Question

a fruit company ships boxes of oranges. the weights of the boxes follow an approximately normal distribution. the mean weight of a box is 85 pounds, and the standard deviation is 4 pounds. using the 68 – 95 – 99.7 rule, what percentage of the boxes weigh at least 81 pounds? enter the answer in the box.

Explanation:

Step1: Identify the mean and standard deviation

The mean ($\mu$) is 85 pounds and the standard deviation ($\sigma$) is 4 pounds. We need to find how many standard deviations 81 is from the mean. Calculate the difference: $85 - 81 = 4$, so 81 is $\mu - \sigma$ (since $85 - 4 = 81$).

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule (empirical rule) states that for a normal distribution:

• Approximately 68% of the data lies within $\mu \pm \sigma$.
• Approximately 95% lies within $\mu \pm 2\sigma$.
• Approximately 99.7% lies within $\mu \pm 3\sigma$.

First, find the percentage of data below $\mu - \sigma$. Since the total area under the normal curve is 100%, and the data is symmetric around the mean, the percentage of data below $\mu - \sigma$ is $\frac{100\% - 68\%}{2} = 16\%$.

Step3: Find the percentage of data at least 81 pounds

"At least 81 pounds" means the percentage of data greater than or equal to 81. Since the percentage below 81 (which is $\mu - \sigma$) is 16%, the percentage at least 81 is $100\% - 16\% = 84\%$.

Answer:

84