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A manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs of solvent. Suppose the amount of solvent in each drum is normally distributed with a mean of 101.3 pounds and a standard deviation of 3.68 pounds. a) What is the probability that a drum meets the guarantee? Give your answer to four decimal places.

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Answer: 0.6381

Explanation:

Given : A manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs of solvent.


\mu=101.3\text{ pounds}\ \ \sigma=3.68\text{ pounds}

Let x be a random variable that represents e the amount of solvent in each drum.

Since ,
z=(x-\mu)/(\sigma)

z-score corresponds x = 100 ,
z=(100-101.3)/(3.68)\approx-0.3533

Required probability :


\text{P-value }: P(x\leq100)=P(z\leq-0.3533)\\\\=1-P(z<-0.3533)\\\\1-(1-P(z<0.3533))\\\\=P(z<0.3533)=0.638068\approx0.6381

[using z-value table.]

Hence, the probability that a drum meets the guarantee = 0.6381

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