time spent waiting in the line is approximately normall distrubuted the mean is 7 min and the variance is 1 min find the probability that a person will wait more than 5 min round to 4 decimal places
Mean =7 min
SD = 1 min
z = (5-7)/1 = -2
P ( z > -2) = 0.97725 = 0.9773 = 97.73%
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