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Assignments2 solutions: Due by Midnight Sunday June 16th, 2013 (drop box of week 2) (Chapters 4, 5, and 6) Total 125 points. True/False (2 points each) Chapter 4 1. If events A and B are independent and A is not an impossible event, then P(A/B) is not equal to zero. TRUE In fact P(A/B) equals P(A) if A and B are independent, which is not zero unless A is an impossible event. 2. If events A and B are mutually exclusive, then P(A/B) is equal to zero. TRUE This is obvious from the definition of mutually exclusive events. If B occurs then A cannot occur at the same time. Therefore P(A/B) = 0. 3. The union of events A and B is given by all basic outcomes common to both A and B FALSE
Chapter 5 4. If the probability of success is 0.4 and the number of trials in a binomial distribution is 150, then its standard deviation is 36. FALSE σ= √(np(1-p)) =√(150*0.4*0.6) = 6 5. If a fair coin is tossed 100 times, then the variance of the random variable defined as the number of heads is exactly five. FALSE σ2 = np(1-p)= 100*0.5*0.5= 25. So the standard dev is 5 not the Variance. 6. If a fair coin is tossed 20 times then the probability of exactly 10 Tails is more than 18 percent. FALSE It is 17.62 percent 7. The probability that a person catches a cold during the cold and flu season is 0.4. If 10 people are chosen at random, the standard deviation for the number of persons catching cold is 1.55. (Hint: convert the problem to a binomial distribution problem). TRUE Here p = 0.4 and n=10. Therefore, standard deviation = sq root of 10*0.4*0.6
Chapter 6 8. The number of defective pencils in a lot of 1000 is an example of a continuous random variable. FALSE It is a result of counting- so discrete. 9. For a continuous distribution, P(X ≤ 100) = P(X < 100). TRUE See my Instructions on this. 10. All continuous random variables are normally distributed. FALSE Continuous random variables can be highly skewed and non-normal. Even if it is symmetrical it may not be normal but other distribution like t-distribution. A normal random variable is a popular example of a continuous random variable, but a continuous r.v. need not be normal.
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11. The mean of a standard normal distribution is always equal to 1. FALSE. Its mean is zero and variance (or std deviation) equal to 1. 12. If the sample size is as large as 1000, we can safely use the normal approximation to binomial even for small p. FALSE (Instructions on Ch6) : For example if p is .001 then np would be only 1 even if sample size is 1000.
Multiple Choice (3 points each) Chapter 4 1. Two mutually exclusive events having positive probabilities are ______________ dependent. A. Never B. Sometimes C. Always They are necessarily dependent because the occurrence of one (seriously) affects the probability of the other (makes it zero). Instructions on Ch 4 page 4 2. If P(A) >0 and P(B) > 0 and events A and B are independent, then: A. P(A) = P(B) B. P((A|B)) = P(A) C. P(A B) = 0 D. P(A B)=P(A)/ P(B/A) E. Both A and C are correct See My Instructions on Ch 4 page 5. Independence does not imply equality of probabilities. So the first choice is clearly wrong. The third choice applies to mutually exclusive events not independent events. The fourth choice is also incorrect because there should be multiplication on the right hand side not division. So the correct answer is B. 3. A recent marketing survey tried to relate a consumer’s awareness of a new marketing campaign with their rating of the product. Consumers rated their awareness as low, medium and high, and rated the product as poor, fair, or good. The results are presented below:
Rating
Poor Fair Good
Low
Awareness Medium
High
0.10 0.06 0.07
0.15 0.11 0.11
0.07 0.06 0.27
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What is the probability that a consumer who ranked the product as fair had a high awareness of the ad campaign? A. B. C. D. E.
0.06 0.26 0.23 0.15 0.40 The completed table is: Low
Awareness Medium
High
0.10 0.06 0.07 0.23
0.15 0.11 0.11 0.37
0.07 0.06 0.27 0.40
Rating Poor Fair Good Total
total 0.32 0.23 0.45 1.00
P(High Awareness/Fair) = 0.06/0.23 = 0.26 rounded to two decimal places. 4. With the data in question 3 above, what is the probability that a randomly selected consumer either has a low awareness or rated the product poor? A. 0.23 B. 0.32 C. 0.55 D. 0.45 E. 0.10 The completed table is: Low
Awareness Medium
High
0.10 0.06 0.07 0.23
0.15 0.11 0.11 0.37
0.07 0.06 0.27 0.40
Rating Poor Fair Good Total
P( Low awareness or Poor rating) = 0.23 + 0.32 - 0.10 = 0.45
total 0.32 0.23 0.45 1.00
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Chapter 5 5. In a study conducted by UCLA, it was found that 25% of college freshmen support increased military spending. If 6 college freshmen are randomly selected, find the probability that: at least 3 support increased military spending A. 0.8306 B. 0.1318 C. 0.1694 D.0 .9624 Remember that at least 3 means 3 or more.(You can use Table on page 855 for n=6 and p=.25) or use computer to get the following: Binomial distribution : n= 6 p= 0.25
X 0 1 2 3 4 5 6
cumulative P(X) probability 0.17798 0.17798 0.35596 0.53394 0.29663 0.83057 0.13184 0.96240 0.03296 0.99536 0.00439 0.99976 0.00024 1.00000 1.00000
6. A fair die is rolled 10 times. What is the probability that an even number (2, 4 or 6) will occur more than 3 times? A. 0.1719 B. 0.1172 C. 0.6230 D. 0.9453 E. 0.8281 Here n =10 and p = 0.5. We need P(X > 3) which is 1- P(X ≤ 3) = 1- [P(X= 0) + P(X = 1) + P(X = 2) + P(X=3)] = 1- 0.1719 = 0.8281. From the Table on page 854 you get the answer. Or you can use computer as follows: Binomial distribution 10 0.5
X 0 1 2 3
P(X) 0.00098 0.00977 0.04395 0.11719
n p cumulative probability 0.00098 0.01074 0.05469 0.17188
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4 5 6 7 8 9 10
0.20508 0.37695 0.24609 0.62305 0.20508 0.82813 0.11719 0.94531 0.04395 0.98926 0.00977 0.99902 0.00098 1.00000 1.00000 7. Which one of the following statements is not a necessary assumption of the binomial distribution? A. Sampling is with replacement B. The experiment consists of n identical trials C. The probability of success remains constant from trial to trial D. Trials are independent of each other E. Each trial results in one of two mutually exclusive outcomes 8. If p = 0.6 and n =10, then the corresponding binomial distribution is A. Right skewed B. Left skewed C. Symmetric D. Bimodal If p = 0.5 it is symmetric for any number of trials. If p is larger than 0.5, then the distribution is left skewed. If p is smaller than 0.5, then it is right skewed. 9. A fair die is rolled 64 times. What is the standard deviation of the even number (2, 4 or 6) outcomes? A. 32 B. 16 C. 4 D. 8 The probability of 2 or 4 or 6 in any trial is 0.5 and there are 64 trials. Therefore, σ = √(64*0.5*0.5) = 4 using the formula for binomial distribution. 10. A fair die is rolled 10 times. What is the probability that an odd number (1, 3 or 5) will occur less than 4 times? A. 0.1719 B. 0.1172 C. 0.2051 D. 0.3770 E. 0.0439 Add P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) from table on page 856 or can use excel to find
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the cumulative probability 11. In the most recent election, 20% of all eligible college students voted. If a random sample of 20 students were surveyed: Find the probability that at least five students voted in the election. A. 0.1746 B. 0.2182 C. 0.3704 D.0.6296 P(X ≥ 5)= 1- P(X 4) = 1- 0.629648 (using excel’s Binomdist) or using table and adding the probabilities up to 4 and subtracting from 1.
Chapter 6 12. The area under the normal curve between z = 0 and z = 1 is ________________ the area under the normal curve between z =1 and z = 2. A. Greater than B. Less than C. Equal to D. A, B or C above dependent on the value of the mean E. A, B or C above dependent on the value of the standard deviation Remember the height of the curve declines as we move farther from the mean value 0. Therefore, the area under the curve (or probability will decrease for the same interval length as we move farther from the mean. 13. The price-to-earnings ratio for firms in a given industry is distributed according to normal distribution. In this industry, a firm with a Z score for its price-to-earnings ratio equal to 0.5: A. Has an average price-to-earnings ratio B. Has a below average price-to-earnings ratio C. Has an above average price-to-earnings ratio D. May have an above average or below average price-to-earnings ratio The mean or average for Z is 0. Therefore the given value is above average. 14. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years 14 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What is the probability that less than 30 accounts will be classified as delinquent? A. 31.86% B. 18.14% C. 81.86% D. 63.72% E. 75.84%
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Here nπ = 250*(0.14) = 35, n(1-π) = 215 and nπ(1-π) = 30.1 all ≥ 10. So, normal approximation without continuity correction can be performed. You can work with X or proportion to get the same answer. Let us work with X first. Mean nπ = 35. Variance = nπ(1-π) = 30. Standard deviation (or standard error) = √30 = 5.4863. The Z score of 30 = (30-35)/5.4863 = -0.91 (rounded to two decimal places). Now I will show you that the same result is obtained working with proportions. The mean of the sample proportion (or its expected value) is 0.14 (the population proportion). The std error is √{π(1-π)/n} = 0.0219 and given value p = 30/250 = 0.12 Therefore Z = (0.12 - 0.14)/0.0219 = - 0.91 (rounded to two decimal places). P(X 30) = P(Z -0.91) = 0.1814 from table or MegaStat Normal distribution
P(lower)
P(upper)
z
0.8186
0.1814
0.91
0.1814
0.8186
-0.91
15. Which of the following statements is not a property of the normal probability distribution? A. The normal distribution is symmetric B. Ninety-five % of all possible observed values of the random variable x are within plus or minus three standard deviations of the population mean C. The mean, median and mode are equal D. The area under the normal curve to the right of the mean is equal to the area under the normal curve to the left of the mean E. All of the above answers are properties of the normal distribution Choice B is selected because 99.73% not 95% corresponds to three standard deviations. This was also included in Chapter 3 as empirical rule.
Essay Type (each question carries 7 points) Chapter 4 1. At a college, 70 percent of the students are women and 45 percent of the students receive a grade of C. About 30 percent of the students are female but not C students. Use this contingency table. C Female Male 0.45
Not C 0.30
0.70
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The completed table is: C 0.40 0.05 0.45
Female Male
Not C 0.30 0.25 0.55
0.70 0.30 1.000
If a randomly selected student is male, what is the probability he is a C student? P(Not C/M) = 0.05/0.30 = 0.167 or 16.7% chance. Some of you just answered 0.05, but that is the probability of "male and C", not the probability of "C" given male.
2. The contingency table about customers of a store who buy cigars and/or beer is given below. Beer Cigars No cigar
No Beer .05
.10
.70
Determine the probability that a customer will buy at least one of these items: cigar or beer. The completed table is:
Cigars No cigar
Beer
No Beer
0.15
.05
0.20
.10
.70
0.80
0.25
0.75
1.00
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Answer: P(C or B) = P(C) + P(B) - P(C and B) = .20 + 0.25 - 0.15 = 0.30 or 30% chance. 3. Four employees who work as drive-through attendees at a local fast food restaurant are being evaluated. As a part of quality improvement initiative and employee evaluation these workers were observed over three days. One of the statistics collected is the proportion of time employee forgets to include a napkin in the bag. Related information is given in the table.
Proportion of Worker Dinners Packed Joe 0.25 Jan 0.20 Cheryl 0.20 Clay 0.35
Proportion of forgetting Napkin when packing Dinner 0.06 0.02 0.10 0.04
You just purchased a dinner and found that there is no napkin in your bag, what is the probability that Jan has prepared your order? Answer: First note that the last column in the above table gives conditional probabilities. For example 0.06 is the probability of forgetting napkin given that Joe packed the dinner or P(No napkin/Joe). In the question we are given that No napkin has occurred and asked to find the probability of Cheryl in light of this result. So here we are asked a reverse conditionality than the one given in the contingency table. According to the Instructions for Chapter 4, this requires Bayesian rule. Therefore, P(Jan/ No napkin) = 0.004/0.053 = 0.0755 or 7.55% The numerator is P(Jan and No napkin)=P(Dinner packed by Jan)*P(No napkin given that Jan packed Dinner) = 0.20*0.02 = 0.004 or 0.4% The denominator is P(No napkin)= P(Joe and no napkin)+ P(Jan and No napkin)+ P(Cheryl and No napkin)+ P(Clay and no napkin) = 0.015 + 0.4 + 0.02 + 0.014 = 0.053 as shown in the table below (everything converted to decimals instead of percentage, because working with percentage is messy): Proportion of Worker Dinners Packed Joe 0.25 Jan 0.20 Cheryl 0.20 Clay 0.35
Proportion of forgetting Napkin Col. 2*Col.3 0.06 0.015 0.02 0.004 0.10 0.020 0.04 0.014 0.053
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This formula is also called the Bayesian rule for probability revision based on the results of an experiment. Here the prior probability of Jan (which is 20%) has been revised downward to 7.55% (called the revised or posterior probability) after noticing that the dinner had no napkins, because Jan is the least forgetful one. If the question were for Cheryl the posterior probability would be higher than the prior probability because she has a very high chance of forgetting napkin (10%).
Chapter 5 4. In a study conducted for the State Department of Education, 30% of the teachers who left teaching did so because they were laid off. Assume that we randomly select 12 teachers who have recently left their profession. Find the probability that at least 4 of them were laid off. We can use Binomial Distribution (because there are only two possible outcomes for each teacher who left their profession (either laid off or not laid off). Here n= 12 and p = 0.3. Question is P(X ≥ 4) which can be easily found from its complement as= 1-P(X ≤ 3). Either using Excel Binomdist formula or the table at the back of the book P(X ≤ 3) = 0.493. Therefore, P(X ≥ 4)= 1-.493= 0.507 or 50.7% Alternatively, using MegaStat: Binomial distribution 12 0.3 X 0 1 2 3 4 5 6 7 8 9 10 11 12
P(X) 0.01384 0.07118 0.16779 0.23970 0.23114 0.15850 0.07925 0.02911 0.00780 0.00149 0.00019 0.00001 0.00000 1.00000 3.600 2.520 1.587
n p cumulative probability 0.01384 0.08503 0.25282 0.49252 0.72366 0.88215 0.96140 0.99051 0.99831 0.99979 0.99998 1.00000 1.00000 expected value variance standard deviation
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5. The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 20%. If 15 calculators are selected at random, what is the probability that 5 or more of the calculators will be defective? Here n =15, p = 0.20 and P(X ≥ 5) = 1-P(X ≤ 4). Now this can be solved using excel or MegaStat or simply by looking at the binomial Table at the end of the book for n = 15 and p= .20: Using Excel Binomdist (4, 15, 0.20, 1) we get P(X ≤ 4) = 0.83577 or 0.836 (rounded). Therefore, P(X ≥ 5) = 1- 0.836 = 0.164 or 16.4% chance. Alternatively, using MegaStat: Binomial distribution 15 0.2 X 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
P(X) 0.03518 0.13194 0.23090 0.25014 0.18760 0.10318 0.04299 0.01382 0.00345 0.00067 0.00010 0.00001 0.00000 0.00000 0.00000 0.00000 1.00000 3.000 2.400
n p cumulative probability 0.03518 0.16713 0.39802 0.64816 0.83577 0.93895 0.98194 0.99576 0.99922 0.99989 0.99999 1.00000 1.00000 1.00000 1.00000 1.00000
expected value variance standard 1.549 deviation
6. A door-to-door sales person for a Household appliance has learned from her past experience that out of twenty demonstrations of her appliance only seven result in actual sales (long run average). This week she needs to make at least four sales. At least how many demonstrations does she need to perform to ensure that the probability of meeting her target is at least 95 percent?
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(Read the chapter instruction posted by me to learn how to use Excel to find probabilities, and try to answer this question playing with cumulative probabilities-. You can also use Tables in the back of any Statistics book. Be careful about the wording of the question) Answer: First note that it is a binomial distribution problem (sale or no sale in each demonstration or trial). The probability of success p = 0.35 (seven out of twenty; from long run experience). We need to find the number of trials n such that: P(X ≥ 4) ≥ 0.95. But Tables and computer deal with cumulative probability in ≤ form not in the ≥ form. Therefore, we will find this probability by using the complement rule as follows: P(X ≥ 4) = 1-P(X ≤ 3). To make this probability greater than or equal to 0.95 we need to make P(X ≤ 3) less than or equal to 0.05 (that is 1-0.95). To find the lowest n which will achieve this we play the following trial and error game: In my Instructions I have indicated how to use Excel to find binomial probabilities. You open Excel, click on formulas (on top row) then select more function, then select statistical functions, then select Binomdist. This opens a dialogue box. We fill 0.35 for p, 3 for the X value for which we want cumulative probability, and put 1 in the logical choice called cumulative (0 there would give probability of only one value) and keep trying increasing values of n. BINOMDIST(3,18,0.35,1) gives Probability P(X ≤ 3)= 0.078267 BINOMDIST(3,19,0.35,1) gives Probability P(X ≤ 3)= 0.05914 BINOMDIST(3,20,0.35,1) gives Probability P(X ≤ 3)= 0.044376 Thus our target Probability is reached when n = 20, or 20 visits. This gives probability of 4 or more successes equal to at least 95 percent. This whole process takes less than two minutes because you only keep clicking at the result cell and in the formula keep changing n and hitting enter to get the above probabilities until you reach the target probability. You could also use the binomial tables at the end of the book to add probabilities for X = 0 to X = 4 (some books even give the cumulative binomial probabilities). But this involves more time in Arithmetic. Computer can save that time.
Chapter 6 7. If x is a binomial random variable where n = 100 and p = .1, find the probability that x is less than or equal to 12 using the normal approximation to the binomial. Answer: Need continuity correction because np(1-p) = 9 < 10. Following my instructions on continuity correction, less than or equal to 12 becomes less than or equal to 12.5 after continuity correction. ). The standard deviation is ((.1)(.9)(100)) = 3. Therefore,
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Z = (12.5 – 10)/((.1)(.9)(100)) = 2.5/3 = .83 P(Z 0.83) = 0.7967 normal distribution P(lower) P(upper) z .7967 .2033 0.83 Some of you did not do the continuity correction and lost one point. 8. The weight of a product is normally distributed with a standard deviation of .8 ounces. What should the average weight be if the production manager wants no more than 10% of the products to weigh more than 10 ounces? Answer: The Z value which has 10% area to the right (or 90% area to the left) is Z = 1.282 or 1.28 rounded to two decimal places. Using this we calculate the required µ which will give the specified probability or area for the specified X equal to 8 ounces. This is obtained by using the relation between X, Z, µ and σ as shown in my Instructions (and rearranging): P(Z ≥ 1.282) = 0.10 μx = X- Zσ = 10 - (1.28)(.8) = 8.974
