Back to the Table of Contents

Name:                                                         

Prob. & Dist. Test Key, 4th 9-weeks, May 20, 2004

  1. DAEHJIBCFG

  2. DGHFAEICJB

  3. (7-1)!/(2!2!)=180. 7 letters, reduce by one since circular, and two letters repeated (OS), each twice.

  4. 8×2×10=160.

  5. Let the test be an air bag trigger and the condition being tested be a collision. Let H0 be the air bag triggered properly and Ha be it did not. Then a type I error or false negative would be our alpha region or there was a collision and the air bag did not trigger. A type II error or false positive would be the beta region or there was not a collision and the air bag did trigger. Both kinds or errors are problematic, but for different reasons.

  6. 1/2 + 1/3 + 1/9 + 1/18 = (9 + 6 + 2 + 1)/18 = 18/18 = 1.000. Yes.

  7. Remember, if you seed the random number generator with zero first, you will always get these values: 1 1 0; 1 0 1; 0 0 1; 0 1 1; 0 0 0; 1 0 0; 1 1 1; 0 0 0. Thus in the 8 families there are 11 boys for an average of 11/8 = 1.375 boys per family.

  8. [3(1-20) + 4(2-20) + 4(5-20) + 6(10-20) + 3(100-20)]/20 = (-57 - 72 - 60 - 60 + 240)/20 = -9/20 = -$0.45. The denominator 20 is part of each probability whereas the numerator 20 is the cost per each win.

  9. Using binompdf(3,.9) or the binomial distribution formula/program one obtains: (0, 0.001), (1, 0.027), (2, 0.243), and (3, 0.729). Here n = 3 and p = 0.9.

  10. Using poissonpdf(1,2) where the first argument is the mean and the second argument is the value whose probability you seek, one obtains 0.1839. Note that this is also the same as the bank example in the text. Let µ = 1 and x = 2. P(2)=e-1/2!=0.3679÷2=0.1839.

  11. The line x = 0.005 intersects the Lorentzian at x = 90.0 and x = 110.0 MHz. Thus the FWHM is 110.0 - 90.0 MHz or 20.0 MHz. (Oops---there are no units given in the problem.)

  12. We compare here, in relative magnitude, the right tail areas of these two distributions. Be sure to include a sketch. tcdf(1.960,9E99,5)÷normalcdf(1.960,9E99) yields the result 2.1459 indicating the area under the probability distribution curve is over twice as big when you have such a small sample and do not know the population standard deviation.

  13. Using the formula (O-E)2/E and summing over all observed (O) and expected (E) we obtain: 0.125 + 0.167 + 3.014 + 1.877 = 5.179.

  14. Using the calculator for a frequency mean: 1-Var Stats L1,L2, we put 0, 1, 2, 3 in L1 and 7, 100, 350, and 543 in L2. This gives a mean of 2.429 and sample standard deviation 0.6982.

  15. The standard error of the mean is 0.6982/(1000)=0.0221. The margin of error corresponding with an alpha of 0.05 is 1.960×0.0221 = 0.0433. The confidence interval then is 2.429±0.043 or (2.386, 2.472). Note 1: the order numbers are given in "interval notation" is critical. Be sure to have the left/least first and second/greatest second. Note also that interval notation for an open interval (endpoints not included) can be ambiguous with an ordered pair. Note 2: one can use invNorm(.975) (one-tail) to find the z value which corresponds to a (two-tail) 95% confidence interval.

  16. Put 0, 31, 59, 90, 120 in L1, 69, 57, 49, 40, 32 in L2, and 20, 14, 10, 5 0 in L3. Do LinReg L1,L2 and obtain y = -0.198x + 63.65 with r = -0.946 and r2 = 0.89. Thus the time and temperature are negatively correlated with 89% of the variation in temperature explained by the variation in time (it cools off in the evening). Do LinReg L2,L3 and obtain y = 0.537x - 16.7 with r = 0.998 and r2 = 0.999. Thus the temperatures in Fahrenheit and Celsius are very well positively correlated with 100% of the variation in Celsius explained by the variation in Fahrenheit. Note 1: it doesn't matter which linear regression you use on the calculator. Note 2: if r and r2 are not displayed, then you need to enable them by doing a Diagnostics On from catalog.

  17. Binomial, Poisson, Lorentzian, Student t (or just t), and Chi Square.

For the May 20, 2005 test students are expected to redo their test for additional test points.
  1. No erasures.
  2. Use a different color (pen vs pencil; black vs blue; etc.).
  3. The redos are due back this week with the following diminishing returns: