According to the U.S. Department of Agriculture, the average American consumed 54.3 pounds (ap- proximately seven gallons) of salad and cook

Question

According to the U.S. Department of Agriculture, the average American consumed 54.3 pounds (ap- proximately seven gallons) of salad and cooking oils in 2008 (www.ers.usda.gov/data/foodconsumption). Suppose that the current distribution of salad and cooking oil consumption is approximately normally distributed with a mean of 54.3 pounds and a standard deviation of 14.5 pounds.
1. What percentage of Americans’ annual salad and cooking oil consumption is less than 10 pounds?
2. What percentage of Americans’ annual salad and cooking oil consumption is between 35 and 60?
3. What percentage of Americans’ annual salad and cooking oil consumption is more than 95 pounds?

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Amity 3 years 2021-08-03T20:04:01+00:00 1 Answers 329 views 0

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  1. King
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    2021-08-03T20:05:19+00:00
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    Answer:

    a) 0.11%

    b) 55.99%

    c) 0.25%

    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.

    Normally distributed with a mean of 54.3 pounds and a standard deviation of 14.5 pounds.

    This means that \mu = 54.3, \sigma = 14.5

    1. What percentage of Americans’ annual salad and cooking oil consumption is less than 10 pounds?

    The proportion is the pvalue of Z when X = 10. So

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

    Z = \frac{10 - 54.3}{14.5}

    Z = -3.06

    Z = -3.06 has a pvalue of 0.0011

    0.0011*100% = 0.11%.

    2. What percentage of Americans’ annual salad and cooking oil consumption is between 35 and 60?

    The proportion is the value of Z when X = 60 subtracted by the pvalue of Z when X = 35.

    X = 60

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

    Z = \frac{60 - 54.3}{14.5}

    Z = 0.39

    Z = 0.39 has a pvalue of 0.6517

    X = 35

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

    Z = \frac{35 - 54.3}{14.5}

    Z = -1.33

    Z = -1.33 has a pvalue of 0.0918

    0.6517 – 0.0918 = 0.5599

    0.5599*100% = 55.99%

    3. What percentage of Americans’ annual salad and cooking oil consumption is more than 95 pounds?

    The proportion is 1 subtracted by the pvalue of Z when X = 95.

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

    Z = \frac{95 - 54.3}{14.5}

    Z = 2.81

    Z = 2.81 has a pvalue of 0.9975

    1 – 0.9975 = 0.0025

    0.0025*100% = 0.25%

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