1. Stats for Engineers Lecture 6 Answers for Question sheet 1 are now online http://cosmologist.info/teaching/STAT/ Answers for Question sheet 2 should be available Friday evening

2. Summary From Last Time Continuous Random Variables Exponential distribution Normal/Gaussian distribution

3. Question from Derek Bruff 1.The graph would be narrower and have a greater maximum value. 2.The graph would be narrower and have a lesser maximum value. 3.The graph would be narrower and have the same maximum value. 4.The graph would be wider and have a greater maximum value. 5.The graph would be wider and have a lesser maximum value. 6.The graph would be wider and have the same maximum value.

4. BUT: for normal distribution cannot integrate analytically. Instead use tables for standard Normal distribution:

5. Why does this work? N(0, 1) - standard Normal distribution

6. Outside of exams this is probably best evaluated using a computer package (e.g. Maple, Mathematica, Matlab, Excel); for historical reasons you still have to use tables.

7. z

8. Example: If Z ~ N(0, 1):

9. = = 0.6915.

10. =

12. - - - = = =

13. = = 0.2417

14. Using interpolation Fraction of distance between 1.35 and 1.36: = 0.6 = 0.4 =0.9125

15. Use table in reverse: Interpolating as before

16. 1.0.0618 2.0.9382 3.0.1236 4.0.0735

18. 0.95 0.05/2=0.0250.025 0.975 0.025+0.95

19. Use table in reverse: 95% of the probability is in the range

20. P=0.025

21. Question from Derek Bruff 1. 2. 3. 4.

22. 1. 2. 3. 4. Too wide CorrectWrong meanToo narrow

23. Example: Manufacturing variability The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). (i) Find the probability that X exceeds 14.99 mm. (ii) Within what range will X lie with probability 0.95? (iii) Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X < Y). XY

24. Example: Manufacturing variability The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). (i) Find the probability that X exceeds 14.99 mm.

25. Example: Manufacturing variability The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). (ii) Within what range will X lie with probability 0.95?

26. 1.0.025 2.0.05 3.0.95 4.0.975 P=0.025

27. Example: Manufacturing variability The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). (iii) Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X < Y).

28. Means and variances of independent random variables just add. Distribution of the sum of Normal variates A special property of the Normal distribution is that the distribution of the sum of Normal variates is also a Normal distribution. [stated without proof] Etc. E.g.

29. Example: Manufacturing variability The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). (iii) Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X < Y). =

30. Which of the following would make a random pipe more likely to fit into a random fitting? The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). 1.Decreasing mean of Y 2.Increasing the variance of X 3.Decreasing the variance of X 4.Increasing the variance of Y XY

31. Which of the following would make a random pipe more likely to fit into a random fitting? The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). XY Common sense. Answer

32. Normal approximations For the approximation to be good, n has to be bigger than 30 or more for skewed distributions, but can be quite small for simple symmetric distributions. The approximation tends to have much better fractional accuracy near the peak than in the tails: don’t rely on the approximation to estimate the probability of very rare events. It often also works for the sum of non-independent random variables, i.e. the sum tends to a normal distribution (but the variance is harder to calculate)

33. Example: Average of n samples from a uniform distribution:

34. [usual caveat: people visiting the Eye unlikely to actually have independent weights, e.g. families, school trips, etc.]

35. Course Feedback Which best describes your experience of the lectures so far? 1.Too slow 2.Speed OK, but struggling to understand many things 3.Speed OK, I can understand most things 4.A bit fast, I can only just keep up 5.Too fast, I don’t have time to take notes though I still follow most of it 6.Too fast, I feel completely lost most of the time 7.I switch off and learn on my own from the notes and doing the questions 8.I can’t hear the lectures well enough (e.g. speech too fast to understand or other people talking) Stopped prematurely, not many answers

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39. p=0.5

41. Approximating a range of possible results from a Binomial distribution

43. Quality control example: The manufacturing of computer chips produces 10% defective chips. 200 chips are randomly selected from a large production batch. What is the probability that fewer than 15 are defective?