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A company claims that the number of defective items manufactured during each run of making 100 of their products is independent of the number from other runs and that the proportion of defectives is not more than 3%. Assuming that the defective rate for each run is 3% Which of the following can be used to determine whether x = 8 is unusually a high number of defective items on the next run of 100 of their products?

A company claims that the number of defective items manufactured during each run of making 100 of their products is independent of the number from other runs and that the proportion of defectives is not more than 3%. Assuming that the defective rate for each run is 3% Which of the following can be used to determine whether x = 8 is unusually a high number of defective items on the next run of 100 of their products?

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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A company claims that the number of defective items manufactured during each run of making

100 of their products is independent of the number from other runs and that the proportion of

defectives is not more than 3%. Assuming that the defective rate for each run is 3%

Which of the following can be used to determine whether x = 8 is unusually a high number of

defective items on the next run of 100 of their products?

 

 

I. Yes. Because the meanu = 100 x 0.03 = 3, the standard deviation o = /npq = V100-0.03. 0.97 %3D - 1.7, and applying the range rule of thumb, x = 8 is over the upper limit: 3 + 2 x 1.7 = 6.4 II. Yes. Because the probability P(x = 8) = 100C8 x 0.03° x 0.972 - 0.00741 is very small III. Yes. Because the probability P(x 2 8) = 1 - P(x< 8) = 1- 0.9968 = 0.0032 is less than 0.05
Transcribed Image Text:I. Yes. Because the meanu = 100 x 0.03 = 3, the standard deviation o = /npq = V100-0.03. 0.97 %3D - 1.7, and applying the range rule of thumb, x = 8 is over the upper limit: 3 + 2 x 1.7 = 6.4 II. Yes. Because the probability P(x = 8) = 100C8 x 0.03° x 0.972 - 0.00741 is very small III. Yes. Because the probability P(x 2 8) = 1 - P(x< 8) = 1- 0.9968 = 0.0032 is less than 0.05
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