Question

Weights of American adults are normally distributed with a mean of 180 pounds and a standard deviation of 8 pounds. What is the probability that a randomly selected individual will be between 185 and 190 pounds?

Answer 1

There is 16.2% probability that a randomly selected individual will be between 185 and 190 pounds

Explanation:

Z score

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (x - μ) / σ

where μ is the mean, x = raw score and σ is the standard deviation.

Given μ = 180, σ = 8.

For x = 185:

z = (185 - 180)/8 = 0.625

For x = 190:

z = (190 - 180)/8 = 1.25

P(185 < x < 190) = P(0.625 < z < 1.25) = P(z < 1.25) - P(z < 0.625) = 0.8944 - 0.7324 = 16.2%

There is 16.2% probability that a randomly selected individual will be between 185 and 190 pounds