Question 1055026
mean of 3.7 minutes and a standard deviation of 1.4 minutes
{{{z =blue (x - mu)/blue(sigma)}}}
0r
{{{blue(sigma)*z + mu= blue (x)}}} 
the 75th percentile for waiting times at this bank: z  = invNorm(.75) = .6745
{{{blue(1.4)*.6745 + 3.7= blue (4.6436)}}} x = 4.7  always round Up for these
|
P(x > 6) = P({{{z =blue (6.0 - 3.7)/blue(1.4)}}} )
Find z  and then find P(z > value found) = normalcdf(value found, 100) 100 a placeholder
z = 1.6429
P(z > 1.6429) = normalcdf(1.6429, 100) = .05  0r 5%