Module 7 examples - Probabilities / Deriving the Binomial Distribution
A medical consultant books appointments to see 20 patients in the course of a
morning. Some of them cancel their appointments at little or no notice. From past
records the following probabilities have been calculated.
P(0 cancellations) = 32%
P(1 cancellation) = 29%
P(2 cancellations) = 22%
P(3 cancellations) = 11%
P(4 cancellations) = 3%
P(5 or more) = 3%
Total 100%
What is the probability that on a particular
morning there will be no more than one cancellation?
Mutually exclusive events The addition rule for
gives:
P(A or B or C or...) = P(A) + P(B) + P( C)...
P(no more than 1 cancellation)
= P(0 or 1
cancellation)
= P(0) + P(1)
= 32% + 29%
= 61%
What is the probability that
on two successive mornings there will be no cancellations?
Conditional Probability for Independent
events of an event under the condition that another event has occurred or
will occur.
P(A)= P(A/B)
The multiplication rule gives:
P( A and B and C) = P(A) x P(B) x P(C)
P(0
cancellations on day 1 and 0 on day 2) = 0.32 × 0.32
= 0.102 i.e. 10.2%
The number of ways of choosing three objects from
eight is:
Combination Probability
nCr
= n!
(r! x (n - r)!)
8C3 =
8!
5! × 3!
=
8 × 7 × 6× 5 x 4 x 3 x 2
x 1
(5 x 4
x 3 x 2 x 1) x 3 x 2 x 1
=
8 × 7 X 6
6
= 8 x 7 = 56
Deriving the Binomial Distribution
The population can be split into two types, those that have heard and those
that have not. A random sample of five is taken from this population. The
underlying distribution is therefore binomial with p = 0.4 and n = 5. The
binomial formula is:
P(r of type 1) = nCr × pr × (1 - p)n-r
Thus:
P(1 person has heard of the
chocolate bar) = 5C1 × 0.4 × 0.64
= 5 × 0.4 × 0.64
= 2 × 0.1296
= 0.26 (approx.)
Using Normal Distribution Table
Mean 250 SD 4
What % falls with
in the range 245 – 255?
255 – 250 = 5
5 / 4 = 1.25
lookup on normal distribution table
|
Z |
.05 |
|
1.2 |
.3944 |
= 39.4%
Approximate Binomial with the Normal
20% of Questionnaires responds received.
What is the probability the less than 50 when 300 questionnaires sent out?
1) Determine if both > 5 rule of thumb to use normal approximate
Mean = np = (300 x .20) = 60 > 5
-n(1-p) = 300(1-.20) = 300 x .80 = 240 > 5
2) Calculate the Deviation of the Distribution
= 6.92
3) Discrete distribution being approx is continuous and 50 adjusted to 49.5
49.5 – 60 = 10.5
10.5 / 6.92 = 1.5 standard deviation from the mean
lookup on normal distribution table
|
Z |
.00 |
|
1.5 |
.4332 |
4) One tailed
.5 - .4332 = .0668
Conclusion: fewer than 50 reply’s = 6.68%