of the people are better than him.\nDan is in the 40% percentile because 60 % of the people are better than him.\nEvan is in the 20% percentile because 80 % of the people are better than him.\nFrank is in the 0% percentile because 100 % of the people are better than him.\n\nPercentiles aren't always a good way to determine skill, because if say everyone gets somewhere between 90 and 100, then the person who got a 90 will look like they are doing poorly when in reality they are not.
855479
For each of these numbers, you want to calculate the z-score and then look at either a table or use a scientific/graphic calculator to find the percentages above or below these numbers.\n\nFor the first one, the z-score is:\n\nz = (x - mean) / stdev\nz = (20000 - 20273) / 4100 = -.0666\n\nNow you just find P(z > -.0666).\n\nThe first link below provides a z-table. Looking at it, we can see that it only gives positive values. Since the normal distribution is symmetric, we can just find the probability that z < 0.67, which is 0.24857 according to the table, and then add .5 to that to get 0.74857. \n\nIf you can't visually see what's going on with this, try looking at the second link. You can perhaps use this to do the problems as well, but it's of the greatest benefit for you to know how to use z-scores and z-tables as well.
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.
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.
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
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!
115395
One way to find the percentage of a value is to convert the percentage into a number then multiply the value by this number.\nYou convert a percentage into a number by dividing the percentage by 100: x% = x/100.\n\nFor example: find 5% of 200.\n1) Convert the percentage to a number:\n5% = 5/100 = 0.05\n2) Multiply the value by the number:\n200*0.05=10\n\nIn your case,\n1) Convert the percentage to a number:\n91&3/8% = 91.375/100 = 0.91375\n2) Multiply the value by the number:\n1000*0.91375 = 913.75\n\nYou would have $913.75
308432
Suppose you have a set of numbers:\n\n0, 1, 1, 2, 3, 4, 6, 8, 9, 11, 15\n\nThe median is the middle number when put in least to greatest: 4\n\nThis breaks your list into two halves: 0, 1, 1, 2, 3 and 6, 8, 9, 11, 15.  Your upper quartile is the median of the greater half: 9.  Likewise, your lower quartile is the median of the lower half: 1.\n\nIn the event you have two numbers for a median, take the average of the two and that becomes your median.  In the above example, if you only had 6, 8, 9, 11 for your upper half, your median would be the average of 8 and 9, or 8.5.\n\nAs for your second question (assuming that x belongs in the numerator):\n\n4/5 = (-9/4) - 7x/10\n4/5 + 9/4 = -7x/10\n16/20 + 45/20 = -7x/10\n61/20 = -7x/10\n61/2 = -7x\nx = -61/14
913451
A box and whisker plot!? No problem! \n\nThe Box is from Quartile 1(25th percentile) to Quartile 3(75th percentile),.\n\nThe Whiskers are Q1 minus 1.5 Interquartile ranges and Q3 plus 1.5 Inter Quartile ranges.\n\nIQR (Interquartile ranges) is defined as Q3-Q1\n\nWell say that Q1 is 12 and Q3 is 25 then the IQR is 13, and 1.5IQR is 19.5\n\nTherefore, Anything below -7.5 (12-19.5) and above 44.5 would be considered an outlier. \n\nAs far as drawing the damn thing, just draw a box with the outside edges being Q1 adn Q3, and then draw the whiskers 1.5IQR long.\n\nAdditional info:\n\nThe box and whisker is only 1 criterion for outliers, + or - 3 standard deviations from the mean is another common way, which can get other results.\n\nHope that helps.
561118
A percent is a hundredth of something, so 12% of something is 0.12 times something. To find 12% of 30, multiply 30 × 0.12 with a calculator. You'll wind up with 3.6.\n\nHope this helps.
584205
put 80 over 50 and set it equal to x/100\n\nx shoudl equal 160\n\ntherefore 160%
59697
typically the measures of central tendency are mean, median, mode\n\nMean--essentially the average\nMedian--essentially the middle number, you find this by listing all numbers in the set in order, then start to cross off from each end equally, if the set has an odd number of items you will end on the median, if it does not (thus even) there will be two numbers which you must take the average of.  if you have recurring numbers then you cant cross out all of them in one cross-out, you need to count each individually\nMode--this is the last measure of central tendency, it is the most recurring number in a set,it is possible to have more than one Mode\n\nhope this helps!!
805977
(75-25)*100/25\n50*100/25\n200%
1118574
Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
358124
http://www.bmcc.org/nish/MathTutorials/Percent/percentintro.htm\n\nhttp://www.masslearns.com/link_library.html?subject=NM&sub_cat=50287&final=50290
446411
Usually when  you pull a sample, you know the size of the sample you pulled.  Otherwise, how would you calculate any of the test statistics much less the margin of error?