A chocolate bar is produced with an advertised mass of 200 g. A consumer organisation finds that the mass of the chocolate bar is norma

Question

A chocolate bar is produced with an advertised mass of 200 g.
A consumer organisation finds that the mass of the chocolate bar is normally distributed with a mean of
200.4 g and a standard deviation of 0.05 g.
a) Find the proportion of the chocolate bars produced that have a) mass within one standard
deviation of the mean.
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Eirian 4 years 2021-09-04T20:27:15+00:00 1 Answers 35 views 0

Answers ( )

  1. mocmien2
    0
    2021-09-04T20:28:18+00:00
    Reply

    Answer:

    The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.

    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 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.

    A consumer organization finds that the mass of the chocolate bar is normally distributed with a mean of 200.4 g and a standard deviation of 0.05 g.

    This means that \mu = 200.4, \sigma = 0.05

    a) Find the proportion of the chocolate bars produced that have mass within one standard deviation of the mean.

    pvalue of Z when X = 200.4 + 0.05 = 200.9 subtracted by the pvalue of Z when X = 200.4 – 0.05 = 199.9.

    X = 200.9

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

    Z = \frac{200.9 - 200.4}{0.05}

    Z = 1

    Z = 1 has a pvalue of 0.8413

    X = 199.9

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

    Z = \frac{199.9 - 200.4}{0.05}

    Z = -1

    Z = -1 has a pvalue of 0.1587

    0.8413 – 0.1587 = 0.6826

    The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.

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