Question

The probability of obtaining a defective 10-year old widget is 68.9%. 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
2
3
4
5
6 or greater

Answer 1

Solution:

For the given geometric distribution, we are given

p = 0.689

q = 1 – p = 1 – 0.689 = 0.311

P(X=k) = p*q^(k – 1)

P(X=1) = 0.689*0.311^(6 - 1) = 0.00200457

P(X=2) = 0.689*0.311^(6 - 2) = 0.006445562

P(X=3) = 0.689*0.311^(6 - 3) = 0.020725279

P(X=4) = 0.689*0.311^(6 - 4) = 0.066640769

P(X=5) = 0.689*0.311^(6 - 5) = 0.214279

P(X≤6) = 1 – P(X≤5) = 1 – (0.00200457 + 0.006445562 + 0.020725279 + 0.066640769 + 0.214279)

P(X≤6) = 1 - 0.31009518

P(X≤6) = 0.68990482

The table is given as below:

k	P(X=k)
1	0.0020
2	0.0064
3	0.0207
4	0.0667
5	0.2143
6 or more	0.6899
Total	1.0000