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.

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 and standard deviation , the z-score of a measure X is given by:

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

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

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

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step-by-step explanation:

95.03 is the answer. hope this !