The document summarizes key concepts relating to probability distributions, including discrete and continuous distributions. It provides examples of binomial probability distributions, such as the distribution of the number of inquiries leading to business proposals on a given day. It also discusses the binomial distribution in the context of the number of successful days in a working week. Key characteristics of binomial distributions such as the mean, variance, and standard deviation are defined. Finally, the document solves an example problem using binomial distribution tables.

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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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