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Session 5

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Session 5

  1. 1. 7/4/2012 Probability Distribution Characteristics of discrete probability distribution Discrete Continuous Expected value, standard deviation of a random variable Binomial Probability Distributions Clear Tone Radios x F(x) 0 0.20 Distribution of Number of separate enquiries leading to Properties 1 2 0.55 0.80 new business proposals on any working day 3 0.90 4 0.95 5 1.00The probability distribution should depict/provide all possibleInformation regarding the random variable in question. • f(x) = P[X=x]; 0 ≤ f(x) ≤ 1; Σ f(x) =1x P[X=x] Prob. dist. of no. of qualifying • CDF F(x) = P[X ≤x] = Σy ≤ x f(y) enquiries in a day0 0.20 • F is a step function with jumps only at the possible1 0.35 0.40 values probability 0.302 0.253 0.10 0.20 • Density to distribution function and vice versa4 0.05 0.105 0.05 0.00 0 1 2 3 4 5 no. of queries 32total 1.00 Expectation and Variance Interpretation of x f xf x2f (x-mean)2f EXPECTED VALUE values prob. 0 0.20 0 0 0.512 1 0.35 0.35 0.35 0.126 2 0.25 0.5 1 0.04 3 0.10 0.3 0.9 0.196 • “mean” in the long run 4 0.05 0.2 0.8 0.288 5 0.05 0.25 1.25 0.578 • not necessarily the most likely value Total 1.00 mean= 1.60 4.30 σ2=1.74 Check: 4.30 - 1.602= 1.74 1
  2. 2. 7/4/2012 Probability distribution of # of successful days in a working week (consisting of six days) Binomial Probabilities A day is deemed to be successful if at least 1 qualifying enquires are made on that day. Possible number of successful days : 0,1,2,3,4,5,6 Probability[0 successful day] = P[FFFFFF]=0.26=0.000064 P(a day is successful) = 0.8 Probability[1 successful day] = P[SFFFFF]+ P[FSFFFF]+ P[FFSFFF]+ P[FFFSFF]+ P[FFFFSF]+ P[FFFFFS] = 6 ×0.25 ×0.8 =0.001536 Probability distribution of # of successful days in a Binomial Distribution working week (consisting of six days): When is it applicable? Binomial distribution • Binomial expt. is one where n independent and identical trials are repeated; each trial No of may result in two possible outcomes(call themsuccessful days Probability 6×5 ‘success’(S) and ‘failure’(F); p=P(S). 0 0.000 6 C 4 = 6C 2 = 1 0.002 1× 2 • In the above context, a random variable X, 2 0.015 denoting the total no. of successes is said to 3 0.082 4 0.246 6 C4 × 0.84 × 0.22 have a Binomial distribution with parameters 5 0.393 n and p. X→B(n,p) 6 0.262 Binomial Distribution(cont.) Exercise • So, in general, for X→B(n,p) Work out the market research problem for the – P[X=x]= nCx px (1-p)(n-x) new product design given in the last week. • Use Excel to calculate probabilities • Mean or the ‘Expected value’ = np Need to feed in P [survey result| Market will do • Variance = np(1-p) well] & P [survey result| Market will NOT do well] • standard deviation = √{np(1-p)} 40 2
  3. 3. 7/4/2012 Redo Problem (using tables) •Let X denote the no. of transmissions with flaws(F). •X→B(10,.02)(a) want P[X>2] = 0.0008 (from AS: Appendix C)(a) P[X= 0] = 0.8171 43 3

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