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the average pieces of candy eaten by children in your neighborhood on halloween has a normal distribution with a mean of 14 and a standard deviation of 5. what percentage eat less than 10 pieces of candy on halloween

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Answer:

21.19% eat less than 10 pieces of candy on halloween

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 = 14, \sigma = 5

What percentage eat less than 10 pieces of candy on halloween

This is the pvalue of Z when X = 10.


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


Z = (10 - 14)/(5)


Z = -0.8


Z = -0.8 has a pvalue of 0.2119

21.19% eat less than 10 pieces of candy on halloween

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