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b) Explanation: Let we should replace the bulb in X hours, if we want not more than 2% of the bulbs. P(X≤x) = 2% P(X-" < = 0.02 P(Z < = 0.02 X-500 50 Using MS Excel: X-500 50 X-500 50 =NORM.S.INV(0.02) NORM.S.INV(probability) -2.054 P(Z < -2.054) = = - 2.054 X = 397.3 0.02 Hence, We should replace the bulb within 397.3 hours

b) Explanation: Let we should replace the bulb in X hours, if we want not more than 2% of the bulbs. P(X≤x) = 2% P(X-" < = 0.02 P(Z < = 0.02 X-500 50 Using MS Excel: X-500 50 X-500 50 =NORM.S.INV(0.02) NORM.S.INV(probability) -2.054 P(Z < -2.054) = = - 2.054 X = 397.3 0.02 Hence, We should replace the bulb within 397.3 hours

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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I asked that you not skip steps. It is hard to follow along when you do that. How did you get 2.054 from a Z of 2%. Show the work. Please don’t skip steps.
6:23 ← Step2 b) Explanation: Let we should replace the bulb in X hours, if we want not more than 2% of the bulbs. P(X ≤ x) = 2% P(X=" P(Z < Using MS Excel: X-500 50 X-500 X-500 50 50 Step3 ⁹⁰0) = 0.02 P(Z ≤ – 2.054) =NORM.S.INV(0.02) NORM.S.INV(probability) = 0.02 -2.054 X = 397.3 = -2.054 0.02 Hence, We should replace the bulb within 397.3 hours ? √x 8
Transcribed Image Text:6:23 ← Step2 b) Explanation: Let we should replace the bulb in X hours, if we want not more than 2% of the bulbs. P(X ≤ x) = 2% P(X=" P(Z < Using MS Excel: X-500 50 X-500 X-500 50 50 Step3 ⁹⁰0) = 0.02 P(Z ≤ – 2.054) =NORM.S.INV(0.02) NORM.S.INV(probability) = 0.02 -2.054 X = 397.3 = -2.054 0.02 Hence, We should replace the bulb within 397.3 hours ? √x 8
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