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Course Documents

Chapter 1 - Intro

Chapter 2 - Methods for Describing Sets of Data

Chapter 3 - Probability

Chapter 4 - Discrete Random Variables

Chapter 5 - Normal Random Variables

Chapter 6 - Sampling Distributions

Chapter 7 - Confidence Intervals

Chapter 8 - Tests of Hypothesis: One Sample

Chapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples

Sample Exam I: Chapters 1 & 2

Sample Exam II: Chapters 3 & 4

Sample Exam III: Chapters 5 & 6

Sample Exam IV: Chapters 7 & 8

Ask the Professor Forum

Hi professor, would you mind doing question 61 from the mixed review? I can't seem to find my mistake. thank you !

Posted to STATS 1 on Tuesday, October 14, 2014   Replies: 1


Professor Mcguckian
10/14/2014
1:23 PM EST

Hi Reina,

Here is the question:

61.   In a production of phones 0.8% are defective. A sample of ten phones is randomly selected without replacement from a set of 20,000 and tested. What is the probability that at least one of those ten tested phones is defective?

To answer a probability of at least one question, we will set the problem using the formula P(at least one) = 1 - P(none).

P(at least one is defective) = 1 - P(none are defective)

                                                 = 1 - ___ * ___ * ___ * ___ * ___ * ___ * ___ * ___ * ___ * ___

Now, we have ten spaces because we are selecting ten phones. We are trying to find the probability that none of the ten are defective because if we subtract that probability from 1, we will find the probability of at least one defective phone. Each space above, represents a phone that is not defective. This means we need the probability that a phone is not defective.

The P(Defective) = 0.8% = 0.008

The P(not Defective) = 1 - P(Defective) = 1 - 0.008 = 0.992

So each of our ten spaces will be filled with 0.992. This will give us: P(at least one defective phone) =  1 - 0.992*0.992*...*0.992

= 1 - 0.992^10 = 0.0772

Hope that makes sense,

Professor McGuckian

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