Ask Question
26 February, 23:07

The weight of adults in a certain state follows an approximately normal distribution with mean 150 pounds and standard deviation 17 pounds.

According to the empirical rule, what percent of adults weigh more than 116 pounds?

+2
Answers (1)
  1. 26 February, 23:35
    0
    97.5%

    Step-by-step explanation:

    Find the z-score:

    z = (x - μ) / σ

    z = (116 - 150) / 17

    z = - 2

    According to the empirical rule, 95% of a population are between - 2 and + 2 standard deviations. That means that half of that, or 47.5%, are between - 2 and 0 standard deviations. Since 50% are greater than 0 standard deviations, the total probability is 47.5% + 50%, or 97.5%.

    P (Z > - 2) = P (-2 < Z 0)

    P (Z > - 2) = 47.5% + 50%

    P (Z > - 2) = 97.5%
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “The weight of adults in a certain state follows an approximately normal distribution with mean 150 pounds and standard deviation 17 pounds. ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers