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

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.

in progress 0
Thành Đạt 5 years 2021-08-25T00:58:39+00:00 1 Answers 290 views 1

Answers ( )

  1. Jezebel
    0
    2021-08-25T00:59:53+00:00
    Reply

    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.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )