Question

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?

Answer 1

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$0.84 = \frac{X - 8.7}{0.5}$

$X - 8.7 = 0.84*0.5$

$X = 9.12$

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