STARTER P = 2A + 3B E(P) = 2 x x 25 = 135

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Presentation transcript:

STARTER P = 2A + 3B E(P) = 2 x 30 + 3 x 25 = 135 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 Find the mean and standard deviation of P P = 2A + 3B 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 Bpd Var x 8 Numtrial 10 p 0.4 Execute P(X = 8) = 0.01062

Sigma Page 67 Exercise 5.1