Question

the weight of oranges growing in an orchard is normally distributed with a mean weight of 4.5 oz. and a standard deviation of 0.5 oz. using the empirical rule, what percentage of the oranges from the orchard weigh between 3 oz. and 6 oz.?

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value. For $x = 3$, $z_1=\frac{3 - 4.5}{0.5}=\frac{- 1.5}{0.5}=-3$. For $x = 6$, $z_2=\frac{6 - 4.5}{0.5}=\frac{1.5}{0.5}=3$.

Step2: Apply the empirical rule

The empirical rule for a normal distribution states that approximately 99.7% of the data lies within $z=-3$ and $z = 3$.

Answer:

99.7%