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Random Variables Discrete Random Variables Continuous Random Variables Discrete Probability Distributions Expected Value for a Discrete Random Variables Variance and Standard Deviation for a Discrete Random Variables Binomial Experiment Binomial Probabilities Expected Value and Variance and Standard Deviation for a Binomial Distribution POISSON Distibution Hypergeometric Distribution Experiment of tossing coin three times Steps in Experiment 3 Outcomes Step 1 2 Outcomes Step 2 2 Outcomes Step 3 2 Total Experimental Outcomes 8 (Sample Points) P(for each) = 0.125 Classical # SP Toss 1 Toss 2 Toss 3 Sample Space 1 H H H (H, H, H) 2 T H H (T, H, H) 3 H T H (H, T, H) 4 H H T (H, H, T) 5 T T H (T, T, H) 6 T H T (T, H, T) 7 H T T (H, T, T) 8 T T T (T, T, T) x = Discrete Random Variable = # of heads in 3 flips Steps: X Frequency P(X) = f(x) Each f(x)>=0? 0 1 0.125 TRUE 1 3 0.375 TRUE 2 3 0.375 TRUE 3 1 0.125 TRUE 8 1 P(X =1) Experiment of tossing coin three P(X > = 2) times P(x = 0 or x = 3) 0.25 0.375 P(X not 2) 0.625 Probability 0.125 . 0 1 # of heads in 3 flips Success = Heads Notation for Heads is H H Probability of # of heads for sample Sample Point points 0.125 3 0.125 2 0.125 2 0.125 2 0.125 1 0.125 1 0.125 1 0.125 0 1 #1 Define Random Variable #2 Build Frequency Distribution #3 Calculate Relative Frequency - P(x) = f(x) #4 Check Requirement #1: f(x) >= 0 #5 Check Requirement #2: Σf(x) = 1 #6 Create a Column Chart for a Discrete Variable (Columns do not touch) #7 Make predictions riment of tossing coin three times 0.375 0.375 0.125 2 3 # of heads in 3 flips Experiment of tossing coin three times Steps in Experiment 3 Outcomes Step 1 2 Outcomes Step 2 2 Outcomes Step 3 2 Total Experimental Outcomes 8 (Sample Points) P(for each) = 0.125 Classical # SP Toss 1 Toss 2 Toss 3 Sample Space 1 H H H (H, H, H) 2 T H H (T, H, H) 3 H T H (H, T, H) 4 H H T (H, H, T) 5 T T H (T, T, H) 6 T H T (T, H, T) 7 H T T (H, T, T) 8 T T T (T, T, T) x = Discrete Random Variable = # of heads in 3 flips Experiment of tossing coin three timesSteps: X Frequency 0.375 P(X) = f(x) Each f(x)>=0? 0.375 0 1 0.125 TRUE Probability 0.125 0.125 1 3 0.375 TRUE 2 3 0.375 TRUE 3 1 0.125 0 TRUE 1 2 3 8 1 # of heads in 3 flips P(X =1) 0.125 P(X > = 2) 0.5 P(x = 0 or x = 3) 0.25 0.25 P(X not 2) 0.625 0.625 Success = Heads Notation for Heads is H H Probability of # of heads for sample Sample Point points 0.125 3 0.125 2 0.125 2 0.125 2 0.125 1 0.125 1 0.125 1 0.125 0 1 oin three #1 Define Random Variable #2 Build Frequency Distribution #3 Calculate Relative Frequency - P(x) = f(x) 0.125 #4 Check Requirement #1: f(x) >= 0 #5 Check Requirement #2: Σf(x) = 1 #6 Create a Column Chart for a Discrete Variable (Columns do not touch) #7 Make predictions Over last 20 days the number of operating rooms used at TG hospital were: Steps to Build Discrete Probability Distribut On 3 days only 1 were used #1 Define Random Variable On 5 days 2 were used #2 Build Frequency Distribution On 8 days 3 were used #3 Calculate Relative Frequency - P(x On 4 days all (4) were used #4 Check Requirement #1: f(x) >= 0 Create a Probability Distribution that can be used to make predictions in the future. #5 Check Requirement #2: Σf(x) = 1 #6 Create a Column Chart for a Discr Discrete Random Variable = x = #7 Make predictions x=# Relative operating Frequency P(x) rooms in use = f(x) = Frequency (# of days) for 1 day Probability Is each f(x) >= 0? P(x) f(x) P() f() P() f() P() f() P() f() Reason we do the whole distribution? Because then it is easy to make predictions check 2 <=2 P(x<=2) f(1) + f(2) 2 <2 P(x<2) f(1) 2 >=2 P(x>=2) f(2) + f(3) + f(4) 2 >2 P(x>2) f(3) + f(4) ete Probability Distribution: ndom Variable uency Distribution Relative Frequency - P(x) = f(x) uirement #1: f(x) >= 0 uirement #2: Σf(x) = 1 olumn Chart for a Discrete Variable (Columns do not touch) Discrete Probability Charts don't have columns that touch. SUMIF function: range = range with all criteria criteria = criteria such as 2, <=2, >=2, >2 sum_range = range with values to add For more about this function, see this video: Excel Magic Trick #203: SUMIF function formula 21 Examples Over last 20 days the number of operating rooms used at TG hospital were: Steps to Build Discrete Probability Distribut On 3 days only 1 were used #1 Define Random Variable On 5 days 2 were used #2 Build Frequency Distribution On 8 days 3 were used #3 Calculate Relative Frequency - P(x On 4 days all (4) were used #4 Check Requirement #1: f(x) >= 0 Create a Probability Distribution that can be used to make predictions in the future. #5 Check Requirement #2: Σf(x) = 1 #6 Create a Column Chart for a Discr Discrete Random Variable = x = X = # of Op. Rooms used in 1 day #7 Make predictions x=# Relative operating Frequency P(x) rooms in use = f(x) = Frequency (# of days) for 1 day Probability Is each f(x) >= 0? P(x) f(x) 3 1 0.15 TRUE P(1) f(1) 5 2 0.25 TRUE P(2) f(2) 8 3 0.4 TRUE P(3) f(3) 4 4 0.2 TRUE P(4) f(4) 20 1 Reason we do the whole distribution? Because then it is easy to make predictions check 2 <=2 P(x<=2) f(1) + f(2) 0.4 0.4 2 <2 P(x<2) f(1) 0.15 0.15 2 >=2 P(x>=2) f(2) + f(3) + f(4) 0.85 0.85 2 >2 P(x>2) f(3) + f(4) 0.6 0.6 ete Probability Distribution: Over last 20 days the number of operating rooms used at TG ndom Variable hospital were: uency Distribution Relative Frequency - P(x) = f(x) 0.4 uirement #1: f(x) >= 0 Probability 0.25 uirement #2: Σf(x) = 1 0.2 0.15 olumn Chart for a Discrete Variable (Columns do not touch) 1 2 3 4 X = # of Op. Rooms used in 1 day Discrete Probability Charts don't have columns that touch. SUMIF function: range = range with all criteria criteria = criteria such as 2, <=2, >=2, >2 sum_range = range with values to add For more about this function, see this video: Excel Magic Trick #203: SUMIF function formula 21 Examples sed at TG Job satisfaction IS Senior Is each f(x) >= IS Middle Is each f(x) >= score Executives 0? Executives 0? 1 5% 4% 2 9% 10% 3 3% 12% 4 42% 46% 5 41% 28% Job satisfaction IS Senior Is each f(x) >= IS Middle Is each f(x) >= score Executives 0? Executives 0? 1 5% TRUE 4% TRUE 2 9% TRUE 10% TRUE 3 3% TRUE 12% TRUE 4 42% TRUE 46% TRUE 5 41% TRUE 28% TRUE 100% 100% Discrete Probability Distribution IS Senior Executives IS Middle Executives 42%46% 41% 28% 9% 10% 12% 5% 4% 3% 1 2 3 4 5 Job satisfaction score Mean and Standard Deviation from Raw Data Mean and Standard Deviation for Discrete Pr x = # operating rooms in use for 1 day Frequency (# of days) Mean 1 3 STDEV 2 5 STDEVP 3 8 Raw Data = Xi X - mu (X-mu)^2 4 4 1 20 1 1 E(x) Σx*f(x) SD of Discrete Random 2 Variable sqrt(Σ(x-mean)^2*f(x)) 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 Must Equal Zero sd Count ndard Deviation for Discrete Probability Distribution Relative Frequency P(x) = f(x) = E(x) Probability =X*f(x) X - mu (X-mu)^2 (X-mu)^2*f(x) 0.15 0.25 0.4 0.2 1 because you are not Var SD Not equal to zero using all the data points Mean and Standard Deviation from Raw Data Mean and Standard Deviation for Discrete Pr x = # operating rooms in use for 1 day Frequency (# of days) Mean 2.650 1 3 STDEV 0.98808693 2 5 STDEVP 0.96306801 3 8 Raw Data = Xi X - mu (X-mu)^2 4 4 1 -1.650 2.7225 20 1 -1.650 2.7225 1 -1.650 2.7225 E(x) Σx*f(x) SD of Discrete Random 2 -0.650 0.4225 Variable sqrt(Σ(x-mean)^2*f(x)) 2 -0.650 0.4225 2 -0.650 0.4225 2 -0.650 0.4225 2 -0.650 0.4225 3 0.350 0.1225 3 0.350 0.1225 3 0.350 0.1225 3 0.350 0.1225 3 0.350 0.1225 3 0.350 0.1225 3 0.350 0.1225 3 0.350 0.1225 4 1.350 1.8225 4 1.350 1.8225 4 1.350 1.8225 4 1.350 1.8225 0.00 Must Equal Zero sd 0.963068 Count 20 ndard Deviation for Discrete Probability Distribution Relative Frequency P(x) = f(x) = E(x) Probability =X*f(x) X - mu (X-mu)^2 (X-mu)^2*f(x) 0.15 0.150 -1.650 2.7225 0.408375 0.25 0.500 -0.650 0.4225 0.105625 0.4 1.200 0.350 0.1225 0.049 0.2 0.800 1.350 1.8225 0.3645 1 2.650 -0.600 because you are not Var 0.9275 SD 0.963068014 Not equal to zero using all the data 2.650 points 0.963068014 0.963068014 0.981360288 0.9814 Unit Demand = x Prob = f(x) (X - Mean)^2 Mean 500 0.1 84100 790 600 0.15 36100 700 0.15 8100 800 0.2 100 900 0.2 12100 1000 0.15 44100 1100 0.05 96100 Mean = monthly order quantity SD 170 Price per unit $125.00 Cost per unit $60.00 Assumed Units sold 1000 Sales Expenses Gross Profit Unit Demand = x Prob = f(x) (X - Mean)^2 500 0.1 84100 600 0.15 36100 700 0.15 8100 800 0.2 100 900 0.2 12100 1000 0.15 44100 1100 0.05 96100 1 Mean 790 = monthly order quantity SD 170 170 Price per unit $125.00 Cost per unit $60.00 Assumed Units sold 1000 Sales $125,000.00 Expenses $47,400.00 Gross Profit $77,600.00 Prob. Of State Stock A Stock B Stock C State of Economy of Economy Return Return Return Boom 0.15 0.15 0.25 0.11 Normal 0.30 0.07 0.13 0.10 Bust 0.55 -0.02 -0.135 0.03 E(Ri) Standard Sd = Proxy for Risk Deviation CV = SD/Mean Prob. Of State Stock A Stock B Stock C State of Economy of Economy Return Return Return Boom 0.15 0.15 0.25 0.11 Normal 0.30 0.07 0.13 0.10 Bust 0.55 -0.02 -0.135 0.03 E(Ri) 0.0325 0.00225 0.063 Standard Sd = Proxy for Risk Deviation 0.0633147 0.156409 0.03662 CV = SD/Mean 1.9481443 69.51499 0.58126 pi = 0.2 1 - pi = 0.8 n= 4 Fixed # of Trials? Yes Each Trial Independent? Yes Success is? Sale S Fail is? No Sale NS Success or fail each trial? Yes pi the same each Trial? Yes Sample Space = 16 Build Discrete Probability Distribution with n = 4, pi = 0.2 Attempted Sale # of Probability Probability of Occurrence Probability of # of Sales (Random Possible Outcomes P(x) P(x) 1st 2nd 3rd 4th Sales 1st 2nd 3rd 4th Calculation Occurrence Variable X) 1 S S S S 4 0.2 0.2 0.2 0.2 0.2*0.2*0.2*0.2 = 0.0016 0 P(0) = 0.4096 2 S S S NS 3 0.2 0.2 0.2 0.8 0.2*0.2*0.2*0.8 = 0.0064 1 P(1) = 0.4096 3 S S NS S 3 0.2 0.2 0.8 0.2 0.2*0.2*0.8*0.2 = 0.0064 2 P(2) = 0.1536 4 S NS S S 3 0.2 0.8 0.2 0.2 0.2*0.8*0.2*0.2 = 0.0064 3 P(3) = 0.0256 5 NS S S S 3 0.8 0.2 0.2 0.2 0.8*0.2*0.2*0.2 = 0.0064 4 P(4) = 0.0016 6 S S NS NS 2 0.2 0.2 0.8 0.8 0.2*0.2*0.8*0.8 = 0.0256 1.0000 7 S NS NS S 2 0.2 0.8 0.8 0.2 0.2*0.8*0.8*0.2 = 0.0256 8 NS NS S S 2 0.8 0.8 0.2 0.2 0.8*0.8*0.2*0.2 = 0.0256 E(x) 0.8 0.8 9 S NS S NS 2 0.2 0.8 0.2 0.8 0.2*0.8*0.2*0.8 = 0.0256 SD 0.8 0.8 10 NS S NS S 2 0.8 0.2 0.8 0.2 0.8*0.2*0.8*0.2 = 0.0256 11 NS S S NS 2 0.8 0.2 0.2 0.8 0.8*0.2*0.2*0.8 = 0.0256 Build Discrete Probability Distribution with n = 4, pi = 0.2 12 S NS NS NS 1 0.2 0.8 0.8 0.8 0.2*0.8*0.8*0.8 = 0.1024 13 NS S NS NS 1 0.8 0.2 0.8 0.8 0.8*0.2*0.8*0.8 = 0.1024 P(x) = Probability of X Sales in 4 14 NS NS S NS 1 0.8 0.8 0.2 0.8 0.8*0.8*0.2*0.8 = 0.1024 0.4096 0.4096 15 NS NS NS S 1 0.8 0.8 0.8 0.2 0.8*0.8*0.8*0.2 = 0.1024 16 NS NS NS NS 0 0.8 0.8 0.8 0.8 0.8*0.8*0.8*0.8 = 0.4096 Attempts 0.1536 0.0256 0.0016 0 1 2 3 4 X Random Variable (# of Sales) cb34a135-5196-4a67-a768-b531ecd4b665.xls BDPD Page 21 of 57 Binomial Experiment 4 Requirements: Experiment consists of a sequence of n identical trials. Random Variable counts the number of successes in a Fixed number of 1 trials, n. Fixed # of Identical Trials = n Only 2 outcomes are possible on each indetical trial. Success or 2 Failure. Each trial only results in S or F Probability of Success = p = (π "pi"). Probability of Failure = 1-p. Probability remains the same on 3 each trial. p remains the same for each trial The trials are independent (one 4 does not affact the next) All events are independent An insurance agent has appointments with 4 clients tomorrow. From past data, the chance of making a sale is 1 in 5. What is likelihood that she will = p (π) = sell 3 policies in 4 tries? Fixed # of Identical Trials = n = Variables P(x) = f(x) Success P(x = 0) Failure P(x >= 0) n P(x > 0) x P(x < 0) p (π) P(x <= 0) X P(x) = f(x) 0 1 2 3 4 = p (π) = Fixed # of Identical Trials = n = P(x) = f(x) SUM #S.P. = COMBIN An insurance agent has appointments with 4 x= clients tomorrow. From past data, the chance n= of making a sale is 1 in 5. What is likelihood s p= that she will sell 3 policies in 4 tries? ns (1-p) = # S.P. C1 C2 C3 C4 P1 P2 P3 1 ns s s s 0.8 0.2 0.2 2 s ns s s 0.2 0.8 0.2 3 s s ns s 0.2 0.2 0.8 4 s s s ns 0.2 0.2 0.2 ent has appointments with 4 . From past data, the chance is 1 in 5. What is likelihood sell 3 policies in 4 tries? P4 P(S.P.) Multiply 0.2 0.2 0.2 0.8 Add Binomial Experiment 4 Requirements: Experiment consists of a sequence of n identical trials. Random Variable counts the number of successes in a Fixed number of 1 trials, n. Fixed # of Identical Trials = n Yes Only 2 outcomes are possible on each indetical trial. Success or 2 Failure. Each trial only results in S or F Yes Probability of Success = p = (π "pi"). Probability of Failure = 1-p. Probability remains the same on 3 each trial. p remains the same for each trial Yes The trials are independent (one 4 does not affact the next) All events are independent Yes An insurance agent has appointments with 4 clients tomorrow. From past data, the chance of making a sale is 1 in 5. What is likelihood that she will Probability of Sale = p (π) = 0.2 sell 3 policies in 4 tries? Fixed # of Identical Trials = n = 4 Variables P(x) = f(x) Sale Success s P(x = 3) Not Sale Failure ns P(x >= 3) Attempts at Sale n 4 P(x > 3) X = # of Sale is made in 4 tries x 3 P(x < 3) Probability of Sale p (π) 0.2 P(x <= 3) X P(x) = f(x) 0 0.4096 1 0.4096 2 0.1536 3 0.0256 4 0.0016 Binomial Distribution, n = 4, p = .2 0.4096 0.4096 0.1536 0.0256 0.0016 0 1 2 3 4 X = # of Sale is made in 4 tries Probability of Sale = p (π) = 0.2 Fixed # of Identical Trials = n = 4 P(x) = f(x) SUM 0.0256 0.0256 0.0272 0.0272 0.0016 0.0016 0.9728 0.9728 0.9984 0.9984 #S.P. = COMBIN 4 An insurance agent has appointments with 4 x= 3 clients tomorrow. From past data, the chance n= 4 of making a sale is 1 in 5. What is likelihood s p= 0.2 that she will sell 3 policies in 4 tries? ns (1-p) = 0.8 # S.P. C1 C2 C3 C4 P1 P2 P3 P4 1 ns s s s 0.8 0.2 0.2 0.2 2 s ns s s 0.2 0.8 0.2 0.2 3 s s ns s 0.2 0.2 0.8 0.2 4 s s s ns 0.2 0.2 0.2 0.8 s appointments with 4 m past data, the chance n 5. What is likelihood policies in 4 tries? P(S.P.) Multiply 0.0064 0.0064 0.0064 0.0064 0.0256 Add Binomial Experiment 4 Requirements: Experiment consists of a sequence of n identical trials. Random Variable counts the number of successes in a 1 Fixed number of trials, n. Fixed # of Identical Trials = n Only 2 outcomes are possible on each indetical trial. 2 Success or Failure. Each trial only results in S or F Probability of Success = p = (π "pi"). Probability of Failure = 1-p. Probability remains the same on each 3 trial. p remains the same for each trial The trials are independent (one does not affact the 4 next) All events are independent A flight from Oakland to Seattle occurs 6 times per day. The probability that any 1 flight is late is 10%. What is the probability that exactly 2 planes are late? Less than 2 are late? What is the mean and standard deviation? Variables Success Failure n x p (π) = p (π) = Fixed # of Identical Trials = n = P(x) = f(x) P(x) = f(x) 2 P(x = 2) 2 P(x >= 2) 2 P(x > 2) 2 P(x < 2) 2 P(x <= 2) Mean Σx*f(x) SD sqrt(Σ(x-mean)^2*f(x)) Mean n*p SD sqrt(n*p*(1-p)) X f(x) 0 0.531441 1 0.354294 2 0.098415 3 0.01458 4 0.001215 5 5.4E-05 6 0.000001 A flight from Oakland to Seattle occurs 6 times per day. The probability that any 1 flight is late is 10%. What is the probability that exactly 2 planes are late? Less than 2 are late? What is the mean and standard deviation? 0.531 0.354 0.098 0.015 0.001 0.000 0.000 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Binomial Experiment 4 Requirements: Experiment consists of a sequence of n identical trials. Random Variable counts the number of successes in a 1 Fixed number of trials, n. Fixed # of Identical Trials = n Yes Only 2 outcomes are possible on each indetical trial. 2 Success or Failure. Each trial only results in S or F Yes Probability of Success = p = (π "pi"). Probability of Failure = 1-p. Probability remains the same on each 3 trial. p remains the same for each trial Yes The trials are independent (one does not affact the 4 next) All events are independent Yes A flight from Oakland to Seattle occurs 6 times per day. The probability that any 1 flight is late is 10%. What is the probability that exactly 2 planes are late? Less than 2 are late? What is the mean and standard deviation? Variables Late Plane Success l Not Late Plane Failure nl # of flights per day n 6 # successes = # of Late Planes x 2 Probability of Late Plane p (π) 0.1 Probability of Late Plane = p (π) = 0.1 Fixed # of Identical Trials = n = 6 P(x) = f(x) P(x) = f(x) 2 P(x = 2) 0.098415 0.6 2 P(x >= 2) 0.114265 0.4 2 P(x > 2) 0.01585 0.2 2 P(x < 2) 0.885735 0 2 P(x <= 2) 0.98415 0 1 2 Mean Σx*f(x) 0.6 SD sqrt(Σ(x-mean)^2*f(x)) 0.734847 Mean n*p 0.6 SD sqrt(n*p*(1-p)) 0.734847 X f(x) 0 0.531441 1 0.354294 2 0.098415 3 0.01458 4 0.001215 5 5.4E-05 6 0.000001 3 4 5 6 n 5 Binomial Distribution, n = 5 p = 0.4 p 0.4 X 3 Binomial Distribution, n = 5 p = 0.4 X p(x) 0.4 0 0.07776 0.3 1 0.2592 0.2 2 0.3456 3 0.2304 0.1 4 0.0768 0 5 0.01024 0 1 2 3 4 5 n 10 Binomial Distribution, n = 10 p = 0.5 P(X<=8) p 0.5 P(X<=8) = 0.9893 X 8 Comparative Operator <= Binomial Distribution, n = 10 p = 0.5 P(X<=8) Result <=8 BINOMDIST 0.989258 0.989258 0.246 P(X<=8) = 0.9893 X p(x) P(X<=8) 0.205 0.205 0 0.000977 0.000977 1 0.009766 0.009766 2 0.043945 0.043945 0.117 0.117 3 0.117188 0.117188 4 0.205078 0.205078 0.044 0.044 5 0.246094 0.246094 0.001 0.010 6 0.205078 0.205078 7 0.117188 0.117188 8 0.043945 0.043945 0 1 2 3 4 5 6 7 8 9 0.009766 10 0.000977 0.5 P(X<=8) 0.044 0.010 0.001 9 10 n 10 Binomial Distribution, n = 10 p = 0.5 P(X<=4) p 0.5 P(X<=4) = 0.3770 X 4 Comparative Operator <= Result <=4 BINOMDIST 0.376953 X p(x) P(X<=4) 0 0.000977 1 0.009766 2 0.043945 3 0.117188 4 0.205078 5 0.246094 6 0.205078 7 0.117188 8 0.043945 9 0.009766 10 0.000977 n 10 Binomial Distribution, n = 10 p = 0.5 P(X<=4) p 0.5 P(X<=4) = 0.3770 X 4 Comparative Operator <= Binomial Distribution, n = 10 p = 0.5 Result <=4 BINOMDIST 0.376953 P(X<=4) 0.3 X p(x) P(X<=4) 0.25 0 0.000977 0.000977 0.2 1 0.009766 0.009766 0.15 2 0.043945 0.043945 0.1 3 0.117188 0.117188 0.05 4 0.205078 0.205078 0 5 0.246094 0 1 2 3 4 5 6 7 6 0.205078 7 0.117188 x 8 0.043945 9 0.009766 10 0.000977 = 10 p = 0.5 8 9 10 Poisson Probability Distribution can be used to estimate the number of occurrences over a specified interval of time or space Properties of a Poisson Experiment: The probability of an occurrence is the same for any two intervals of equal length The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval No upper limit for X: 0,1,2,3… , but as x increases past the mean, the probability decreases and gest quite small. Mean = Var Excel function: POISSON POISSON(x,mean,cumulative (0 = exact, 1 = cumulative)) x = Number of Web visitors 7 per minutes Business = Your Cool Web Site x = Number of Web visitors arrive at a rate of 7 per minute at Your Cool Web Site Assumption 1 Probability is the same for any 1 minute interval Arrival or nonarrival of a web visitor in a 1 minute period is independent of the arrival or nonarrival in any other 1 Assumption 2 minute period Mean 7 visits per 1 minute a x 0 visits per 1 minute Mean P(0 visits per 1 minute) P(x) = (mu^x*e^-mu)/X!) b x 2 P(2 or more in 1 minute) c x 1 0.5 minutes Mean P(1 or more in 30 sec.) d x 5 P(5 or more in 1 minute) x p(x) x = Number of Web visitors arrive at a rate of 7 pe 0 0.000911882 Cool Web Site 1 0.006383174 0.16 2 0.022341108 0.14 0.14 3 0.052129252 0.12 4 0.091226192 0.1 5 0.127716668 0.08 6 0.14900278 0.06 7 0.14900278 0.04 8 0.130377432 0.02 9 0.10140467 0 10 0.070983269 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 11 0.045171171 x = Number of Web visitors 12 0.02634985 13 0.014188381 14 0.00709419 15 0.003310622 16 0.001448397 17 0.000596399 18 0.000231933 19 8.5449E-05 20 2.99071E-05 21 9.96904E-06 22 3.17197E-06 23 9.65382E-07 24 2.8157E-07 25 7.88395E-08 26 2.1226E-08 27 5.50304E-09 28 1.37576E-09 29 3.3208E-10 30 7.74854E-11 rs arrive at a rate of 7 per minute at Your Cool Web Site 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 = Number of Web visitors Poisson Probability Distribution can be used to estimate the number of occurrences over a specified interval of time or space Properties of a Poisson Experiment: The probability of an occurrence is the same for any two intervals of equal length The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval No upper limit for X: 0,1,2,3… , but as x increases past the mean, the probability decreases and gest quite small. Mean = Var Excel function: POISSON POISSON(x,mean,cumulative (0 = exact, 1 = cumulative)) x = Number of Web visitors 7 per minutes Business = Your Cool Web Site x = Number of Web visitors arrive at a rate of 7 per minute at Your Cool Web Site Assumption 1 Probability is the same for any 1 minute interval Arrival or nonarrival of a web visitor in a 1 minute period is independent of the arrival or nonarrival in any other 1 Assumption 2 minute period Mean 7 visits per 1 minute a x 0 visits per 1 minute Mean 7 P(0 visits per 1 minute) 0.000911882 =POISSON(C16,C17,0) P(x) = (mu^x*e^-mu)/X!) 0.0009118820 =(C17^C16*EXP(1)^-C17)/FACT(C16) b x 2 P(2 or more in 1 minute) 0.992704944 c x 1 0.5 minutes Mean 3.5 P(1 or more in 30 sec.) 0.969802617 d x 5 P(5 or more in 1 minute) 0.827008392 x p(x) x = Number of Web visitors arrive at a rate of 7 pe 0 0.000911882 Cool Web Site 1 0.006383174 0.16 2 0.022341108 0.14 0.14 3 0.052129252 0.12 4 0.091226192 0.1 5 0.127716668 0.08 6 0.14900278 0.06 7 0.14900278 0.04 8 0.130377432 0.02 9 0.10140467 0 10 0.070983269 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 11 0.045171171 x = Number of Web visitors 12 0.02634985 13 0.014188381 14 0.00709419 15 0.003310622 16 0.001448397 17 0.000596399 18 0.000231933 19 8.5449E-05 20 2.99071E-05 21 9.96904E-06 22 3.17197E-06 23 9.65382E-07 24 2.8157E-07 25 7.88395E-08 26 2.1226E-08 27 5.50304E-09 28 1.37576E-09 29 3.3208E-10 30 7.74854E-11 rs arrive at a rate of 7 per minute at Your Cool Web Site 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 = Number of Web visitors

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Success as an Insurance Agent document sample

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