Stats.. problem - stuck?
969083
That statement is false.\n\nIf those statements above are true and assuming n is large, then the standard error would be 5, because by the empirical rule, it would take two standard errors above and below expected to get 95% in the middle.\n\nThe standard error for the sum is equal to sqrt(n)*SD, so in order to double the standard error, you would have to actually multiply the sample size by 4.  That is because\n\nsqrt(4n)*SD = sqrt(4)sqrt(n)*SD\n= 2*sqrt(n)*SD\n\nSo doubling the sample size would not be enough to double the standard error for the sum.  Since the standard error for the second part of the problem would be 10 (using the same reasoning as before, and using the fact that doubling the sample size would double the expected sum), then doubling the sample size would not be enough.\n\nIn other words, in order for that statement to be true, you would have to quadruple the sample size, not double.\n\nedit: The population should be large as compared to the sample size also.  That probably should have been stated in the problem, but, given the wording of the problem, it is probably safe to assume that.\n\nUm, rev, 400 is the expected sum, not the number of draws.  There's a difference.  Way to pretend you know what is going on here.
212031
Well, the question doesn't have all the required details, but why not:\n- make sure that you ar eusing the correct type of vaiable (string/number/etc)\n- make a simple if statement that says, IF the number of tries is less than 10 then continue (where number of tries is lets say "x")\nOnce an answer is chosen, if the answer is corret, then thats it, ELSE, if x < 10 then x = x + 1 \n(this makes sure that x is less than 10...and adds 1)\nIF x = 10 THEN MsgBox "HARD LUCK" (like that, if the previous number was 9, by adding 1 it became 10...this line finishes the game up)\n\nAt the end, make sur eto reset the value of "x"\nx = 0
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
873308
OK - I will try and explain it, but forgive me if I get myself confused...  hehehe\n\nFirst step - make a bell curve.  Put a mark in the center and label that "7 hours" - that's your mean.  Next, put three marks on either side of that middle point - those will be your standard deviation points.  For the points on the right, ADD 5.3 to the mean for the first one, then add 5.3 again for the 2nd, etc...  Points on the left side of the mean have the 5.3 SUBTRACTED.  (Am I still making sense?  lol)\n\nNow - the 68-95-99.7 rule says that 68% of the freshmen will fall into the area between the first line on EITHER side of the mean, 95% between the second lines and 99.7% between the third lines on either side of the mean.\n\nI think you should be able to figure it from there - I hope so...  Right now I am so confused I think I gave myself a headache!  lol\n\nGood luck!
822655
How is the problem phrased?  Most statistical questions are ok with a 22.8 answer.  Like in the US, most families have 2.4 kids... they don't really have .4 of a kid, but that it is an appropriate way to give a stat.  \n\nOtherwise, if you just need to know something like "how many times could the car have possibly stopped," then 22.\n\nIt would help to know what the problem says and what class and grade you're in. =)
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
541248
Yes, I faced exactly the same problem a few days back. Instead of using the normal chart, use a scatterplot instead. Cheers!
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.
574687
Range Questions:\n1) max value is 154, min value is 2: 154-2 = 152\n2) max value is 189, min value is 13: 189-13 = 176\n3) max value is 155, min value is 2: 155-2 = 153\n\nMean/Median/Mode Questions:\n\nRemember that the mean is another way of saying "average" -- that is, add up all the values you have, and then divides that sum by the number of data points you have.\n\nThe median is the "center point" of your data -- if you sorted your data from smallest to largest, there are as many data points that are larger as there are that are smaller. \n\nThe mode of a data set is the value that is repeated most often. \n\n3) Mean = 106, Median = 113, Mode = 113\n4) Mean = 73, Median = 79.5, Mode = 81\n5) Mean = 102, Median = 121, Mode = 65 and 121\n\nIn problem 5 you have two modes, because both 65 and 121 are present twice, while all the other values are only present once.
17841
You might try DAP. You can find it here along with a lot of other free statistical software.\nhttp://members.aol.com/johnp71/javasta2.html
611456
answer A LOT of questions. to get ur stats, click my Q&A
112971
There's a problem with equation. Go back and check it.\nAs it stands, no x would satisfy this.\nInfity is an abstraction, not a number.
571247
i wish i could help you...it looks like you need a stats book...i was tryin to find this website i used to use in my stats class...but of course when you are lookin for somethin u cant find it...sorry about that
806376
an enormous number..!!!\ncalculator gave up
123406
Try Chi square testing, it starts out complicated but once you understand it, it is way faster than punnett squares! You can find all kinds of stuff about it on the internet, it's a statistical analysis.
408817
X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.
700836
http://72.14.209.104/search?q=cache:IQ-eljx9Fi4J:www.mathimp.org/downloads/resources/impressions/IMPressions_fall02.pdf+POW+1+Growth+Of+Rat+Populations&hl=en&gl=us&ct=clnk&cd=3\n\nThis link will show you how to answer the problem.