Question

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?

Answer 1

97.5%

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 > 0)

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

P(Z > -2) = 97.5%