Question

A chocolate bar is produced with an advertised mass of 200 g.

Answer 1

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%.