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: Calculate the z - score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x = 81$, $\mu=85$, and $\sigma = 4$. So $z=\frac{81 - 85}{4}=\frac{- 4}{4}=-1$.

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule for a normal distribution states that about 68% of the data lies within 1 standard - deviation of the mean ($\mu\pm\sigma$), 95% within 2 standard - deviations ($\mu\pm2\sigma$), and 99.7% within 3 standard - deviations ($\mu\pm3\sigma$). The area within 1 standard - deviation of the mean ($z=-1$ to $z = 1$) is 68%. The area to the left of $z=-1$ is $\frac{100 - 68}{2}=16\%$.

Step3: Find the percentage of boxes weighing at least 81 pounds

The percentage of boxes weighing at least 81 pounds is $100\%-16\% = 84\%$.

Answer:

84