Several Probability and Statistics Techniques
Date: 04/13/98 at 01:59:01 From: Gareth Evans Subject: Probability 1) An investment corporation is formed with enough capital to finance 9 independent ventures. If the chance of being successful on any one venture is 25%, what are the corporation's chances of having: a) exactly one successful venture? b) at least one successful venture? c) all nine ventures unsuccessful? I have no idea how to approach this question. 2) A safety supervisor at a factory believes the average number of accidents per month to be 3.5. What is the probability of: a) exactly 2 accidents next month? b) three or more accidents? I also have no idea with this one.
Date: 04/13/98 at 07:24:27 From: Doctor Anthony Subject: Re: Probability The first question is binomial probability with: n = 9 p = probability of success = 1/4, and q = probability of failure = 3/4 The formula for r successes in n trials is P(r) = C(n,r)p^r*q^(n - r), where C(n,r) is the binomial coefficient: n! ---------- r!(n - r)! Thus, the probability of exactly one success is: Prob(exactly 1 success) = C(9,1)(1/4)(3/4)^8 = 0.225254 The probability of at least one successful venture includes 1, 2, 3, ... through 9 successes, and is best calculated from: 1 - P(0) = 1 - (3/4)^9 = 0.924915 The probability that all nine ventures are unsuccessful is: P(0) = (3/4)^9 = 0.075085 Question 2 is a Poisson probability with mean 3.5 for a period of a month. So: P(0) = e^(-3.5) = 0.0301974 P(1) = 3.5/1 * P(0) = 0.1056908 P(2) = 3.5/2 * P(1) = 0.1849590 P(3) = 3.5/3 * P(2) = 0.2157855 So the probability of exactly 2 accidents is 0.18496. To find probability of 3 or more we calculate: 1 - [P(0) + P(1) + P(2)] = 1 - 0.320847 = 0.679153 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 05/06/98 at 07:25:33 From: Gareth Evans Subject: Re: Probability Dear Dr. Maths, I am currently studying probability and am having a bit of trouble with the following questions due on Friday. I was wondering if you would be able to help me. If you can, thank you very much. The questions are: 1.) An office manager wishes to estimate the mean time required to handle customer complaints. A sample of 38 complaints shows a mean time of 28.7 minutes with a standard deviation of 12 minutes. a) Construct a 90% confidence interval for the true mean time required to handle customer complaints. b) How large a sample should be taken, if the manager wishes to estimate the mean time to handle a complaint to within +/- one minute? Assume that the confidence interval is to be 90%. 2.) A hospital claims that the average length of stay is 5 days. A study of the length of stay for a random sample of 25 patients found the mean stay to be 6.2 days with a standard deviation of 1.2 days. Do these data present sufficient evidence to support the hospital's claim? Use a 0.05 significance level. 3.) A survey was conducted to determine the proportion of registered nurses in a particular state that are actively employed. A random sample of 400 nurses selected from the state registry showed 274 actively employed. Find a 95% confidence interval estimate for the true proportion of registered nurses actively employed in the state. Thanks for your help. George Evans
Date: 05/06/98 at 11:26:56 From: Doctor Anthony Subject: Re: Probability In part b of the first question, the 90% confidence limits are given by: 28.7 - xm ----------- = +/- 1.645 12/sqrt(38) where xm = population mean. Thus: xm = 28.7 +/- 12/sqrt(38)*1.645 = 28.7 +/- 3.20225 And so: 25.498 < xm < 31.902 For question (1a), you require: 12/sqrt(n)*1.645 = 1 sqrt(n) = 12*1.645 = 19.74 n = 389.66 n = 390 For the second question, find the z-statistic: 6.2 - 5 1.2 * 5 ------------ = --------- = 5 1.2/sqrt(25) 1.2 This value must be checked against z = 1.645 for a one-tailed 5% test. It is a VERY significant result, and we reject the null hypothesis that the mean is 5 days. For the third question, we have: p = 274/400 = 0.685 q = 0.315 n = 400 pq/n = .0005394 sqrt(pq/n) = 0.023226 Then the true proportion of nurses employed is: .685 - pm ---------- = +/- 1.96 0.023226 pm = .685 +/- 0.023226*1.96 = .685 +/- 0.04552 So: 0.63948 < pm < 0.7305 This is the 95% confidence interval for the proportion of nurses actively employed in the state. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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