A Caribbean resort has a nightly limbo contest on the beach, Participants must be less than 64 inches tall. The distribution of heights of adult American men is approximately normal with mean 69 inches and standard deviation 2.5 inches. About what percent of adult American males could enter this contest?

2.23%

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = < 64 inches

μ is the population mean = 69

σ is the population standard deviation = 2.5

Since: Participants must be less than 64 inches tall.

For x < 64 inches

z = 64 - 69/2.5

z = -2

Probability value from Z-Table:

P(x<64) = 0.02275

Converting to percentage

= 0.02275 × 100

= 2.275%

Approximately = 2.23%

Because participants must be less than 64 inches tall to participate in this limbo contest, the percent of adult American males thag could enter this contest is 2.23%