Question

When a new machine is functioning properly, only 3% of the items produced are defective. Assume that we will randomly select ten parts produced on the machine and that we are interested in the number of defective parts found. Compute the probability associated with:

a. [4 points] No defective parts

b. [6 points] At least 1 defective parts.

2. [12 points] Over 500 million tweets are sent per day (Digital Marketing Ramblings website, December 15, 2014). Bob receives on average 9 tweets during his lunch hour. What is the probability that Bob receives no tweets during the first 15 minutes of his lunch hour?

Answer 1

Solution:

Question 1)

Given:

p = probability of the items produced are defective = 3% = 0.03

q =probability of the items produced are non-defective = 1 - p = 1 - 0.03 = 0.97

n =Number of parts produced on the machine are selected = 10

Part a) Find:

P( No defective parts ) = P( All 10 part are non-defective parts)

P( No defective parts ) = 0.97¹⁰

P( No defective parts ) = 0.737424

P( No defective parts ) = 0.7374

Part b) Find:

P( At least 1 defective parts ) =..........?

P( At least 1 defective parts ) = 1 -P( No defective parts )

P( At least 1 defective parts ) = 1 - 0.737424

P( At least 1 defective parts ) = 0.262576

P( At least 1 defective parts ) = 0.2626

.

.

Question 2)

Given: Bob receives on average 9 tweets during his lunch hour.

X = Bob receives number of tweets during the his lunch hour follows a Poisson distribution with mean = per hour.

We have to find:

P( Bob receives no tweets during the first 15 minutes of his lunch hour) =................?

P( X = 0 ) =..............?

We have per hour, so for 15 minutes , mean =

Thus probability mass function is: