Question

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

Answer 1

Answer:

21.19% eat less than 10 pieces of candy on halloween

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:

$\mu = 14, \sigma = 5$

What percentage eat less than 10 pieces of candy on halloween

This is the pvalue of Z when X = 10.

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

$Z = \frac{10 - 14}{5}$

$Z = -0.8$

$Z = -0.8$ has a pvalue of 0.2119

21.19% eat less than 10 pieces of candy on halloween