advertisement

ENGRD 241: Engineering Computation PS06S, Fall 2002 SOLUTION TO PROBLEM SET NUMBER 6 1. C&C Problem 17.3, page 471. 28 (a) The mean is given by, x x i 1 i = 27.6786 28 28 (b) The standard deviation is given by, sx (x x ) i 1 2 i 28 1 = 10.8014 (c) The variance is the square of the standard deviation, and is found to be 116.6706 (d) The coefficient of variance, in percent, is given by, 100*(Std. Deviation)/(Mean) = 39.024% (e) The upper and lower limits of the 90% confidence interval for the mean, are given by, Lx sx s t 2,n 1 and U x x t 2,n 1 n n The t statistic has to be found at t0.05,27 , which could be found either from a table in a book on statistics or by using the command TINV(0.1,27) in EXCEL. Its value is 1.703. By applying the appropriate values, we find the lower limit = 24.2023, and upper limit = 31.1549 (f) The histogram is, Number Histogram for C&C Prob. 17.3 8 7 6 5 4 3 2 1 0 1-5 6-10 11-15 16-20 21-25 26-30 31-35 36-40 41-45 Intervals (g) It is known theoretically that for a normal distribution, the range limited by x S x and x S x would encompass 68% of the total measurements. In this problem, this range is Page 1 of 2 ENGRD 241: Engineering Computation PS06S, Fall 2002 from 16.8772 to 38.48. From the data given, 20 of the 28 measurements (i.e., 71.43%) fall within this range. This shows that it is a very conservative estimate for this set of measurements. 2. Suppose the force acting on a column that helps to support a building is normally distributed with mean 15.0 kips and standard deviation 1.25 kips. What is the probability that the force: (a) Is at most 17 kips? (b) Is between 12 and 17 kips? (c) Differs from 15.0 kips by at most 2 standard deviations? [Acknowledgement: J. L. Devore, Probability and Statistics for Engineering & the Sciences, 4th Ed., 1995, Prob. 4.31, p. 164.] Given, = 15 and = 1.25 (a) The required probability is, P(X 17) = P(Z 17 15 ) = P(Z 1.60) = (1.60) = 0.9452 1.25 (b) The required probability is, P(12 X 17) = P( 12 15 17 15 ) = 0.9370 Z 1.25 1.25 (c) The required probability is, P(|X-15| 2(1.25), which can be rearranged as, P(-2.5 X-15 2.5) = P(12.5 X 17.5) = P(-2 Z 2) = 0.9544 3. Before agreeing to purchase a large order of polyethylene sheets for a particular type of highpressure oil-filled submarine poser cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than 0.05 mm. What hypotheses should be tested, and why? In this context, what are the type I and type II errors? [Acknowledgement: J. L. Devore, Probability and Statistics for Engineering & the Sciences, 4th Ed., 1995, Prob. 8.5, p. 313.] One of the hypotheses could be that the standard deviation is less than 0.05 and the other hypothesis could be that the standard deviation is equal to 0.05. If denotes the standard deviation, the 2 hypotheses can be represented as: H o : = 0.05, and H a : < 0.05 These 2 are exclusive hypotheses, and the sheets won’t be used unless H o is rejected in favor of H a . The 2 types of errors are: Type 1: Conclude that the standard deviation is less than 0.05mm when it is really equal to 0.05mm. Type 2: Conclude that the standard deviation is equal to 0.05mm when it is really less than 0.05mm. Page 2 of 2