of University Ottawa
NAME: Time: 3 hours Total marks: 50 S.N.
Faculty of Administration
Section:
ADM 2303: STATISTICS FOR MANAGEMENT I FINAL EXAMINATION April 20, 1997 1400-1700
ALL ANSWERS (INCLUDING BRIEF EXPLANATIONS) GO ON THE ANSWER SHEET. THE EXAM QUESTION SHEETS WILL NOT BE MARKED, though space is provided here for your rough work. The question sheets must be deposited in the box provided. NOTE THAT THERE ARE MARKS FOR EXPLAINING YOUR ANSWERS, SO MAKE SURE YOU INCLUDE BRIEF EXPLANATIONS ON THE ANSWER SHEET. THERE ARE MARKS FOR IDENTIFYING PROBABILITY DISTRIBUTIONS. Calculators, 1 sheet of notes, and approved statistical tables permitted (MBB or signed by J C Nash). You do not need to interpolate, but take the nearest table value. 1. Acme Soap Co. markets its bars of soap under two different brand names that pretend to be fierce competitors: Clean-your-Clock (CYC) and Wedge 1000. Both are made on the same production line with simply different packaging. Sales of CYC average 2350 boxes per week with standard deviation 402 boxes. Wedge sales average 1500 boxes per week with standard deviation 350 boxes. Acme sells Wedge for $25 per box (there are 200 bars of soap in a box) and CYC to $18 per box. For some reason, soap sales rise and fall with the business cycle, so the correlation between the sales of the two brands has a correlation of 0.55. a) [ 5 ] What is the standard deviation of the combined weekly revenue from both types of soap? b) [ 2 ] What is the mean combined weekly revenue from both types of soap?.
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Q 2. A Revenue Canada auditor is told by his/her director that 15% of self-employed people make claims that are exaggerated, that is, go beyond the allowances of the Tax Act, but that only 0.1% are clearly fraudulent. Assume these figures are correct, and that the auditors are good enough to catch all errors and frauds. The auditor investigates 6 accounts drawn at random from the files. a) [ 3 ] What is the probability no accounts have exaggerated claims?
b) [ 3 ] What is the probability two or more accounts have exaggerated claims?
The whole office examines a total of 1234 accounts for a particular taxation period. c) [ 4 ] What is the probability they find more than 2 fraudulent claims.
d) [ 4 ] What is the probability they find more than 200 exaggerated claims.
e) [ 3 ] To train new auditors, the director prepares a set of 20 files, 6 of which contain exaggerated claims. A novice auditor is given 4 files at random. What is the probability he/she gets at least 1 file with an exaggeration?
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Q 3. Farmaco, a drug manufacturer, is packaging a new pill that helps students to get high marks in statistics. Each pill is supposed to contain 10 mg. of the active ingredient. The manufacturer decides that pills with between 9 mg. and 11 mg. of active ingredient will be acceptable. You are hired to help manage the pill production. a) [ 2 ] You decide to test the distribution of active ingredient in the pills. Data for 250 pills is given below. Based on this data, is the SHAPE of the distribution Gaussian?
MTB > desc c2 C2 C2 N 250 MIN 7.8064 MEAN 9.9926 MAX 12.2335 MEDIAN 10.0162 Q1 9.5980 TRMEAN 9.9949 Q3 10.3632 STDEV 0.6683 SEMEAN 0.0423
MTB > stem c2 Stem-and-leaf of C2 Leaf Unit = 0.10 2 8 15 46 121 (84) 45 15 4 2 7 8 8 9 9 10 10 11 11 12 N = 250
89 022344 6888999 0011111112222333333333444444444 55555555555555555666666666666777777777777777888888888889999999999+ 00000000000000000000000000011111111111111111112222222222333333333+ 555555556666666667778888888899 00112223334 78 02
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b) [ 4 ] Regardless of your answer in (a), for this part of the question assume that you have decided that the data is NOT Gaussian. What is the probability that a "sample" of 250 observations would have the observed mean ingredient content or more if it came from a parent population having mean active ingredient content of 9.9 mg per tablet and standard deviation of 0.6 mg. Be sure to state any assumptions used.
c) [ 6 ] Regardless of your answer in (a), you decide that the data can be accepted as Gaussian. A further sample of pills is taken and the content of active ingredient is as follows: C5 9.85
9.93
9.20 11.10
9.85 10.61 10.92 10.54 10.95 10.86
What is the approximate probability that the sample will contain as much or more of the active ingredient as it does, assuming that the parent population of pills has mean active ingredient of 10 mg. but the standard deviation is NOT given.
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d) [ 2 ] How would you modify your answer in (c) if the population standard deviation were given? [DESCRIBE the change in approach. You do NOT have to complete the calculation.]
Regardless of your answers above, you decide to use 10 mg and 0.5 mg as the specifications for the mean and standard deviation of active ingredient in the population of pills and to assume an approximate Gaussian distribution of active ingredient content. [For the rest of the question you may use this information.] e) [ 2 ] Approximately what proportion of pills should be expected to be "out of spec." i.e., having more than 11 mg or less than 9 mg of active ingredient? Justify BRIEFLY.
f) [ 2 ] You decide to check the variability of active ingredient and take samples of ten pills every few minutes from the line. This yields the chart:
MTB > schart c2 10; SUBC> sigma .5. S Chart for C2 + 1.050+ + + + ---------------------------------------------------UCL=0.8347 + + 0.700+ + + + + + + + + + + + --+---+---+---------+-----------------------+------S=0.4863 + + 0.350+ + ---------------------------------------------------LCL=0.1380 0.000+ +---------+---------+---------+---------+---------+ 0 5 10 15 20 25 Sample Number
S a m p l e S t d e v
What is your conclusion about the pill making process based on this chart?
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g) [ 2 ] Regardless of your answer in (f), assume that pill production is carried out and monitored. What do you conclude from the following chart?
MTB > SUBC> SUBC> SUBC> xbarchart c2 10; mu 10; sigma .5; tests 1:8.
S a m p l e M e a n
X-bar Chart for C2 10.500+--------------------------------------------------UCL=10.47 + -+ + + + + 10.150+ + + + + + ----------------+-+-------------------+------------MU=10.00 + + + + 9.800+ + + + + + + + ---------------------------------------------------LCL=9.526 9.450+ +---------+---------+---------+---------+---------+ 0 5 10 15 20 25 Sample Number
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Q 4. You run a company makes that laminated sheets used for kitchen and bathroom counters. The process is generally very reliable, but there is a small probability that the materials do not bind strongly enough, resulting in minor flaws in the laminate surface. These occur at a rate of 1.5 per 1000 square metres of laminate surface. The rate appears to be constant over your production. You produce 1200 square metres of laminate per day. a) [ 3 ] What is the probability you will have no flaws in a full day's production?
b) [ 3 ] A counter will use just 3.5 square metres of laminate. What is the probability such a counter will have at least one flaw in the laminate?
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ADM 2303 - STATISTICS FOR MANAGEMENT I NAME:
1. a) [ 5 ]
April 20, 1997 1400-1700 Section:
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| b) [ 2 ] | | | | | | | | | | | | | | | | | | | -------------------------------------------------------------------------------------------------Q 2. a) [ 3 ] | b) [ 3 ] | | | | | | | | | | | | -------------------------------------------------------------------------------------------------2 c) [ 4 ] | d) [ 4 ] | | | | | | | | | | | | | | | | | | | | | | -------------------------------------------------------------------------------------------------2e) [ 3 ]
Q 3.a) [ 2 ]
-------------------------------------------------------------------------------------------------b) [ 4 ] | c) [ 6 ] | | | | | | | | | | | | | | | | | | | --------------------------------------------------| d) [ 2 ] | | | | | | | | | -------------------------------------------------------------------------------------------------e) [ 2 ] | f) [ 2 ] | | | | | | | | | | | -------------------------------------------------------------------------------------------------g) [ 2 ]
-------------------------------------------------------------------------------------------------Q 4. a) [ 3 ] | b) [ 3 ] | | | | | | | | |
ADM 2303 - STATISTICS FOR MANAGEMENT I NAME:
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April 20, 1997 1400-1700 Section:
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| b) [ 2 ] | C=sales CYC per week, W=sales of Wedge per week |E(Revenue) = E($18*2350 + $25*1500) E(C) = 2350 boxes, E(W) = 1500 boxes, | (1 marks) SD(C)=402 boxes, SD(W)=350 boxes | = $42,300 + 37,500 = 79,800 $ Prices C = $18 per box, W = $25 per box, | (1 mark) CORR(C, W) = 0.55 | | V(Revenue) = V( $18 * C + $25 * V) | = 18^2 * 402^2 + 25^2 * 350^2 | + 2*18*25*402*350*0.55 $^2 (2 marks) | = 198568696 $^2 (1 mark) | SD(Revenue) = $14,091.44 (1 mark value, 1 units) | | | | | | | -------------------------------------------------------------------------------------------------Q 2. a) [ 3 ] | b) [ 3 ] Binomial situation n=6, p=.15 (1 mark) | P( 2 or more exaggerated) P(0 exaggerated | 6 accounts p=0.15) | = 1 - P(0) - P(1) = 1 - .37715 - 6*.15*.85^5 = .85^6 = .37715 (1 work, 1 answer) | (1 mark) (1 mark) | | = 1 - P(0) - P(1) = 1 - .37715 - .3993 | = 1 - .60067 | = .22352 (1 mark) | | | | -------------------------------------------------------------------------------------------------2 c) [ 4 ] | d) [ 4 ] Binomial n=1234, p= 0.001 (1 mark) | Binomial n=1234, p=0.15; P(>2 frauds | n=1234, p=.001) | Use Gaussian approximation (1 mark) = 1 - ( .999^1234 + 1234*.001*.999^1233 | P(K>200 | n,p) + (1234*1233*.001^2*.999^1232)/2 ) | ~= P(z>(200.5-.15*1234)/sqrt(1234*.15*.85)) = 1 - .999^1232 * (.999^2 + 1234*.001*.999 | + 1234*1233*.001^2 / 2 ) | = P(z > (200.5 - 185.1)/sqrt(157.335) ) = 1 - .87212 = .12788 | (3 more marks if correct) | = P(z > (15.4/12.54332) ) = P(z > 1.227) | Easier via Poisson approx | ~= P(z > 1.23) = 0.5 - 0.3907 P(> 2 frauds | 1.234 expected) | = 1 - exp(-1.234)*(1 + 1.234 + 1.234^2/2) | = .1093 (roughly 11%) = 1 - .29113 * (2.99538) | (3 marks for this, = 1 - .87203 = .12797 | 1 off if no Correction for Cont) (3 more marks if correct) | | | | | | | -------------------------------------------------------------------------------------------------2e) [ 3 ] Hypergeometric (N = 20, S = 6, n=4) K = # with exaggeration P(K>=1) = 1 - P(K=0) = 1 - C(14, 4) C(6, 0) / C(20,4) = 1 - (14!/(10! 4!)) * 1 /(20!/(16! 4!) = 1 - 14*13*12*11/4*3*2 / (20*19*18*17/4*3*2) = 1 - 14*13*12*11/(20*19*18*17) = 1 - .2066 = .7934 Check by combs: 1 - 1001/4845 = 1 - .2066
Q 3.a) [ 2 ] Approximately Gaussian from Stem & Leaf (1 mark) and straight line in Normal Probability plot (1 mark), though some evidence of heavy tails. (NOT skewed -- 1 mark off if this mentioned)
-------------------------------------------------------------------------------------------------b) [ 4 ] | c) [ 6 ] We use the Central Limit Theorem since n=250 is | Since population is Gaussian but no populatio "large" and the data is not skewed. (1 mark) | standard deviation is given, we use the P(X_bar > 9.9926 mg | mu=9.9 mg, sigma=0.6 mg) | Student t distribution. However, we first (1 mark) | need the mean and standard deviation of the ~= P(z > (9.9926 - 9.9)/(0.6/sqrt(250)) ) | sample. X_bar = 10.381 mg, s = 0.6325 mg, = P(z > 2.44) | n = 10 (3 marks) = .5 - .4927 = 0.0073 (0.73 %) | P(X_bar > 10.381 mg | mu=10 mg, s=.6325 mg) (2 marks to finish) | = P( t>(10.381-10)/(.6325/sqrt(10)) | | 9 degrees of freedom) | = P(t>1.90487 | 9 d of f) (2 marks) | Tables give t(0.050,9) = 1.833 and | t(.025,9)=2.262, so this probability is | greater than 2.5% but less than 5% (1 mark) | | | | | --------------------------------------------------| d) [ 2 ] | The distribution of X_bar would now be EXACTLY | Gaussian (1 mark) and we substitute sigma for s | in the calculation and use the Gaussian tables. | (1 mark) | | | | | -------------------------------------------------------------------------------------------------e) [ 2 ] | f) [ 2 ] By Empirical rule, approximately 5% | The VARIABILITY of the process is too (i.e., beyond 2 sigma) | great, given the specification. We need to | revise the process to get the variability OR work it out via the Gaussian table | under control before looking at the level. P(9 mg < X < 11 mg) | = P( (9-10)/.5 < z < (11-10)/.5) ) | = 2 * 0.0228 = 4.56 % | | | | | -------------------------------------------------------------------------------------------------g) [ 2 ] Here the level is IN CONTROL. Pill production is meeting specifications (of mu and sigma), though because we are NOT looking at individual pills, they may not satisfy the bounds of 9 mg and ll mg as we saw earlier.
-------------------------------------------------------------------------------------------------Q 4. a) [ 3 ] | b) [ 3 ] POISSON model with expected flaws | P(0 | 3.5 * 1.5 / 1000 expected) = 1200 sq. m * 1.5 / 1000 sq. m = 1.8 | = exp(-.00525) = 0.99476 | P(0 flaws | 1.8 expected) = .1653 | Therefore P(flaw) = .00524 | | | |