A soft drink machine outputs a mean of 2323 ounces per cup. The machine’s output is normally distributed with a standard deviation of 22 oun

Question

A soft drink machine outputs a mean of 2323 ounces per cup. The machine’s output is normally distributed with a standard deviation of 22 ounces. You have been asked to calculate the probability of putting less than 2424 ounces in a cup. Using the normal distribution tables or appropriate technology, find the area under the standard normal curve.

in progress 0
Kaylee 2 days 2021-09-13T17:10:02+00:00 1 Answer 0

Answers ( )

    0
    2021-09-13T17:11:44+00:00

    Answer:

    Area under the normal curve: 0.6915.

    69.15% probability of putting less than 24 ounces in a cup.

    Step-by-step explanation:

    Problems of normally distributed samples are solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

    In this problem, we have that:

    \mu = 23, \sigma = 2

    You have been asked to calculate the probability of putting less than 24 ounces in a cup.

    pvalue of Z when X = 24. So

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

    Z = \frac{24 - 23}{2}

    Z = 0.5

    Z = 0.5 has a pvalue of 0.6915

    Area under the normal curve: 0.6915.

    69.15% probability of putting less than 24 ounces in a cup.

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )