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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 200.4, \sigma = 0.05[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{200.9 - 200.4}{0.05}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
X = 199.9
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{199.9 - 200.4}{0.05}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] 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%.
