Statistics Question help?
You will accept the lot if there are 0, 1, or 2 defective units out of 25. The probability of any one unit being defective is .01 (1%=.01, not .001, so the probability that the unit *isn't* defective is .99) P(0 defective units) = (0.99)^25 P(1 defective unit) = 25 * (0.01) * (0.99)^24 ------This is because there are 25 ways to get one defective unit: in the first one you check, the second, the third, etc. P(2 defective units) = 25_C_2 * (0.01)^2 * (0.99)^24 ------25_C_2 is read "25 choose 2" and represents the number of ways to get 2 defective units out of 25. You can find this in your calculator as the nCr command, which is under the MATH->PRB menu if you have a TI-83 or 84 So, the probability of getting at most 2 defective units is the sum of P(0)+P(1)+P(2)= .7778+.1964+.0238=.9980 So your answer is choice (A) For the record, this is an example of the Binomial Probability Model (any example of finding x successes out of n trials is). You can do this calculation really quickly by typing binomcdf(25,.01,2) into your calculator (the command is under the DISTR menu on a TI-83/84). That command automatically calculates the probability of at most 2 successes out of 25 trials, where the probability of success is .01. The similar binompdf command will give you the probability of *exactly* the number of successes you request.
What's the expected number of successes for this statistics problem?
A new phone answering system installed by the Ohio power company is capable of handling 5 calls every ten minutes. Prior to installing the new system, company analysts determined that the incoming calls to the system are Poisson distributed with a mean equal to two every 10 minute. If this incoming call distribution is what the analysts think it is, whats the probability that in a 10 minutes period more calls will arrive that the system can handle?
Statistics question, How do I do this?
A) ANSWER: PROBABILITY = 0.13 (13%) probability 13 graduate in 4 years Why??? BINOMIAL DISTRIBUTION, POPULATION PROPORTION n = NUMBER OF TRIALS [ 30 ] (sample size) k = NUMBER OF SUCCESSES [ 13 ] (exactly k NUMBER OF SUCCESSES) p = POPULATION PROPORTION [ 39 %] significant digits 3 "Look-up" value of PROBABILITY = 0.13 =BINOMDIST ( 13 , 30 , 39/100 ,FALSE ) "Using Excel function: BINOMDIST(number_s, trials, probability_s, cumulative) Number_s is the number of successes in trials. [ 30 ] Trials is the number of independent trials. [ 13 ] Probability_s is the probability of success on each trial. [ 39]" Cumulative is a logical value that determines the form of the function. If cumulative is TRUE, then BINOMDIST returns the cumulative distribution function, which is the probability that there are at most number_s successes; if FALSE, it returns the probability mass function, which is the probability that there are number_s successes. B) ANSWER: PROBABILITY = 0.92 (92%) probability 15 or fewer students graduate in 4 years Why??? BINOMIAL DISTRIBUTION, POPULATION PROPORTION n = NUMBER OF TRIALS [ 30 ] (sample size) k = NUMBER OF SUCCESSES [ 15 ] (from 0 up to and including k NUMBER OF SUCCESSES) p = POPULATION PROPORTION [ 39 %] significant digits 2 "Look-up" value of PROBABILITY = 0.92 =BINOMDIST ( 15 , 30 , 39/100 , TRUE ) "Using Excel function: BINOMDIST(number_s, trials, probability_s, cumulative) Number_s is the number of successes in trials. [ 30 ] Trials is the number of independent trials. [ 15 ] Probability_s is the probability of success on each trial. [ 39]" Cumulative is a logical value that determines the form of the function. If cumulative is TRUE, then BINOMDIST returns the cumulative distribution function, which is the probability that there are at most number_s successes; if FALSE, it returns the probability mass function, which is the probability that there are number_s successes.
Help with statistics part of question?
Brooke Consider a random sample of the size n=1800 from a binomial probability distribution with P=.40, and X= number of successes. a) Find the mean and standard deviation of the number of successes mean = np = 1800(0.40) = 720 std dev = √(npq) = √[(1800)(0.40)(1 - 0.40)] = 20.7846 b) Find the probability that the number of successes is greater than 775. Using the Normal approximation: First the continuity adjustment ... P(X > 775) =P(X > 775.5) Now use the Normal ... P(X > 775.5) = P[z > (775.5 - 720) / 20.7846] = P(z > 2.67) = 0.0038 [from the Standard Normal table] Note: If you have a binomial calculator, the "exact" binomial solution is 0.0039, so you see the "Normal Approximation" is very very accurate. Hope that helped
Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test?
Let's see... You created your profile ten days ago and have asked 27 other questions, (all of which are statistics problems), yet you've only answered 4 questions. So, if you're asking three questions per day, (on average), the probability that you need a tutor or additional help in your chosen subject exceeds three standard deviations. I suggest you seek other sources of help. Good luck in your studies, ~ Mitch ~
Statistics question PLEASE HELP ME IN DETAIL?
A professional basketball player makes 85% of the free throws he tries. Assuming this percentage will hold true in future attempts, find the probability that in the next 8 tries, the number of free throws he will make is: (a) exactly 8; (b) exactly 5; (c) at least 1; (d) at most 7 (use result in part (a)). (e) Find the mean and standard deviation of X, the number of free throws that the basketball player makes.
Math statistics question binomial cumulative distribution?
A door to door canvasser tries to persaude people to have a certain double-glazing installed. Probability that his canvassing at a house is successful is 0.05. Use tables of cumulative binomial probabilities or otherwise to find to 4 decimal places the probability that he will have at least 2 successes out of first 10 houses he canvasses. Next, find the number of houses he should canvass per day to average 3 successes per day. Next, calculate the least number of houses that he must canvass in order that the probability of his getting at least one success exceeds 0.99 Help with the last question. The first two is quite easy, but the last is hard and I can't do :( answer is 90 for last ques
What will be the answer if a multiple choice test consisting of 10 questions with four choices each, and the student guesses the answer to each question? What is the probability that he gets 8 questions correctly?
For each question we have 4 choices. So if you have 10 on such questions, you ll have :4*4*4*4*4*4*4*4*4*4= 4^10 = 2048Getting 8 correct means=> 8 corrects and two wrongs3/4 is the probability if a question is answered wrong1/4 is the probability if a question is answered correctlyso we can consider this probability :P=(3/4)*(3/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)=> P=0.00000858306But as we can select any combination of 2 from 10 for being wrong, so we need to multiply P by (10,2)=10!/(2!*8!) =45so the answer is P*45 = 0.00038623809So not very lucky to answer randomly!By the way, is it a question from your home work assignment? ;) ;)
I need help Statistics Problem?
Binomial distribution is to be used p=80%=0.8 q=1-p=1-0.8=0.2 n=10 P(x)=nCx*q^(n-x)*p^x Find out the probabilities corresponding to x=0, x=1, x=2..........x=10 by using the above formula and list out the probabilities to get the required sampling distrbution. As n =10, the probabilities may be very veery small. It is better to make approximation of the probabilities obtained to the extent desired. for ex. P(x=3) = 10C3*(0.2)^7*(0.8)^3 = 120*0.0000128*0.512 = 0.000786432 This probability can be approximated to 0.0008 (This is same as the Pr. given in the book answer) Another ex. P(x=9) = 10C9*(0.2)^1*(0.8)^9 = 10*0.2*0.134217728 = 0.268435456 This probability can be approximated to 0.2684 ( This is same as the Pr. given in the book answer) P(x=10) can be taken as the balancing figure as the total of probabilities must be equal to 1 In this case some difference can be observed either upwards or downwards if compared with the calculated value due to approximation of so many probabilities.