Question

A psychologist regularly runs rats through a maze and records the amount of time it takes them to finish. One hundred rats were run through a maze and they averaged 20 seconds with a standard deviation of 5 seconds. Assume that run-times follow a mound-shaped distribution. What can you say about the percentage of rats that took less than 10 seconds to finish?

Answer 1

Explanation:

• Given that μ = 20seconds and SD-standard deviation = 5seconds

• To get the percentage of rats that took less than 10 seconds to finish is;

• P( X < 10) = P( z < x -μ/SD)

• = P( z< -2.00) = 0.028 = 2.28%

• Hence, 2.28% of the rats took less than 10seconds to finish