Question

# The probability of obtaining a defective 10-year old widget is 68.9%. For our purposes, the random...

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

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

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