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Order the data. ........................................... 1 2.68 2 2.76 3 2.78 4 2.81 5 2.82 6 2.86 7 2.92 8 3.10 9 3.13 10 3.14 11 3.18 12 3.38 13 3.38 14 3.40 15 3.94 16 3.99 17 4.13 18 4.16 19 4.19 20 4.20 21 4.29 Range = 4.29 - 2.68 = 1.61 Number of intervals = 5 Interval amplitude = 1.61/5 = 0.322 [.................) fi..Midpoint 2.680 - 3.002 7...2.841 3.002 - 3.324 4...3.163 3.324 - 3.646 3...3.485 3.646 - 3.968 1...3.807 3.968 - 4.290 5...4.129 Assume the last interval is closed. Download Graph 4.4 from www.padowan.dk for free. On "Axes", change the range of edge X from 2.5 to 4.3 Choose "Insert function", "Insert number of points", type X ans Y, then "OK". You'll see the function!.
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Among U.S. cities with a population of more than 250,000 the mean one-way commute to work is 24.3 minutes. The longest one-way travel time is New York City, where the mean time is 38.3 minutes. Assume the distribution of travel times in New York City follows the normal probability distribution and the standard deviation is 7.5 minutes. What percent of the New York City commutes are for less than 30 minutes? I know this is probably a very simple question but I can't seem to figure it out. Please show all steps involved. Thanks
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For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.) You can translate into standard normal units by: Z = ( X - μ ) / σ Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed with mean μ and standard deviation σ /√(n) An applet for finding the values http://www-stat.stanford.edu/~naras/jsm/... calculator http://stattrek.com/Tables/normal.aspx how to read the tables http://rlbroderson.tripod.com/statistics... In this question we have X ~ Normal( μx = 232000 , σx² = 4.9e+07 ) X ~ Normal( μx = 232000 , σx = 7000 ) Find P( X < 232000 ) P( ( X - μ ) / σ < ( 232000 - 232000 ) / 7000 ) = P( Z < 0 ) = 0.5 an easier thing to do with this is just to see that for any normal random variable, the probability of observing a value greater than the mean is 50% and so is the probability of observing a value less than the mean. This is because the normal distribution is symmetric about it's mean. Find P( 232000 < X < 239000 ) = P( ( 232000 - 232000 ) / 7000 < ( X - μ ) / σ < ( 239000 - 232000 ) / 7000 ) = P( 0 < Z < 1 ) = P( Z < 1 ) - P( Z < 0 ) = 0.8413447 - 0.5 = 0.3413447 Find P( X < 239000 ) P( ( X - μ ) / σ < ( 239000 - 232000 ) / 7000 ) = P( Z < 1 ) = 0.8413447 Find P( X < 245000 ) P( ( X - μ ) / σ < ( 245000 - 232000 ) / 7000 ) = P( Z < 1.857143 ) = 0.9683546 Find P( X > 225000 ) P( ( X - μ ) / σ > ( 225000 - 232000 ) / 7000 ) = P( Z > -1 ) = P( Z < 1 ) = 0.8413447
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Combinations. (If it mattered which juror was selected 1st, etc, then it would be permutations.)
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P( at least one) = 1 - P(zero people ) Use binomial probability P(x = k successes) = (n k) * p^k*(1-p)^(n-k) where (n k ) = n! / ( (n-k)! * k!) so k =0 , p = 0.00114 1-p = 1- 0.00114 = 0.99886 (n k ) = 30! / ( 30!* 0!) = 1 P(zero people) = 1 * (1)* (0.99886)^ 30 = 0.966359357 === answer to part 1 P(at least 1 ) = 1 - 0.966359357 = 0.033640643 Referring to Question #2, is it likely that such combined samples test positive? Will the company have to perform many individual blood tests? for every 30 people, equation 2 says the the probability of testing all 30 equals 0.033640643 rate = 1/ p = 1/ 0.033640643 = 29.72594782 so this means 1 in every 29.72594782 group test of 30 you will end testing all 30 individuals if you have a 900 employees company so that means out of 900 people (which is 30 tests) you will only have to individually test 30 or 60 people Out of 900 people testing 30 or 60 is not alot. === my first answer to part 2 so I would say they would not have to perform many individual blood test.
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So my daughter gave me this question and she has more like it but I cannot find the solution anywhere online while searching probability. If anyone knows the steps it would be much appreciated so I can help her. 1) Data collected over a long period of time show that the length of time x to complete a particular college entrance test is normally distributed with an average of 125 minutes and a standard deviation of 18 minutes. What is the probability that a student taking this test will finish in 100 minutes or less? Round your answer to 4 decimal places. Remember to round all z values to 2 decimal places. A. 0.9177 B. 0.7538 C. 0.0823 D. 0.1566 E. None of The Above
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Let D be the event of having the disease Let ~D be the event of not having the disease Let + be a positive test Let - be a negative test. We know that: P( + | D) = 0.934 P( - | ~D) = 0.968 P(D) = 1/500 = 0.002 From this info we can also find that: P( - | D) = 1 - 0.934 = 0.066 P( + | ~D) = 1 - 0.968 = 0.032 P( ~D) = 1 - 0.002 = 0.998 == this is an exercise in conditional probability. You can think of it as and exercise in Bayes' rule as well. This is an exercise in conditional probability. For any two events A and B, where P(B) ≠ 0, you have the conditional probability: P( A | B ) = P( A ∩ B ) / P( B ) = P( B | A) * P(A) / P(B) the above is read as: the probability of A given B is equal to the probability of A and B divided by the probability of B. We need to find: P( D | + ) P( D | + ) = P( + | D ) * P(D) / P(+) Let's start by finding P(+) Use The Law of Total Probability For a set of events A1, A2, A3, ... , An where the Ai's are mutually exclusive and exhaustive events and for any other event B P(B) = P(B and A1) + P(B and A2) + ... + P(B and An) = P(B | A1) * P(A1) + P(B | A2) * P(A2) + ... + P(B | An) * P(An) P(+) = P(+ | D) * P(D) + P( + | ~D) * P(~D) P(+) = 0.934 * 0.002 + 0.032 * 0.998 P(+) = 0.033804 Now we can find the solution. P( + | D ) = P( + | D ) * P(D) / P(+) P( + | D ) = 0.934 * 0.002 / 0.033804 P( + | D ) = 0.05525973 Not great, the probability of having the disease given one positive test is 0.05525973 ====== ========= =========== = = = = = = = = = What if you had two positive test? P( ++ | D) = 0.934 ^ 2 = 0.872356 P( ++ | ~D) = 0.032 ^ 2 = 0.001024 P(D) = 0.002 P( D | ++ ) = P(++|D) * P(D) / P(++) P(++) = P(++|D) * P(D) + P(++|~D) * P(~D) P(++) = 0.872356 * 0.002 + 0.001024 * 0.998 P(++) = 0.002766664 P( D | ++ ) = 0.872356 * 0.002 / 0.002766664 P( D | ++ ) = 0.6306194 this is why many times when a test for a rare disease comes back positive that a second or third test will be run to confirm. To have the disease after multiple positive tests increases the probability drastically.
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How do i answer this? A relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution with a mean of 93 months and a standard deviation of 3.5 months. When this computer-relay microchip malfunctions, the entire satellite is useless. A large London insurance company is going to insure the satellite for 50 million dollars. Assume that the only part of the satellite in question is the microchip. All other components will work indefinitely. (a) For how many months should the satellite be insured to be 99% confident that it will last beyond the insurance date? (Round your answer to the nearest month.)
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Multiply the number of names on the list, in this case "10" by the sample "4". 40 is the outcome of that. Therefore, your answer is 40 different samples.
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1.) From the binomial distr. equation: P(X=x) = (nCx) * p^x * q^(n-x) in this case, x=2, p= 0.3, n=5, q = 1 - p = 1 - 0.3 So substituting... P(X=2) = (5C2) * 0.3^2 * (1-0.3)^3 = 5!/(2!*3!) * 0.09 * 0.7^3 .... dont have a calculator but think its... = 10 * 0.09 * 0.49 * 0.7 = 10 * 0.09 * 0.343 = 3.43 * 0.09 = 3.087 ....should be right, just verify 2.) Many ways to do this, but ill take the shortest way, we just need to make sense of this first If we flip a (balanced!)coin, we get either a head or a tail. this means the probability of heads is 0.5, right? they give us a number of times the coin is tossed, this implies we working with the binomial distribution. For variance(binomial), var = npq var = 72 * 0.5 * 0.5 var = 18 std dev = sqrt(var) = sqrt (18) = 4.something ...again, calculator work 3.) Now im not american (no idea what SAT is), and the question was written down a bit funny, but i think you want the who are in the top 5% ? So... this is not a sample, since they not providing us with a number of applicants to test, so we use "population" equations... We now try to find the test score which would put us in the top 5%. Other words, the people who beat at least 95% of the people taking the test... We'll use the normal table for the corresping z-score for 0.95 (or 0.05 depending on the table). Cant remember but im pretty sure its 1.645, anycase i'll just call it z (x - 1000) / 200 > z x-1000 > 200z x> 200z + 1000 substitute z in, and get the score for x (again look in normal table and do calculator work) x > 200 * 1.645 + 1000 = 329 + 1000 = 1329 ... if my guess for z is right so a minimum score of 1329 would put an applicant in the top 5% I adivse you to work along with these answers i've given, just make sure u understand where everything is coming from, and what you want to do with them. Irrespective if your question is right or wrong, your procedure and understanding of the work is what would give you an A in statistics (hopefully a bursary and deduction of class fees as well :)