Question

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

A consumer organisation 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.
a) Find the proportion of the chocolate bars produced that have a) mass within one standard
deviation of the mean.
​

Answer 1

The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.

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

`Z = (200.9 - 200.4)/(0.05)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

X = 199.9

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

`Z = (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%.