Question

During a manufacturing process, 15 units are randomly selected each day

from the production line to check the percent defective. From historical

information it is known that the probability of a defective unit is 0.05.

Any time 2 or more defectives are found in the sample of 15, the process

is stopped. This procedure is used to provide a signal in case the

probability of a defective has increased.

• What is the probability that on any given day the production process will

be stopped? (Assume 5% defective.)

Answer 1

Let x = number of defective items in in the sample of of 15 units.

P= probability of success in each trial = 0.05

Since it follows the probability of exactly x successes on n repeated trials, and X can only have two outcomes x has binomial distribution with n= 15 and p =0.05

To find: The production will be stopped if there are two or more defective units, so we have to find P(x$\geq$ 2)

$P(x\geq 2)$ =1-P(X<2)

=1- P(x$\leq$ 1)

=1-0.8290(using binomial distribution table for n=15, x=1, p= 0.05

=0.1709

There is a 17.09% probability that on any given day the production process will be stopped