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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?

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.CR: Chapter 13 Review
Problem 31CR
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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 mean u = 100 x 0.03 = 3, the standard deviation o = /npg = V100-0.03. 0.97 %D - 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 %3D (A) I only (B) || only (C) I and II (D) All can be used
Transcribed Image Text:I. Yes. Because the mean u = 100 x 0.03 = 3, the standard deviation o = /npg = V100-0.03. 0.97 %D - 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 %3D (A) I only (B) || only (C) I and II (D) All can be used
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