Lesson 6: Binomial Distribution Download

Transcript
STARTER
A manufacturing procedure is made up to two
separate processes. The time to complete process A
has mean of 30 secs and standard deviation of 3
secs. The time to complete process B has mean of
25 secs and standard deviation of 6 secs. To
produce a required component (P) the manufacturer
must undertake process A twice and process B
three times. Assume process A is independent of
process B.
Find an equation linking A, B and P P = 2A + 3B
Find the mean and standard deviation of P
E(P) = 2 x 30 + 3 x 25 = 135
var(P) = 22 x 32 + 32 x 62 = 360
SD(P) = √360 = 18.97
Note 7: Binomial Distribution
Any experiment in which there are only two
outcomes (success or failure) is called a
Bernoulli trial.
If the process is repeated n times, and the
random variable, X is the total number of
successes, then X has a binomial distribution.
The conditions for X to have a binomial
distribution are:
There
are a fixed number of identical trials
There are only two possible outcomes for each trial
Probability of success at each trial must be constant
Each
trial is independent of the other trials
There are two parameters n and π.
The random variable X is the total number
of successes in n trials
 π is the probability of success at an
individual trial
The probability formula is:
Example: A duck shooter has a probability
of 0.4 of hitting any duck that he shoots at
during a hunt. In a hunt he fires 10 shots.
Explain why the duck shooter can be modelled
by a binomial distribution:
There
are a fixed number of identical trials
There are 10 shots
There are only two possible outcomes for
each trial
Each shot – hit or miss
Probability of success at each trial must
constant
Probability of hitting a duck remains
constant at 0.4
Each trial is independent of the other trials
Each shot is independent of the other
Find the probability that he shoots exactly 8
ducks
STAT
DIST
BINM
x
8
Numtrial 10
p
0.4
Execute
P(X = 8)
= 0.01062
Bpd
Var
Sigma
Page 67
Exercise 5.1