Question

The probability of obtaining a defective 10-year old widget is 73.2%. For our purposes, the random variable will be the number of items that must be tested before finding the first defective 10-year old widget. Thus, this procedure yields a geometric distribution.
Use some form of technology like Excel or StatDisk to find the probability distribution.
(Report answers accurate to 4 decimal places.)
k P(X = k)
1 Correct
2
3
4
5
6 or greater

Answer 1

Answer:

P(X=0)=0.732

P(X=1)=0.1962

P(X=2)=0.0526

P(X=3)=0.0141

P(X=4)=0.0038

P(X=5)=0.001 0

P(X≥6)=0.0004

Step-by-step explanation:

We want to find the probability distribution of X, where X is the number of items that must be tested before finding the first defective 10-year old widget.

This means that if X=2, the first two items are not defective and the third item is defective.

The probability of finding a defective item is p=0.732 for each trial.

Then, the probability for X=k can be written as:

$P(X=k)=(1-p)^k\cdot p=(1-0.732)^k\cdot0.732=0.278^k\cdot 0.732$

Then, we can write:

$P(X=0)=0.268^0\cdot 0.732=0.732\\\\P(X=1)=0.268^1\cdot 0.732=0.1962\\\\P(X=2)=0.268^2\cdot 0.732=0.0526\\\\P(X=3)=0.268^3\cdot 0.732=0.0141\\\\P(X=4)=0.268^4\cdot 0.732=0.0038\\\\P(X=5)=0.268^5\cdot 0.732=0.001\\\\P(X\geq6)=0.0004$