Statistics questions - help!!!?
915221
- How many links should be strung together to form a chain?\n\nNumber of links = 54/0.5 = 108\nsd = 108 * 0.01 = 1.08\n\nSo mean length (μ) of 108 links = 54" with a sd (σ) of 1.08"\n\n- What proportion of the chains would meet BD's standards?\n\nz = (x - μ)/σ + 0.5\n= (54.25 - 54)/1.08 + 0.5\n= ~0.7315\n\nFrom a table of z-scores establish the proportion that would lie above the mean and double it for those that lie below the mean.\n\nP(-∞ to 0.7315) ≈ 0.617 (Ref Table 1)\nTherefore P (-0.7315 to 0.7315)\n≈ (0.617 - 0.5)*2\n= 0.117 *2\n= 0. 234\nie 23.4% of the chains will meet the standards and 76.6% of chains fail\n\n- Supposed BJ Company wished to reduce the proportion of chains that failed to meet Bridgestone’s standards to one third the current value to what value would they have to change the standard deviation of the length of the individual chain links?\n\nif only 1/3 fail (1/6 too long and 1/6 too short) then 2/3 pass and so want to establish z for p ≈ 0.5 + 0.333 = 0.833\n\nFrom table (See Ref 1)\nz ≈ 0.7975\n\nSo 0.7975 = (54.25 - 54)/σ + 0.5\n0.25/σ = 0.2975\nσ = 0.25/0.2975\n= 0.84 inches for the 108 links\n= 0.0039 inches per link
776173
1) this operation is made up of two steps, that means the probablility of the whole operation = the multiplication of the probablility of every step\n a) probaility of a 1 from the first box = 1/6, the other box = 1/7 , so P(Both will be 1's) = 1/7 * 1/6\n\nb) p(Exactly one of the will be a 2) = 1/6 * 6/7 + 1/7 * 5/6 \n\nc) P(Both will be even) = 3/6 * 3/7 \nd) P( Exactly one of them will be even) = 3/6 *4/7 + 3/6 * 3/7\n\ne) P(At least one of them will be even) = 1- both of the odd = 1- (3/6 * 4/7)\n\n>> as for the sample space it would take me forever to write it down! sorry\n\n2) Not sure about this one\n3) a) 5%\n    b) 16% >> dont know wut details to include its just a law
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. =)
663334
check out awstats or http://google.com/analytics
907438
I could tell you the answer or I could give you an idea of what to consider when you do it yourself.\n\na) Binomial numbers and p = 0.7, q= 0.3 might help here.  You do not know the size of the population of the mailing list so you might need to make an assumption that it is large enough to assume a Normal distribution.  I think the central limit theorem might help here.\n\nb) Binomial theory, and conditional probabilities.  Look it up in your text book!  Again if the size of the population is large enough you might be able to simplify the problem.
765221
CASE 1:\nThe probability that 2 green balls are selected from box A:\n1/2 (chance of picking box A)\nChance of picking first ball green (3/6)\nChance of picking second ball green (2/5)\n\nSo the combined odds are: 1/2 x 3/6 x 2/5 = 1/4 x 2/5 = 1/10\n\nCASE 2:\nThe probability that 2 red balls are selected from box B:\n1/2 (chance of picking box B)\nChance of picking first ball red (4/9)\nChance of picking second ball red (3/8)\n\nSo the combined odds are: 1/2 x 4/9 x 3/8 = 12/72 = 1/6
808142
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.
883654
If the claim is correct, then the mean of samples of size 25 has a normal distribution with a mean of 10 (the population mean) ounces and a standard deviation of 0.5/sqrt(25) = 0.1.\n\nz = (10.15-10)/0.1 = 1.5\n\nAccording to the standard normal table, the probability is 0.0669.\n\nThis is due to the fact that the population is normal.\n\nedit: You guys, it is not three sd's above.  You have to think in terms of standard error.  It is 1.5 standad errors above.  Do the math.
1180637
I was answering another similar question, and I came across a website that has a downloadable sample spreadsheet for doing statistical distributions.  Play with this and see if it helps you out.
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
331044
The College Board website has some sample multiple choice questions, as well as past free responses:\nhttp://collegeboard.com/student/testing/ap/statistics/samp.html?stats\n\nGood luck!
971574
This is a proportioal problem\n\nMultiply the means\n\nmultiply the extremes\n\n- - - - - - -\n\nx = the amount of men\n\n4 / 7 = x / 28\n\nthe means = 7x\n\nThe extremes = 4 x 28\n\n- - - - - - - - - - - - - - -\n\n7x = 4 x 28\n\n7x = 112\n\n7x / 7 = 112 / 7\n\nx = 16\n\nThe answer there are 16 men\n\n- - - - - - s-
652897
Here are some statistics:
1025917
i should know this, I did a semester of statistics....\n\n hmmmm.....\n\n I guess I should have went to class more...... I hope I passed the exam.....\n\n good luck....\n\n..
188021
As you can see the answers are all over the place.  I have provided a link to a website which shows statistics on a graph from the year 2001.  Although it's not the most current of information, it is a good starting point.
407360
yea, multiplying the single event probabilities of two independent events to get the prob of the independent events conjunction is correct. In other words (to avoid you going to grab an Advil already): p(A&B)=p(A)*P(B) assuming A and B two independent events. That's BASIC stats.\n\nIn excel, use the function POWER. First argument the probability, second argument the number of occurence in a row.
292525
several classes (100 students)of 6th graders went on a field trip to the zoo. 1/5 of them went to c the monkeys. 1/2 of them visited the penguins. 1/4 went 2 c the bats. and the rest went to the snack bar. how many students went to the snack bar? if u hav a heart. u will help \n\nwell...i got 5 students went to the snack bar...\n1/5 of 100=20\n1/2 of 100=50\n1/4 of 100=25\nso \n  20+50+25=95\n  100-95=5 students\n~Brendon