MATH 118 Class Notes For Chapter 5 By: Maan Omran

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1 MATH 118 Class Notes For Chapter 5 By: Maan Omran Secton 5.1 Central Tendency Mode: the number or numbers that occur most often. Medan: the number at the mdpont of a ranked data. Ex1: The test scores for a test were: 78,81,8,76,84,81,76. Fnd the mode and the medan. The mode s 81 and 76, both of them repeated twce The medan must be found after the data s ranked from smallest to largest. For the above data: 76,76,78,81,81,8,84 the medan s 81 whch s located n the mddle. Ex: The test scores for a test were: 78,81,8,76,84,86. Fnd the mode and the medan. There s no mode, no score s repeated more than once The medan must be found after the data s ranked from smallest to largest. For the above data: 76,78,81,8,84,86. There are two values n the mddle 81 and 8, then the medan s average of those two values or (81+8)/ 81.5 Ex3: The test scores for a test were: 78,78,78,81,81,95,95,95,100. Fnd the total. As you notced, there are repeated scores and t s easer to fnd the total of those scores ths way: Total 3(78) + (81) + 3(95) + 1(100) 781 Or: total f x where: s the symbol for sum x s the score; f s the frequency of each score Ex4: Fnd the average score for the tests n example 3. The average score s the total dvded by the number of tests, there are 9 tests, The average s (781)/ fx Or: x n where n s number of tests, n f (sum of frequences) Ex5: A student obtaned the followng grades for one semester, what s the grade pont average GPA? Course Grade Pont Value Frequency (# of credts) Product Math A Englsh B Physcs F Accountng C 3 6 Sum The GPA or the mean s 38/16.375

2 Secton 5. Expected Value and Standard Devaton Random Varable: A functon X that assgns to every outcome exactly one real number. Probablty Densty Functon: A lst of all possble values of the random varables and the assocated probabltes. Outcomes (events) Random Varable (X) Prob. Densty (P) all possbltes value of each possblty prob. of each possblty Sum 1 Ex1: An unfar con n whch P(H) /3 s flpped twce. The random varable X s defned to be the number of heads. Fnd the densty functon. X the number of heads Outcomes X P HH /3. /34/9 HT or TH 1 /3. +. /3 4/9 TT 0. 1/9 Usng the Tree: Use ths method when the problem s wrtten s way that the selecton s not smple, for example: the experence stops when certan condton s met. We could have used the tree method n example whch would make t easer to solve. Ex: An experment conssts of flppng an unfar con where P(H) /3 untl a total of heads occur or 3 flps. The random varable s defned to be the number of tals. Fnd the expected value of the random varable. 0T, H 0 4/9 0 1T, H 1 4/7 + 4/7 8/7 8/7 T, 1 H /7 + /7 + /7 6/7 1/7 3T, 0 H 3 1/7 3/7 Sum 1 E(X) 3/ H 0T,H 4/9 /3 H /3 T /3 H 1T,H 4/7 T T,1H /7 /3 H 1T,H 4/7 T /3 T H /3 T T,1H /7 H T,1H /7 (page ) T 3T,0H 1/7

3 Bnomal & Non-Bnomal Dstrbuton (Usng Tables) Expected Value E[X], m : The mean, the average value of a random varable n whch: m E[X] P X X 1 P 1 + X P + X 3 P Varance : P ( X µ ) Standard Devaton: Bnomal Dstrbuton (Usng Formula) Expected Value E[X], m : µ n.p Varance : n.p.q Standard Devaton: Bnomal Dstrbuton: A dstrbuton that descrbes the probablty of all possble outcomes of a a seres s dentcal to every other and has two possble outcomes. (Bernoull trals of Sec. 4.4) P C(n,r). pr. q n-r Ex3: Stereo speakers manufactured wth probablty of 0% beng defectve. Three are selected off contnuous assembly lne, defne the random varable X as the number of the defectve parts. Fnd: a) the densty functon and the expected value for the defectve parts b) the expected value for the good parts c) the varance, the standard devaton. a) X number of defectve parts 0D, 3G 0 C(3,0).(0.80) 3.(0.0) D, G 1 C(3,1).(0.80).(0.0) D, 1G C(3,).(0.80) 1.(0.0) D, 0G 3 C(3,3).(0.80) 0.(0.0) Sum Expected value for the defectve part s 0.6 b) Expected value for the good part s c) X P ( X µ ) ( X µ ) P( X µ ) (0-0.6) -0.6 (-0.6) (0.36) (1-0.6) 0.4 (0.4) (0.16) (-0.6) 1.4 (1.4) (1.96) (3-0.6).4 (.4) (5.76) 0.05 Sum 0.48 Varance : P ( X µ ) 0.48 Standard Devaton: (page 3)

4 Ex4: Solve example 3 agan but wthout tables. Ths problem s Bnomal, frst fnd n,p & q: n 3, p 0., q 0.8 Expected Value: µ n.p 3(0.) 0.6 for the defectve parts Varance: n.p.q 3(0.)(0.8) 0.48 Standard Devaton: Ex5: A box wth 6 good parts and 4 defectve n whch 3 are selected. The random varable X s defned as the number of defectve parts selected. Fnd: a) the densty functon and the expected value for the defectve parts b) the expected value for the good parts c) the varance, the standard devaton. a) Ths problem s not Bnomal. X number of defectve parts 0D,3G 0 C( 4, 0). C( 6, 3) D,G 1 C( 4, 1). C( 6, ) D,1G C( 4, ). C( 6, 1) D,0G 3 C( 4, 3). C( 6, 0) Sum 1 1. Expected value for the defectve part s 1. b) Expected value for the good part s c) X P ( X µ ) ( X µ ) P( X µ ) 0 10 (0-1.) -1. (-1.) 0 (1.44) (1-1.) -0. (-0.) 60 (0.44) (-1.) 0.8 (0.8) 36 (0.64) (3-1.) 1.8 (1.8) 4 (3.4) Sum 0.56 Varance : P ( X µ ) 0.56 Standard Devaton: (page 4)

5 Ex6: A class wth 00 males and 100 females. One student was selected and replaced and that was repeated 6 tmes. Fnd: a) the probablty that men wll be selected. b) the expected value for men c) the expected value for women d) the varance, the standard devaton. Ths problem s a Bnomal, fnd n,p & q: n 6, p /3, q. a) P C(6,).(/3).() 4 b) Expected Value: µ n.p 6(/3) 4 for men c) Expected Value: µ n.p 6() for women d) Varance: n.p.q 6(/3)() 1.33 Standard Devaton: Ex7: A multple-choce test contans 10 questons wth 4 choces for each answer. If a student guesses the answers, fnd: a) the probablty that he wll get 4 correct answers. b) the expected value for the correct answers c) the expected value for the wrong answers d) the varance, the standard devaton. Ths problem s a Bnomal, fnd n,p & q: n 10, p 1/4 0.5, q a) P C(10,4).(0.5) 4.(0.75) 6 b) Expected Value: µ n.p 10(0.5).5 for the correct answers c) Expected Value: µ n.p 10(0.75) 7.5 for the wrong answers d) Varance: n.p.q 10(0.5)(0.75) Standard Devaton: Ex8: By rollng a par of dce, a game s played n whch you wn $ f the sum s, 3, 4 or 5. You wn $3 f the sum s 6, 7 or You loose $5 f the sum s 9, 10, 11 or 1. If you pay $ to play the game, fnd the expect gan or loss.,3,4,5 + 6+/36+3/36+4/3610/36 0/36 6,7,8 +3 5/36+6/36+5/3616/36 48/36 9,10,11,1-5 4/36+3/36+/36+510/36-50/36 Sum 1 E(X)$ 0.5 Expected gan s $0.5, but you pad $ to play the game, then there s a loss of $1.5 (page 5)

6 Secton 5.3 Normal Random Varable Z X µ Where Z s the Z-score: Ex1: Let Z be a random varable wth normal dstrbuton. Usng the table, fnd: a) P(Z < 1.87) d) P(Z > 1.87) b) P(0.49 < Z < 1.75) e) P(-1.66 < Z < -1.00) c) P(-1.77 < Z <.53) f) P(0.00 < Z <.17) g)p(-1.76 < Z < 0) Ex: Suppose that for a certan populaton the brth weght of nfants n pounds s normally dstrbuted wth mean 7.75 pounds and standard devaton of 1.5 pounds. Fnd the probablty that an nfant's brth s more than 9 pounds. Ex3: Bolts produced by a machne are acceptable provded that ther length s wthn the range 5.95 to 6.05 nches. Suppose that the length of the bolts produced are normally dstrbuted wth mean of 6 nches and standard devaton of 0.0 nches. what s the probablty that a bolt wll be an acceptable length? Ex4: The dstrbuton of low daly temperature n wnter n central Florda s random varable wth a mean of 50 degrees and a standard devaton of 8 degrees. If the daly temperature falls below 30 degrees, the orange crop wll suffer frost damage. What s the probablty that on a wnter day the orange crop wll suffer frost damage? Ex5: The number of daly order receved by mal order frm s normally dstrbuted wth a mean 50 and standard devaton 0. The company must ether hre extra help or pay overtme on those days when the number of order receved s 300 or hgher. What percentage of these days must the company ether hre extra help or pay overtme? Secton 5.4 Normal Approxmaton To The Bnomal RULES: To approxmate bnomal probablty by normal curve area: Step 1) determne n, p, q Step ) check that both n.p > 5 and n.q > 5 Step 3) fnd the expected value and the standard devaton µ n.p n p q.. Step 4) fnd the new ponts by: * subtractng 0.5 from the startng pont * addng 0.5 to the fnsh pont examples: P(3 < X < 6) wll be P (.5 < X < 6.5) P( X 7) wll be P (6.5 < X < 7.5) P(X > 8) wll be P ( X > 7.5) P( X < 8) wll be P (X < 5) Step 5) fnd the Z-scores and the area under the normal curve usng the table (page 6)

7 Example : Accordng to the Department of Health and Human Servces, the probablty s about 80% that a person aged 70 wll be alve at the age of 75. Suppose that 500 people aged 70 are selected at random. Fnd the probablty that: a) exactly 390 of them wll be alve at the age of 75 b) between 375 and 45 of them wll be alve at the age of 75. a) Step 1) n 500, p 0.8, q 0. Step ) check f both n.p and n.q are more than 5: n.p (500).(0.8) 400 n.q (500).(0.) 100 Step 3) fnd the expected value and the std. devaton: µ n.p (500).(0.8) 400 n. p. q ( 500).( 0. 8).( 0. ) Step 4) fnd the new pont: P(X 390) wll be P( < X < 390.5) Step 5) fnd the Z-score: X 389.5, Z X 390.5, Z and now by usng the table: P(-1.17 < Z < -1.06) b) for P( 375 < X < 45), we use the nformaton of steps 1, and 3 then: P(375 < X < 45) wll be P(374.5 < X < 45.5) X 374.5, Z X 45.5, Z. 85 and now by usng the table: P(-.85 < Z <.85) Note: If you try to fnd P(X390) of the frst queston by applyng the Bernoull formula, you wll have: C(500,390).(0.8) 390.(0.) 110 whch cannot be found by most calculators, and that s why the approxmaton method s very useful. (page 7)

benefit is 2, paid if the policyholder dies within the year, and probability of death within the year is ).

benefit is 2, paid if the policyholder dies within the year, and probability of death within the year is ). REVIEW OF RISK MANAGEMENT CONCEPTS LOSS DISTRIBUTIONS AND INSURANCE Loss and nsurance: When someone s subject to the rsk of ncurrng a fnancal loss, the loss s generally modeled usng a random varable or