use the information to answer the question.\na fruit company ships boxes of oranges. the weights of the…
use the information to answer the question.\na 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.\nusing the 68 - 95 - 99.7 rule, what percentage of the boxes weigh at least 81 pounds? enter the answer in the box.\n%
Answer
Explanation:
Step1: Calculate the z - score
The z - score formula 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 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 to the left of $z = - 1$ is $(100 - 68)\div2=16%$. The percentage of data to the right of $z=-1$ (boxes weighing at least 81 pounds) is $100 - 16=84%$.
Answer:
84