A machine at a food-distribution factory fills boxes of rice. The distribution of the weights of filled boxes of rice has an
approximately Normal distribution, with a mean of 28.2 ounces and a standard deviation of 0.4 ounces. What
percentage of filled boxes of rice have a weight between 28 and 29 ounces?
Find the z-table here.
O 33.1%
O 36.7%
O 63.3%
66.9%

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by: [tex]\mu = 28.2, \sigma = 0.4[/tex].

The proportion of filled boxes of rice have a weight between 28 and 29 ounces is the p-value of Z when X = 29 subtracted by the p-value of Z when X = 28, hence:

X = 29:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{29 - 28.2}{0.4}[/tex]

Z = 2

Z = 2 has a p-value of 0.977.

X = 28:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{28 - 28.2}{0.4}[/tex]

Z = -0.5

Z = -0.5 has a p-value of 0.308.

0.977 - 0.308 = 0.669 = 66.9%.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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