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May 29, 2009 Math 244 Spring 2009 Quiz 7 Name: 1. Suppose sample data is collected for a particular measurement. The resulting 15 measurements are: 0.68, 2.1, 3.6, 5.3, 7.1, 9.1, 11.4, 13.9, 16.7, 20, 24.1, 29.1, 35.8, 46, 68 Construct two probability plots, one comparing the data with a normal distribution and another comparing it to a exponential distribution. You may do this by hand or with a computer but show your calculations clearly and make your plot as accurate as possible. Then decide which distribution better describes the data. For the normal probability plot you could construct a table with the percentiles and the corresponding values from the standard normal distribution. I used a more sophistocated calculation tool to get the values rather than the table but your values should agree with the normal values to two decimal digits if you used the table. i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Percentile in Decimal From 0.033333333 0.1 0.166666667 0.233333333 0.3 0.366666667 0.433333333 0.5 0.566666667 0.633333333 0.7 0.766666667 0.833333333 0.9 0.966666667 Stnd. Norm Values -1.833914636 -1.281551566 -0.967421566 -0.727913291 -0.524400513 -0.340694827 -0.167894005 -1.39146E-16 0.167894005 0.340694827 0.524400513 0.727913291 0.967421566 1.281551566 1.833914636 Observation 0.68 2.1 3.6 5.3 7.1 9.1 11.4 13.9 16.7 20 24.1 29.1 35.8 46 68 The resulting probability plot plots the third column against the fourth column and gives: Normal Probability Plot Normal Probability Plot 80 70 60 50 40 30 20 10 0 ‐2.5 ‐2 ‐1.5 ‐1 ‐0.5 0 0.5 1 1.5 2 2.5 For the exponential comparison, the only column that changes is the Target. You can utelize the fact that the cdf for an exponential random variable with λ = 1 is easily written down as F (x) = 1 − e−λx . i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Percentile in Decimal From 0.033333333 0.1 0.166666667 0.233333333 0.3 0.366666667 0.433333333 0.5 0.566666667 0.633333333 0.7 0.766666667 0.833333333 0.9 0.966666667 The resulting probability plot looks like: Exponential Values 0.033901552 0.105360516 0.182321557 0.265703166 0.356674944 0.456758402 0.567984038 0.693147181 0.836248024 1.003302109 1.203972804 1.455287233 1.791759469 2.302585093 3.401197382 Observation 0.68 2.1 3.6 5.3 7.1 9.1 11.4 13.9 16.7 20 24.1 29.1 35.8 46 68 Exponen'al Probability Plot Exponen2al Probability Plot 80 70 60 50 40 30 20 10 0 0 0.5 1 1.5 2 2.5 3 3.5 4 It is clear that the second plot is much more linear then the first plot, thus the data fits closely to what would be expected with an exponential random variable when sampled 15 times. 2. Answer problem number 41 from chapter 5 (page 212). ”Let X be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the disribution of X is as follows: x 1 2 3 4 p(x) .4 .3 .2 .1 (a) Consider a random sample of size n=2 (two customers), and let X be the sample mean number of packages shipped. Obtain the probability density distribution of X We could start by building a table, here I will also include the values of r used for part c as well. x1 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 x2 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 p(x1 , x2 ) .16 .12 .08 .04 .12 .09 .06 .03 .08 .06 .04 .02 .04 .03 .02 .01 x 1 1.5 2 2.5 1.5 2 2.5 3 2 2.5 3 3.5 2.5 3 3.5 4 R 0 1 2 3 1 0 1 2 2 1 0 1 3 2 1 0 From this it is pretty clear that the pmf for the mean X will be: x 1 1.5 2 2.5 3 3.5 4 pX (x) .16 .24 .25 .2 .1 .04 .01 (b) Refer to part (a) and calculate P (X ≤ 2.5). For this probability we just need the to add up the probability for all the cases where x ≤ 2.5. Namelu: P (X ≤ 2.5) = .16 + .24 + .25 + .2 = .85 (c) Again consider a random sample of size n = 2, but no focus on the statistic R = the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of R. Using the table from (a) we have that the pmf for R is: r 0 1 2 3 pR (r) .3 .4 .22 .08 (d) If a random sample of size n = 4 is selected, what is P (X ≤ 1.5? To answer this question I am going to make a table for the values of X1 , X2 , X3 , and X4 , but only include cases where the mean is less than or equal to 1.5. x1 1 2 1 1 1 2 2 2 1 1 1 3 1 1 1 x2 1 1 2 1 1 2 1 1 2 2 1 1 3 1 1 x3 1 1 1 2 1 1 2 1 2 1 2 1 1 3 1 x4 1 1 1 1 2 1 1 2 1 2 2 1 1 1 3 p(x1 , x2 , x3 , x4 ) (.4)4 (.4)3 (.3) (.4)3 (.3) (.4)3 (.3) (.4)3 (.3) (.4)2 (.3)2 (.4)2 (.3)2 (.4)2 (.3)2 (.4)2 (.3)2 (.4)2 (.3)2 (.4)2 (.3)2 (.4)3 (.2) (.4)3 (.2) (.4)3 (.2) (.4)3 (.2) x 1 1.25 1.25 1.25 1.25 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 Since these are the only cases then the probability P (X ≤ 1.5) = (.4)4 + 4(.4)3 (.3) + 6(.4)2 (.3)2 + 4(.4)3 (.2) = .24 3. Answer problem number 50 from chapter 5 (p 218) The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. (a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200? Since we are taking a simple random sample with n = 40 > 30 then the Central Limit Theorem applies and the sample mean X√is approximated well by a normal random variable with µX = 10, 000 and σX = 500/ 40 ≈ 79.0569. We use this fact to carry out the calculation: P (9900 ≤ X ≤ 10, 200) ≈ P = Φ 9900 − 10000 500 √ 40 ≤Z≤ 10200 − 10000 √ ! √ ! 40 2 40 −Φ − 5 5 ! 500 √ 40 = Φ (2.53) − Φ (−1.26) ≈ .9943 − .1038 = .8905 (b) If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information. No, in this case since the distribution of an individual rivet is not given then we can not compute the behavior of X unless we have n large enough to invoke the Central Limit Theorem. Unfortunately we can not do that here since n = 15 < 30.