Question

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. 33.1% 36.7% 63.3% 66.9%

Explanation:

Step1: Calculate the z - score for \(x = 28\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 28.2\) (mean) and \(\sigma=0.4\) (standard deviation).
For \(x = 28\):
\(z_1=\frac{28 - 28.2}{0.4}=\frac{- 0.2}{0.4}=-0.5\)

Step2: Calculate the z - score for \(x = 29\)

For \(x = 29\):
\(z_2=\frac{29 - 28.2}{0.4}=\frac{0.8}{0.4}=2\)

Step3: Find the probabilities from the z - table

From the standard normal table (\(z\) - table):
The probability \(P(Z\lt - 0.5)\) is \(0.3085\)
The probability \(P(Z\lt2)\) is \(0.9772\)

Step4: Calculate the probability \(P(-0.5\lt Z\lt2)\)

Using the formula \(P(a\lt Z\lt b)=P(Z\lt b)-P(Z\lt a)\)
\(P(-0.5\lt Z\lt2)=P(Z\lt2)-P(Z\lt - 0.5)\)
\(P(-0.5\lt Z\lt2)=0.9772 - 0.3085=0.6687\approx0.669\)

Answer:

\(66.9\%\) (Option D)