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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 5, \sigma = 0.4[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.34 = \frac{X - 5}{0.4}[/tex]
[tex]X - 5 = 0.4*1.34[/tex]
[tex]X = 5.536[/tex]
The dye amount that represents the 91st percentile of the distribution is 5.536 ml.
