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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 1.00
The probability distribution should depict/provide all possible
Information regarding the random variable in question.
• f(x) = P[X=x]; 0 ≤ f(x) ≤ 1; Σ f(x) =1
x P[X=x]
Prob. dist. of no. of qualifying • CDF F(x) = P[X ≤x] = Σy ≤ x f(y)
enquiries in a day
0 0.20 • F is a step function with jumps only at the possible
1 0.35
0.40
values
probability
0.30
2 0.25
3 0.10 0.20
• Density to distribution function and vice versa
4 0.05 0.10
5 0.05 0.00
0 1 2 3 4 5
no. of queries 32
total 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
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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 them
successful 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)}
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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
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