Normal Distribution Problem?
494209
are u kidding, just use the dam formulas for confidence interval and z scores it is not hard just plug them in
522796
a) Sensitivity = P(Test + | Has disease), so out of the people who have the disease, 93.4% would test positive for it.\n\nSpecificity = P(Test - | Doesn't have disease), so out of the people who don't have the disease, 98.6% would test negative for it.\n\nb) This is again a Bayes' Rule problem\nP(T+|D) = 0.934\nP(T-|ND) = 0.968\nP(D) = 0.002\n\nP(D|T+) = P(D)P(T+|D)/(P(D)P(T+|D) + P(ND)P(T-|ND))\n= (0.002)(0.934)/(0.002(0.934) + 0.998(0.968))
554599
4.2\n\nfor a binomial distribution the mean is the number * the probability
339708
Use a Normal table to find that the value of z so that Probability that Z < z = .90 is z = 1.2816.\n\nNow set z = 1.2816 = (x-100)/15, and solve for x:\n\nx = 1.2816*15 + 100 = 119.224
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
877981
For the first part, per hour the rate is 30.  so we have lambda = 30, so we know the mean and variance (look at the answer before mine).  To find the Probability the 240th person enters in the 8th hour, find the probability 240 or more people arrive in an 8 hour day, then times that by the probability less than 240 arrive in 7 hours.\n\nso (1-P(<240 in 8))*P(<240 in 7)\n\nRemember, The poisson process is memoryless.\n\nSteve
570786
First of all, if the population is normal, the distribution of sample means is normal regardless of sample size.  So I. is true.\n\nSecond, if the population is not normal, the distribution of sample means is approximately normal if the sample size is large.  So III. is also true.\n\nII. cannot be determined to be true unless we know what the population looks like.\n\nSo I would go with C.\n\nUPDATE: Apparently ghakh hasn't heard of the Central Limit Theorem.  You know, the "Fundamental Theorem of Statistics"?  In layman's terms, it states that if the sample size is large enough, then the sample means will be approximately normal if the sample size is large enough REGARDLESS OF WHAT THE POPULATION LOOKS LIKE!!!!!!  All you need is a finite variance, which for elementary statistics students is not something they need to be concerned with.\n\nMaybe you should take my statistics class, ghakh.  You might learn something!
380813
1-i) P(h>200)=P(S>(200-178)/10)=P(S>2.2)=0.5 - 0.4861 = 0.0139 = 1.39 %   (Where S denotes Standard normal distribution value, h the non standard value).\n1-ii) P(h<158)=P(S<(158-178)/10)=p(S<-2)=0.5 - 0.4772 = 0.0228 = 2.28 %\n2)\n  i)P(39.5 < h <40.5)= P((39.5-37)/3.5 < S < (40.5-37)/3.5)\n                   = P( 0.71 < S < 1)\n                   = P( 0 < S < 1 ) - P( 0 < S < 0.71)\n                   = 0.3413 - 0.2611 = 0.0802 = 8.02 %\n  ii)The limits are 38.5 : 39.5\n iii)P(38.5 < h <39.5)= P((38.5-37)/3.5 < S < (39.5-37)/3.5)\n                   = P( 0.43 < S < 0.71)\n                   = P( 0 < S < 0.71 ) - P( 0 < S < 0.43)\n                   = 0.2611 - 0.1664 = 0.0947 = 9.47 %\nNote : S = (h - mean)/sd.
458174
Actually, statistics books are notorious for being written in such a way that no normal human can understand them. The trick, generally, is trying to find a way to relate the problems to something you're interested in. For example, if you like basketball: one possible sampling distribution would be the number of free throws individual players make over a season. You could take a sample of, say, 10 of the players and find the mean and standard deviation. You could compute a 95% confidence interval with that information, which would allow you to say that you're 95% confident that the average number of free throws made by NBA players in a season will be between this number and that number. You could find the probability that a given player would make at least a certain number of free throws in a season by finding the z-score and looking it up on the normal probability distribution table, etc. If you're really struggling with the terminology, I recommend visiting your math help center on campus (most of them have one), or contacting a tutor.
1015585
P(X<6) = Probability that X is less than 6\nTransform the X value of 6 to Z\n\nZ= 6 - 8 / 2 = -2/2 = -1\n\nLook up the Z value of -1 in the standard normal table and we get a probability of 0.1587 or 15.87%.\n\nI think this is right.
278272
You'll use Z-scores and the Z-table. Z-scores convert deviations to standard deviation units. \n\nIn this case, you're looking for the range of scores that makes up the middle 88% of the distribution - 44% on either side of the mean. \n\nThe z-score is 1.56 (look in the table for the z-score that cuts off 44% of the distribution.) This tells you that 44% of the scores fall between the mean and 1.56 standard deviation units above the mean - and 44% of all scores fall between the mean and 1.56 standard deviation units below the mean. \n\nNow, since each standard deviation unit in this particular distribution is 4ml (as per your question), 1.56 standard deviation units is equal to 1.56 x 4, which is equal to 6.24ml.\n\nBack to the problem - your range is 6.24 ml above and below the mean - which is 250ml - therefore the range of volumes is from 6.24ml less than 250 to 6.24ml more than 250 - or in other words - 243.76ml to 256.24 ml.\n\nHope that helped!
341707
The RAND() function is not the appropriate way to solve this.\n\nThe number of cars can be determined a couple of ways.\n\n1.  One is to use NORMINV(.99,2600000,100955) = 2834856\n2.  Or you could use the standard normal distribution:\n\n    x= 2600000 + 100955 NORMSINV(.99)\n     = 2600000 + 100955(2.326348) = 2834856\n\nRAND() just gives you a random number between 0 and 1.  I don't see how that would help you in this problem.\n\nThe second problem is also easily solved using the the standard normal distribution:\n\nFirst, the 2.4 million is 200000 below the mean, \nso z=-200000/100955= -1.98108\n\n2.7 million is 100000 above the mean, so \nz=100000/100955=.99054\n\nSo the probability the number of cars will be between 2.4 and 2.7 million is\n\n= NORMSDIST(.99054) - NORMSDIST(.023791)\n=.839045 - .023791 = .815254
220083
I am pondering this question a bit. I might have a better answer for you, but here is a start:\n- - - - - - - - - - - -  - \n\nYour data is not exactly amenable to standard quantitative analysis. You can't really form an average, for example. \n\nYou can find the mode. The most common answer is, diet. So the mode of your data is "diet". \n\nThe data categories, as they stand, are not represented by ordinal numbers. So it is a bit hard to even describe the distribution, let alone the dispersion of the distribution. What you could do, is say that 35% preferred diet, and 65% did not. Then you could try to estimate the error bounds on these figures, given some assumptions. Or do a robust estimate using a jackknife or bootstrap.\n\n- - - - - -  - - \n\nHere is another idea. You might want to know the average number of people that are in a given category. That is, you might want to know how many people like a given type of soda. You have 5 categories, so the average is \n\n(20)/5=4\n\nThere are an average of 4 people in each category.\n\nNow you can find the standard deviation.\n\nFirst find an estimate of the variance V using the standard formula:\n\nV=(1/4)*[(1-4)^2 + (3-4)^2 +(3-4)^2 +(7-4)^2 +(6-4)^2 ]\nV=(1/4)*[9 + 1 + 1 + 9 +4 ]\nV=(1/4)*[24]\nV=6\n\nThen take the square root\n\nS=sqrt(V)= sqrt(6)\nS=sqrt(6) = 2.44948974\n\nThe average soft drink is liked by 4 people, but there is a standard deviation of about 2.4.\n\nSo one can say a soft drink is liked by 4 += 2.4 people (if we set the error bounds to be 1 standard deviation, but some people set it at 2).\n\nYou could even convert this to percentages. Then you could say that a soft drink is liked by 20% of the people, +- 12%.\n\nThe standard deviation assumes that the distribution is normal, that is a bell-shaped curve (aka Gaussian). However, it probably is not. A robust estimate of the dispersion of the dispersion can be formed just resampling methods, like the bootstrap or the jackknife. I have placed a link to these if you are interested, but I think it might be too advanced for your needs.\n\nAnother thing to do is to look at the median, and the median average deviation, or the median average deviation away from the median. A reference discusses these and other measures of dispersion if you are interested.\n\nI still feel that one might be able to do more with this. I will think some more and if I have any good ideas I will post them. I am not a statistician, so it is likely that one of them might have a much better suggestion right away.
851484
well if you happen to have a graphics calculator then this is easy...\njust key in normalcdf(30,50,42,6)= answer\nnormalcdf(lower boundry, upper boundry, mean, standard deviation)
874605
In a normal distribution:\n68% of the points are within 1 standard deviation\n95% of the points are within 2 standard deviations\n\nYou'll have to look at your normal distribution tables.  You want to find the point where the area is 90%.  This happens around z = 1.645 standard deviations.\n\nSo they should advertise the life of the calculators as 2,150 +/- 450 * 1.645\n\nThat works out to a range of between 1,410 and 2,890 hours, or 2,150 +/- 740 hours.
215721
The previous answer looks good.\nI do have a question on the statement of the problems.\n\nWhen you say 90% are good do you mean that exactly 90% are good or that at least 90% are good?\n\nThe answer above covers exactly 90%.\nFor "at least 90%" you need to add up up p(0)+...+p(10).
905827
If a random variable has a standard deviation of s and you take N samples, then the standard deviation of the sample mean is s/sqrt(N).\n\nIn your case you assume that the standard deviation of the random variable is close to 50 because you have such a huge number of samples. Now if you turn around and only look at 30 samples then the standard deviation will be\n50/sqrt(30).