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The amount of energy associated with a production of bottled water is approximately Normal with a mean of 8.7 million Joules and a standard deviation 0.5 million Joules. How much energy should be required for the bottom 80% of bottled water?

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Answer:

At most 9.12 joules should be required for the bottom 80% of bottled water

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:


\mu = 8.7, \sigma = 0.5

How much energy should be required for the bottom 80% of bottled water?

At most the 80th percentile.

The 80th percentile is X when Z has a pvalue of 0.8. So it is X when Z = 0.84.


Z = (X - \mu)/(\sigma)


0.84 = (X - 8.7)/(0.5)


X - 8.7 = 0.84*0.5


X = 9.12

At most 9.12 joules should be required for the bottom 80% of bottled water

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