Question The weight of oranges growing in an orchard is normally distributed with a mean weight of $8 \mathrm{oz}$. and a standard deviation of $1.5 \mathrm{oz}$. Using the empirical rule, determine what interval would represent weights of the middle $99.7 \%$ of all oranges from this orchard. Answer Artempt 1 out of 2
Solution
Step 1 ：Given that the weight of oranges growing in an orchard is normally distributed with a mean weight of 8 oz. and a standard deviation of 1.5 oz.
Step 2 ：Using the empirical rule, also known as the 68-95-99.7 rule, we know that for a normal distribution, almost all values lie within 3 standard deviations of the mean.
Step 3 ：More specifically, 68% of the data falls within the first standard deviation from the mean, 95% fall within two standard deviations, and 99.7% fall within three standard deviations.
Step 4 ：Therefore, to find the interval that represents the middle 99.7% of all weights, we need to calculate the mean minus 3 standard deviations and the mean plus 3 standard deviations.
Step 5 ：Let's denote the mean weight as \(\mu\) and the standard deviation as \(\sigma\). So, \(\mu = 8\) oz. and \(\sigma = 1.5\) oz.
Step 6 ：The lower bound of the interval is \(\mu - 3\sigma = 8 - 3 \times 1.5 = 3.5\) oz.
Step 7 ：The upper bound of the interval is \(\mu + 3\sigma = 8 + 3 \times 1.5 = 12.5\) oz.
Step 8 ：Final Answer: The interval that represents the middle 99.7% of all weights of the oranges from this orchard is \(\boxed{[3.5, 12.5]}\) oz.
