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 6 milliliters (ml) and a standard deviation of 0.4 ml. Find the dye amount that represents the 90th percentile (i.e. 90%) of the distribution.
a. 4.464 ml.
b. 4.836 ml.
c. 4.936 ml.
d. 4.964 ml.
e. 5.536 ml.

Answer 1

Answer:

The 90th percentile of the distribution is 6.512 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 6 milliliters (ml) and a standard deviation of 0.4 ml.

This means that $\mu = 6, \sigma = 0.4$

Find the dye amount that represents the 90th percentile (i.e. 90%) of the distribution.

This is X when Z has a p-value of 0.9, so X when Z = 1.28. Then

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

$1.28 = \frac{X - 6}{0.4}$

$X - 6 = 1.28*0.4$

$X = 6.512$

The 90th percentile of the distribution is 6.512 ml.