Question

A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is known to have a normal distribution with a mean of 5 milliliters (ml) and a standard deviation of .4 ml. Answer the following questions based on this information. Find the dye amount that represents the 91st percentile of the distribution.

Answer 1

Answer:

The dye amount that represents the 91st percentile of the distribution is 5.536 ml.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 5 milliliters (ml) and a standard deviation of .4 ml.

This means that $\mu = 5, \sigma = 0.4$

Find the dye amount that represents the 91st percentile of the distribution.

This is X when Z has a p-value of 0.91, so X when Z = 1.34. So

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

$1.34 = \frac{X - 5}{0.4}$

$X - 5 = 0.4*1.34$

$X = 5.536$

The dye amount that represents the 91st percentile of the distribution is 5.536 ml.