In a study of facial behavior, people in a control group are timed for eye contact in a 5-minute period. their times are normally distributed with a mean of 184 and standard deviation of 55. find the probability that the eye contact time is greater than 230.0 which is the mean for paranoid

The probability is 0.2005.

We compute the z-score for this time:
z = (X-μ)/σ = (230-184)/55 = 0.84

Looking this up in a z-table (http://www.z-table.com) we see that the area to the left of this (less than this) is 0.7995.  We want the area greater than this, so we subtract this from 1:
1 - 0.7995 = 0.2005.

Answer is negative 2

Idon't know and i'm a teacher