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The mean weight of a brand of cereal is 0.297 kg with a standard deviation of .024kg. Assuming a normal distribution, find the percentage of data that falls below 0.274kg.

User Ashg
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3 votes

Answer:

16.85% of data that falls below 0.274kg.

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 = (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 = 0.297, \sigma = 0.024

Assuming a normal distribution, find the percentage of data that falls below 0.274kg.

This is the pvalue of Z when X = 0.274. So


Z = (X - \mu)/(\sigma)


Z = (0.274 - 0.297)/(0.024)


Z = -0.96


Z = -0.96 has a pvalue of 0.1685.

16.85% of data that falls below 0.274kg.

User Jtlowe
by
8.2k points
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