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The lifetime of a certain make of car battery is found to be normally distributed with a mean of 1250 days and a standard deviation of 175 days. How many months warranty could the manufacturer give so that no more than 6% need replacement. (Take 1 month = 30 days) -3 6% -2 d₁ - 1250 175 44% -1 0 f(x)=√exp = -1.555 50% 1 2 3 f(z) dz = 0.44 gives If d1 is the number of days by which 6% of the batteries fail, (and therefore need replacing under guarantee), then : z = 1.555 and d₁ = 977.9 days or 32.6 months

The lifetime of a certain make of car battery is found to be normally distributed with a mean of 1250 days and a standard deviation of 175 days. How many months warranty could the manufacturer give so that no more than 6% need replacement. (Take 1 month = 30 days) -3 6% -2 d₁ - 1250 175 44% -1 0 f(x)=√exp = -1.555 50% 1 2 3 f(z) dz = 0.44 gives If d1 is the number of days by which 6% of the batteries fail, (and therefore need replacing under guarantee), then : z = 1.555 and d₁ = 977.9 days or 32.6 months

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Tutorial Q1.1 The lifetime of a certain make of car battery is found to be normally distributed with a mean of 1250 days and a standard deviation of 175 days. How many months warranty could the manufacturer give so that no more than 6% need replacement. (Take 1 month = 30 days) -3 6% -2 d₁ - 1250 175 44% -1 0 f(z)=√2/exp=2²2 50% 1 2 3 0 f(z) dz = 0.44 gives z = 1.555 If d1 is the number of days by which 6% of the batteries fail, (and therefore need replacing under guarantee), then : = = -1.555 and d₁ = 977.9 days or 32.6 months
Transcribed Image Text:Tutorial Q1.1 The lifetime of a certain make of car battery is found to be normally distributed with a mean of 1250 days and a standard deviation of 175 days. How many months warranty could the manufacturer give so that no more than 6% need replacement. (Take 1 month = 30 days) -3 6% -2 d₁ - 1250 175 44% -1 0 f(z)=√2/exp=2²2 50% 1 2 3 0 f(z) dz = 0.44 gives z = 1.555 If d1 is the number of days by which 6% of the batteries fail, (and therefore need replacing under guarantee), then : = = -1.555 and d₁ = 977.9 days or 32.6 months
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