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Math 321 Test 1 (take home) BSAD 560 Assignment 3a Please highlight your answers with a color different than black. Please do not highlight any explanations you provide. Last Name First Name Use the formulas, and examples on page 266 - 269 for problems 1, and 2. 1) How many ways can we select a committee of size 4 members from among 9 candidates? 2) How many distinguishable permutations can be made from the word sausage? 3) Determine whether or not the following is a probability distribution. If it is not give the reason why it is not. X P(X = x) 1 0.15 2 0.25 3 0.5 4 0.3 4) Solve problem # 8 page 298. For good examples on this please see examples on pages 293 – 297 in your textbook. Please make sure you find both the mean ( Expected value) , and then the standard deviation. 5) Suppose a random variable has a binomial distribution with n = 10 and p = 0.1 a Find the expected value of X b Find the standard deviation of X c Find the probability that X equals 2 d Find the probability that X is at most 3 e Find the probability that X is a least 1. 6) A sales representative has a chance of 10% of making a sale on every independent phone call she makes. She is about to call 20 different independent people. a. What is the chance that she will make at least 3 sales? Give the answer to 4 decimal places. b. What is the average number (expected value) for the number of sales when she calls 30 people? Give the answer to one decimal places. Math 321 Test 1 (take home) BSAD 560 Example 3a V2 Please highlight your answers with a color different than black. Please do not highlight any explanations you provide. Last Name First Name Use the formulas, and examples on page 266 - 269 for problems 1, and 2. 1) How many ways can we select a committee of size five members from among 7 candidates? Solution: 21 EXP: This is a combination problem since the order in which committee members are selected does not affect decisions made by that committee! So you should use he formula 7 C 5 . 2) How many distinguishable permutations can be made from the word Malayalam? Solution: 3780 EXP: the word has 2 m’s, 4 a’s, 2 l’s and one y. that will make the denominator in the fraction equal to (2!)(4!)(2!)(1!), but the numerator is 9! So the answer is 9!/[ (2!)(4!)(2!)(1!)] 3) Solve problem # 12 page 291 Solution: NO, it is not a distribution EXP: since P( x = 100 ) is negative. Further, even if we mistakenly ignore the fact that one of the numbers is negative, the sum of probabilities do not add up to one. 4) Solve problem # 6 page 298. For good examples on this please see examples on pages 293 – 297 in your textbook. Please make sure you find both the mean ( Expected value) , and then the standard deviation. Solution: Expected value = - 0.09 Standard deviation = 1.8873 ( to 4 decimal places ) Explanation: Please note I used the letter m to denote he mean (or the expected value) in table below. x -5 -2 0 2 5 P(X) 0.06 0.15 0.58 0.18 0.03 x*p(x) -0.3 -0.3 0 0.36 0.15 (x-m)^2*p(x) 1.446486 0.547215 0.004698 0.786258 0.777243 5) Solve problem # 10 on page 312. a. 9.6 since 12*0.8 = 9.6 b. 1.3856 Since SQRT(12*0.8*0.2) = 1.3856 c. 0.0531 Since P(X = 7) = BINOMDIST(7,12,0.8,0) d. 0.0006 Since P(X<=4) =binomdist(4,12,0.8,1)="" e.="" 1.000="" since="" p(x="">=1) =1-BINOMDIST(0,12,0.8,1) = 0.999999996 f. 0.2749 Since P(X > 9) =1-BINOMDIST(0,12,0.8,1) 6) A sales representative has a chance of 6% of making a sale on every independent phone call she makes. She is about to call 50 different independent people. a. What is the chance that she will make at least 3 sales? Give the answer to 4 decimal places. Solution: 0.5838 EXP: in this problem n=50, and p = 0.06, and they are asking for P( X >=3) =1-BINOMDIST(2,50,0.06,1) b. What is the average number (expected value) for the number of sales when she calls 50 people? Give the answer to one decimal places. Solution: 3.0 EXP: Mean = 50*0.06 = 3.0 Math 321 Test 1 (take home) BSAD 560 Assignment 3a Please highlight your answers with a color different than black. Please do not highlight any explanations you provide. Last Name First Name Use the formulas, and examples on page 266 - 269 for problems 1, and 2. 1) How many ways can we select a Basketball team of six players from among 8 candidates? Assume anybody can paly any position. Solution: 28 Explanation: since the order of selections do not matter this is a combination case and the answer is found by computing nCk in this case 8C6 = 8!/[6! X 2!] = 1x2x…x6x7x8/[(1x2…x6)(1x2)] = 28 2) How many distinguishable permutations are there for the word Statistics? Solution: 50400 EXP: We should use the formula for distinguishable permutation. Statistics has 10 letters so he numerator is 10! However, the letter S is repeated 3 times, and T is repeated 3 times also, and I is repeated twice so we need to ivied the numerator by the products of 3! And 3!, and 2! I am not mentioning division by 1! For he letter am since division by 1 does not make any difference. Answer is found by 10!/[(3!)(3!)(2!)]= 50400 3) Solve problem #11 Page 291 Solution: It is a Probability Distribution All we have to do is to make sure that a) non of the probabilities are negative. And b) that the sum of all probabilities equal to one. 0.18+0.39 + 0.08 + 0.35 = 1.00 On another note: Problem 10 on page page 291 is not a probability distribution since sum of the probabilities add up to 0.96 only. Neither is problem 9 since P(X = 3) is negative. 4) I am not providing any examples for mean, and standard deviation of a probability distribution since pages 293 – 297 of your text have many good examples in there already For problem 5 please read the file Extra Examples for 3a.doc. In that File I show you a very quick and easy way (Using EXCEL) to compute binomial probabilities. 5) According to the IRS 10% of all tax returns have mistakes in them. Suppose an auditor randomly selects 10 tax returns. Let X be the number of tax returns that will have mistakes in them. Find the following probabilities. Use 4 decimal places. a. Probability that exactly 2 tax returns have mistakes in them. P( X = 2 ) Solution: 0.1937 EXP: I chose a blank cell in EXCEL and in that cell I typed = BINOMDIST (2,10,0.1,0) and pressed return. b. Probability that At most 3 tax returns have mistakes in them. P( X <= 3="" )="" solution:="" 0.9872="" exp:="" i="" typed="BINOMDIST(3,10,0.1,1)" c.="" probability="" that="" less="" than="" 3="" tax="" returns="" have="" mistakes="" in="" them.="" p(="" x="">< 3="" )="" solution:="" 0.9298="" exp:="" i="" typed="BINOMDIST(2,10,0.1,1)" remember="" that="" p(="" x="">< k)="1-BINOMDIST(k-1,n,p,1)" d.="" probability="" that="" more="" than="" 5="" tax="" returns="" have="" mistakes="" in="" them.="" p(="" x=""> 5 ) Solution: 0.0002 EXP: I typed =1–BINOMDIST(5,10,0.1,1) = 0.00015 e. Probability that at least 5 tax returns have mistakes in them. P( X >= 5 ) Solution: 0.0016 EXP: =1–BINOMDIST(4,10,0.1,1) Remember: P(X >= k ) = 1-BINOMDIST(k-1,n,p,1) Math 321 Test 1 (take home) You should read pages 302 – 311. Pay special attention to the definition on page 305 as it is useful for solving the word problems. Examples 6.15, and 6.16, and 6.17 are very important. However, I insist you use EXCEL formulas instead of the Binomial formula on page 306: No need to torture yourself ( It is best ( much much easier ) to use my instructions below for EXCEL as opposed to the binomial formula. Also forget about the tables on pages 1008 – 1013. How to use EXCEL: Open your textbook and look at Example 6.15 on pages 307, 309. For P( X = 0 ) all you have to do is pick an empty cell in EXCEL and in it you type =BINOMDIST(0,4,1/6,0) AND PRESS RETUN. you will get 0.482253086 If you round that to 4 decimal places you will get the same answer as your book. So Practice with another Example in your textbook. For