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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)=√//exp-22²2 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

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)=√//exp-22²2 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

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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how has the z score been calculated in this, please don't use z table 

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)=√//exp-22²2 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)=√//exp-22²2 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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