2011-Psych 60-Lecture 8

2011-Psych 60-Lecture 8 - 4/13/11 Announcements Exam Next...

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Unformatted text preview: 4/13/11 Announcements Exam Next Monday Scanton will be PROVIDED Z-Table will be PROVIDED Can have two pages (8.5x11) of notes, both sides, typed or hand wriQen ANY calculator is fine As long as it can not communicate with anyone else (e.g. a cell phone or laptop calculator is not okay) Announcements Excel funcUon for Standard DeviaUon of a PopulaUon =stdev.p() for the newest version of Excel (2010 and 2011) =stdevp() for all versions of Excel Excel funcUon reference on WebCT WebCT Excel Materials Excel funcUons (by category) 1 4/13/11 Last Time FR O FO BE M RE Z-Score The number of standard deviaUons a parUcular score deviates from its corresponding mean Describing the LocaUon of Scores FR OM FO BE RE Z-Score FR OM FO BE RE Raw Scores Quiz 1 12 10 Quiz 2 12 14 Mean 12 12 zi = Xi ! " Student Tom Mike Sign: whether score is above or below mean of the populaUon Value: number of standard deviaUons the score is from the mean 2 4/13/11 FR O FO BE M RE Raw Scores Quiz 1 12 10 Quiz 2 12 14 Mean 12 12 FR O FO BE M RE ConverUng between Raw and Z-scores Student Tom Mike Raw score to Z-Score zi = Z-Score to Raw Score Z Scores Student Tom Mike Quiz 1 +.5 -0.166 Quiz 2 -1 +0.333 Mean Z -0.25 +.083 Xi ! " X i = + zi! This Week This Week Describing the LocaUon of Scores Probability The Normal DistribuUon Describing the LocaUon of Scores Probability The Normal DistribuUon 3 4/13/11 Probability 50 Total Marbles 40 White 10 Blue Probability of Event A Number of A outcomes p(A) = Number of possible outcomes 4 4/13/11 Random SelecUon Check Your Understanding p(Ace of Spades) = A B C D E 1/13 13/52 1/52 .52 1 Each individual in the populaUon has an equal chance of being selected Check Your Understanding p(Ace of Spades) = A B C D E 1/13 13/52 1/52 .52 1 5 4/13/11 1 52 p(Any Spade) = Check Your Understanding p(Any Spade) = A B C D E 1/13 13/52 1/52 .52 1 Check Your Understanding p(Any Spade) = A B C D E 1/13 13/52 1/52 .52 1 6 4/13/11 13 52 0.25 Number of A outcomes p(A) = Number of possible outcomes 0 p(A) 1 7 4/13/11 p(Win LoQery) 1. Win LoQery 2. Lose LoQery Win Win Lose 1 2 8 4/13/11 0.50? Win Win Lose Win Win Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, Lose, 1 16,000,000 9 4/13/11 p(Card > 5) 10 4/13/11 Probability of Event A Number of A outcomes p(A) = Number of possible outcomes 36 52 Probability of Event A Number of A outcomes p(A) = Number of possible outcomes p(A) = Area of A outcomes Total Area of DistribuUon 11 4/13/11 p(2 < Card < 5) 8 52 12 4/13/11 p(X > 60) 13 4/13/11 5 + 11 + 7 + 2 35 25 35 ~ 0.71 14 4/13/11 p(X > 110)? p(X > 110)? p(X > 110)? 15 4/13/11 1+1+0+2+1+3+8+21+42+62+96+157+239+385+546+78 6+1013+1523+1979+2370+3015+3575+4557+4909 100,000 25,291 100,000 p(X > 110) = .253 16 4/13/11 Probability and the Normal DistribuUon The Normal DistribuUon is Everywhere... Carl Friedrich Gauss Pierre-Simon Laplace Most physical characterisUcs of plants and animals Height and Weight of people Performance of stocks Errors in measurement 17 4/13/11 The Normal DistribuUon aka Gaussian Distribu<on 1 2!" 2 # ( X # )2 2" 2 The Normal DistribuUon aka Gaussian Distribu<on 1 2!" 2 # ( X # )2 2" 2 e e = 0 = 1 -2.5 -2 -1.5 -1 -.5 .5 " ( X " 0)2 2*12 1 1.5 2 2.5 1 2!" 2 e # (X #) 2" 2 2 1 2! 1 2 e 18 4/13/11 = 100 = 15 = 10 = 2 -2.5 62.5 -2 70 -1.5 77.5 -1 85 -.5 92.5 .5 107.5 " ( X "100)2 2*15 2 1 115 1.5 122.5 2 130 2.5 137.5 -2.5 62.5 5 -2 70 6 -1.5 77.5 7 -1 85 8 -.5 92.5 9 .5 107.5 11 " ( X "10)2 2*2 2 1 115 12 1.5 122.5 13 2 130 14 2.5 137.5 15 1 2! 15 2 e 1 2! 2 2 e = 0 = 1 = 0 = 1 -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 (aka Unit Normal DistribuUon) The Z DistribuUon 19 4/13/11 = 0 = 1 = 0 = 1 ?% ?% -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 (aka Unit Normal DistribuUon) The Z DistribuUon (aka Unit Normal DistribuUon) The Z DistribuUon Area Under The Curve Area Under The Curve = 0 = 1 = 0 = 1 50% 50% 50% ?% ?% -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 (aka Unit Normal DistribuUon) The Z DistribuUon (aka Unit Normal DistribuUon) The Z DistribuUon Area Under The Curve Area Under The Curve 20 4/13/11 = 0 = 1 = 0 = 1 50% 50% -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 -2.5 -2 -1.5 -1 -.5 .5 1 1.5 2 2.5 (aka Unit Normal DistribuUon) The Z DistribuUon (aka Unit Normal DistribuUon) The Z DistribuUon Area Under The Curve Area Under The Curve = 0 = 1 = 0 = 1 50% 34% 16% 16% 2 2.5 -2.5 -2 -1.5 -1 34% 34% 16% -2.5 -2 -1.5 -1 -.5 .5 1 1.5 -.5 .5 1 1.5 2 2.5 (aka Unit Normal DistribuUon) The Z DistribuUon (aka Unit Normal DistribuUon) The Z DistribuUon Area Under The Curve Area Under The Curve 21 4/13/11 = 0 = 1 = 0 = 1 34% 16% -2.5 -2 -1.5 -1 -.5 34% 16% .5 1 1.5 2 2.5 -2.5 -2 -1.5 -1 34% 34% 14% -.5 .5 1 1.5 2 2% 2.5 (aka Unit Normal DistribuUon) The Z DistribuUon (aka Unit Normal DistribuUon) The Z DistribuUon Area Under The Curve Area Under The Curve = 0 = 1 = 0 = 1 34% 14% -1.5 -1 -.5 34% 14% .5 1 1.5 34% 14% -1.5 77.5 7 -1 85 8 -.5 92.5 9 34% 14% .5 107.5 11 1 115 12 1.5 122.5 13 -2.5 2% -2 2 2% 2.5 (aka Unit Normal DistribuUon) The Z DistribuUon -2.5 62.5 5 2% -2 70 6 2 130 14 2% 2.5 137.5 15 Area Under The Curve 22 4/13/11 = 0 = 1 p(X > 100)? = 100 = 15 34% 14% -1.5 77.5 7 -1 85 8 -.5 92.5 9 34% 14% .5 107.5 11 1 115 12 1.5 122.5 13 -2.5 62.5 5 2% -2 70 6 2 130 14 2% 2.5 137.5 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 p(X > 100)? = 100 = 15 p(X > 100)? p(z > 0)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 23 4/13/11 p(X > 100)? p(z > 0)? 50% = 100 = 15 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 p(85 < X < 115)? = 100 = 15 p(85 < X < 115)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 24 4/13/11 p(85 < X < 115)? p(-1 < z < 1)? = 100 = 15 p(85 < X < 115)? p(-1 < z < 1)? = 100 = 15 34% 34% 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 p(85 < X < 115)? p(-1 < z < 1)? = 100 = 15 34% 34% 68% Total 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 1 1.5 2 2.5 107.5 115 122.5 130 137.5 25 4/13/11 Check Your Understanding Find p(X < 115)? = 100 = 15 Check Your Understanding Find p(X < 115)? = 100 = 15 A B C D E .16 .50 .34 .84 .68 A B C D E .16 .50 .34 .84 .68 p(X < 115)? = 100 = 15 p(X < 115)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 26 4/13/11 p(X < 115)? p(z < 1)? = 100 = 15 p(X < 115)? p(z < 1)? 50% 34% = 100 = 15 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 p(X < 115)? p(z < 1)? 50% 34% = 100 = 15 Finding Area Under the Normal Curve 1. 2. 3. 4. Sketch distribuUon Shade in area to be found Restate problem in terms of z Use known proporUons to find area 84% Total 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 1 1.5 2 2.5 107.5 115 122.5 130 137.5 27 4/13/11 = 0 = 1 z 0 1 2 Propor6on in Body Propor6on in Tail Propor6on between mean and Z 34% 14% -1.5 -1 -.5 34% 14% .5 1 1.5 -2.5 2% -2 The Z DistribuUon 2 2% 2.5 z 0 1 2 Propor6on in Body .50 Propor6on in Tail Propor6on between mean and Z z 0 1 2 Propor6on in Body .50 Propor6on in Tail .50 Propor6on between mean and Z 28 4/13/11 z 0 1 2 Propor6on in Body .50 Propor6on in Tail .50 Propor6on between mean and Z 0 z 0 1 2 Propor6on in Body .50 Propor6on in Tail .50 Propor6on between mean and Z 0 z 0 1 2 Propor6on in Body .50 .84 Propor6on in Tail .50 Propor6on between mean and Z 0 z 0 1 2 Propor6on in Body .50 .84 Propor6on in Tail .50 .16 Propor6on between mean and Z 0 29 4/13/11 z 0 1 2 Propor6on in Body .50 .84 Propor6on in Tail .50 .16 Propor6on between mean and Z 0 .34 z 0 1 2 Propor6on in Body .50 .84 Propor6on in Tail .50 .16 Propor6on between mean and Z 0 .34 z 0 1 2 Propor6on in Body .50 .84 Propor6on in Tail .50 .16 Propor6on between mean and Z 0 .34 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 Propor6on between mean and Z 0 .34 30 4/13/11 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 p(70 < X < 85)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 31 4/13/11 p(70 < X < 85)? = 100 = 15 p(70 < X < 85)? p(-2 < z < -1)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 32 4/13/11 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 .48 33 4/13/11 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 .48 .34 .48 .34 .14 p(70 < X < 85)? p(-2 < z < -1)? = 100 = 15 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 p(-2 < z < -1) = .14 34 4/13/11 p(X > 110)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 p(X > 110)? p(X > 110)? = 100 = 15 p(X > 110)? p(z > ?)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 35 4/13/11 = 100 = 15 X = 110 zi = Xi ! " = 100 = 15 X = 110 zi = zi = Xi ! " 110 !100 15 = 100 = 15 X = 110 zi = zi = Xi ! " p(X > 110)? p(z > .667)? = 100 = 15 110 !100 15 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 1 1.5 2 2.5 107.5 115 122.5 130 137.5 10 2 zi = = = .667 15 3 36 4/13/11 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 Unit Normal Table 37 4/13/11 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 38 4/13/11 z 0 1 2 Propor6on in Body .50 .84 .98 Propor6on in Tail .50 .16 .02 Propor6on between mean and Z 0 .34 .48 p(X > 110)? p(z > .667)? = 100 = 15 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 39 4/13/11 p(X > 110)? p(z > .667)? = 100 = 15 25.3% 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 1 1.5 2 2.5 107.5 115 122.5 130 137.5 p(z > .667) = .253 1+1+0+2+1+3+8+21+42+62+96+157+239+385+546+78 6+1013+1523+1979+2370+3015+3575+4457+4909 100,000 40 4/13/11 Finding Area Under the Normal Curve 1. 2. 3. 4. Sketch distribuUon Shade in area to be found Restate problem in terms of z Use proporUons from Unit Normal Table to find area in shaded region p(X > 110) = .253 p(77.5 < X < 92.5)? = 100 = 15 p(77.5 < X < 92.5)? = 100 = 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 41 4/13/11 p(77.5 < X < 92.5)? p(? < z < ?)? = 100 = 15 = 100 = 15 X = 77.5 zi = zi = Xi ! " 77.5 !100 15 !22.5 = !1.5 15 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 zi = = 100 = 15 X = 92.5 zi = zi = Xi ! " p(77.5 < X < 92.5)? p(-1.5 < z < -.5)? = 100 = 15 92.5 !100 15 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 1 1.5 2 2.5 107.5 115 122.5 130 137.5 !7.5 zi = = !0.5 15 42 4/13/11 z -0.5 -1.5 Propor6on in Body Propor6on in Tail Propor6on between mean and Z z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 ? .43 ? 43 4/13/11 z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 .43 .19 ? .43 .19 .24 p(77.5 < X < 92.5)? p(-1.5 < z < -.5)? = 100 = 15 z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 .43 .19 .24 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 107.5 .5 115 1 122.5 1.5 130 2 137.5 2.5 .24 p(-1.5 < z < -.5) = .24 .24 44 4/13/11 z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 z -0.5 -1.5 Propor6on in Body .6915 .9332 Propor6on in Tail .3085 .0668 Propor6on between mean and Z .1915 .4332 .43 .19 .24 .43 .19 .24 .93 .69 .24 .93 .69 .24 .24 .31 .07 .24 Finding Area Under the Normal Curve 1. 2. 3. 4. Sketch distribuUon Shade in area to be found Restate problem in terms of z Use proporUons from Unit Normal Table to find area in shaded region 45 4/13/11 p(X > ?) = .20 = 100 = 15 p(X > ?) = .20 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 p(X > ?) = .20 = 100 = 15 p(X > ?) = .20 p(z > ?) = .20 = 100 = 15 20% 62.5 70 77.5 85 92.5 =100 107.5 115 122.5 130 137.5 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 107.5 20% 115 122.5 1.5 130 2 137.5 2.5 ?? ?? 1 46 4/13/11 47 4/13/11 p(X > ?) = .20 p(z > .84) = .20 = 100 = 15 20% 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 1 1.5 2 2.5 107.5 115 122.5 130 137.5 z = .84 p(z > .84) = .20 p(X > ?) = .20 = 100 = 15 = 100 = 15 z = .84 Z-score to Raw Score X i = + zi! 20% 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 1 1.5 2 2.5 107.5 115 122.5 130 137.5 z = .84 48 4/13/11 Z-score to Raw Score = 100 = 15 z = .84 Z-score to Raw Score = 100 = 15 z = .84 X i = + zi! X i = 100 + .84(15) X i = + zi! X i = 100 + .84(15) Xi = 100 + 12.6 Z-score to Raw Score = 100 = 15 z = .84 p(X > 112.6) = .20 p(z > .84) = .20 = 100 = 15 Xi = 112.6 62.5 -2.5 70 -2 77.5 -1.5 85 -1 92.5 =100 -.5 0 .5 107.5 20% 112.6 115 1 122.5 1.5 130 2 137.5 2.5 49 4/13/11 Finding Scores from Area 1. Sketch distribuUon 2. Shade in area to be found 3. Use proporUons from Unit Normal Table to find z - boundaries 4. Restate problem in terms of X ProporUons are correct only for normally distributed data! Quiz 1 = 10.5 = 3 TOM -2.5 Quiz 2 = 13.5 = 1.5 TOM 50 4/13/11 Quiz 1 TOM Quiz 1 = 10.5 = 3 = 10.5 = 3 TOM -2.5 -1.5 -1 -.5 0 .5 1 1.5 -2.5 -1.5 -1 -.5 0 .5 1 1.5 Quiz 2 TOM Quiz 2 = 13.5 = 1.5 = 13.5 = 1.5 TOM -5 -4 -3 -2 -1 0 1 -5 -4 -3 -2 -1 0 1 ProporUons are correct only for normally distributed data! 51 4/13/11 For Next Time (Re)Read: Chapter 6 Do Review Homework 2 SoluUons Start Studying for Midterm 1 Work on notes for midterm (Two 8.5"x11" pages, both sides, any content desired) Exam covers Chapters 1 6 and lecture material through Friday Download Study Guide from WebCT 52 ...
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