1. Athens Technology – p. 58 This problem clearly involves uncertainty. The experiment: testing a sample of chips from a single production run. random variable: the mean temperature at which the chips in the sample failed, call it the sample mean failure temperature (SMFT). sample space: all temperatures at which the chips can fail

2. Athens Technology – p. 59 OutcomeProbability 4.20221564378796 0 5.02017273476363 0 5.15823847163305 0 3.37565233314005 0 2.60585345011750 0 2.93325601977599 5.90240180669576 0.12616351817377 2.10541093173009 3.08322397534104 3.79918820764794............

3. Athens Technology – p. 73 Using the Excel commands we get: Mean = 180.5 Standard deviation = 20.9 And consequently a PDF that looks like: P(x < 170 ) = area of curve to left of 170 = NORMDIST (170,180.5,20.9,true) =.309

4. Athens Technology – p. 74 What would be the impact of promising a vendor chips that are guaranteed to function at temperatures up to 170 degrees? On average have to discard 30.9% of production Currently the chips are intended to function at temperatures up to 150 degrees. What is the probability that the SMFT will be less than 150? P(x < 150) = NORMDIST(150,180.5,20.9,true) =.073

5. Athens Technology – p. 75 What options should the marketing and production manager consider? Charge more for the chips in order to cover the increases in production discards Try to convince the customer to consider a lower failure temperature Look for ways to improve the production process so that there is a lower probability of failing at temperatures below 170. Raise the mean or lower the standard deviation

6. Athens Technology – p. 76 Suppose the two managers agree that it would be acceptable to discard 25% of the production runs. What failure temperature could the company promise? That is find the value such that (P x < value) = 0.25. Instead of finding the area of the curve to the left of some value, we need to find a value such that the area to the left equals 0.25. This is the inverse of what we have been doing and is accomplished by the NORMINV function. NORMINV(.25,18.5,20.9) = 166.3

7. Quick Examples– p. 80 A company manufactures plastic sheets that must not exceed 0.22 mm in thickness. Recent records suggest that the thickness of the sheets of plastic is normal random variable with a mean of 0.20 and a standard deviation of.01. Find P(x >.22). P ( x >.22) = 1 – NORMDIST(.22,.20,.01,true) = = 1 -.97725 =.02275

8. Quick Examples – p. 81 The diameters of certain manufactured ball bearings must fall between.91 cm and.95 cm. The diameters are a normal random variable with a mean of.93 cm and a standard deviation of.012cm. Find P (.91 < x <.95 ) P (.91 < x <.95 ) = NORMDIST(.95,.93,.012,true) – NORMDIST(.91,.93,.012,true) =.95221 -.04779 =.904419

9. Averson Distributors – p. 82 Averson Distributors keeps track of weekly sales for all items in their warehouses for three years. Marketing analysts are expected to look for any significant changes in sales that might require an adjustment in inventory levels and marketing strategies. The sales for the most recent week increased from 991 to 1092. What is the percent increase? (1092 – 991)/991 =.102 or 10.2% Calculate the probability that the weekly sales increase would be this high or higher. P(x >.102) Are the sales increases normally distributed? (See averson.xls) Checking frequency histogram of sales increases suggests they are normal. mean =.0079 or 0.79% st dev =.1242 or 12.42% P(x>.102) = 1 – NORMDIST(.102,.0079,.1242,true) =.2246 An increase that high or higher could be expected over 22% of time. It is not an unusual increase.

10. Averson Distributors – p. 84 What is the probability that the increase will be more than one standard deviation away from the mean? In our case, mean =.0079 and stdev =.1242. One standard deviation above the mean =.0079 +.1242 =.1321 P(x >.1321) = 1 – NORMHIST(.1321,.0079,.1242,true) =.1587 What is the probability that the decrease will be more than one standard deviation away from the mean? One standard deviation below the mean =.0079 -.1242 = -.1163 P(x < -.1163) = NORMHIST(-.1163,.0079,.1242,true) =.1587 What is the probability that increase will fall in between the mean minus 1 stdev and the mean plus 1 stdev? = 1 -.1587 -.1587 =.6827 (rounding)

11. Averson Distributors – p. 85 What is the probability that the increase will be more than two standard deviations away from the mean? Two standard deviation above the mean =.0079 + 2*.1242 =.2564 P(x >.2564) = 1 – NORMDIST(.1321,.0079,.1242,true) =.2275 What is the probability that the decrease will be more than two standard deviations away from the mean? Two standard deviation below the mean =.0079 – 2*.1242 = -.2406 P(x < -.2406) = NORMDIST(-.2406,.0079,.1242,true) =.02275 What is the probability that increase will fall in between the mean minus 2 stdev and the mean plus 2 stdev? = 1 -.02275 -.02275 =.9545