The probability of obtaining a defective 10-year old widget is 66.6%. 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 .666 Correct 2 3 4 5 6 or greater

For k = 1:

=NEGBINOMDIST(0, 1, 0.666) = 0.6660

For k = 2:

=NEGBINOMDIST(1, 1, 0.666) = 0.2224

For k = 3:

=NEGBINOMDIST(2, 1, 0.666) = 0.0743

For k = 4:

=NEGBINOMDIST(3, 1, 0.666) = 0.0248

For k = 5:

=NEGBINOMDIST(4, 1, 0.666) = 0.0083

For k = 6:

=NEGBINOMDIST(5, 1, 0.666) = 0.0028

Step-by-step explanation:

The probability of obtaining a defective 10-year old widget is 66.6%

p = 66.6% = 0.666

The probability of obtaining a non-defective 10-year old widget is

q = 1 - 0.666 = 0.334

The random variable will be the number of items that must be tested before finding the first defective 10-year old widget.

The geometric distribution is given by

Solving manually:

For k = 1:

For k = 2:

For k = 3:

For k = 4:

For k = 5:

For k = 6:

Using Excel function:

NEGBINOMDIST(number_f, number_s, probability_s)

Where

number_f = k - 1 failures

number_s = no. of successes

probability_s = the probability of success

For the geometric distribution, let number_s = 1

For k = 1:

=NEGBINOMDIST(0, 1, 0.666) = 0.6660

For k = 2:

=NEGBINOMDIST(1, 1, 0.666) = 0.2224

For k = 3:

=NEGBINOMDIST(2, 1, 0.666) = 0.0743

For k = 4:

=NEGBINOMDIST(3, 1, 0.666) = 0.0248

For k = 5:

=NEGBINOMDIST(4, 1, 0.666) = 0.0083

For k = 6:

=NEGBINOMDIST(5, 1, 0.666) = 0.0028

As you can notice solving manually and using Excel yields the same results.

113.1

step-by-step explanation:

v=4

3πr3=4

3·π·33≈113.09734