Question 918930
mean of 4 hours and standard deviation 1.25 hours .
a) {{{z=blue (x - mu)/blue(sigma)}}} z(5) = 1/1.25 = .8, 
P(z > .8) = normalcdf(.8,100) = .2119 0r 21.19%
.........
b) n = 4,  {{{z =blue (x - mu)/blue(sigma/sqrt(n))}}} 
z(4.95) =  -.05/.625 = -.08  and z(5.05) = .08
P(x=5) = P(z < .08) - P(z < .08)
P(x=5) = 53.19 - 46.81 = 6.38%
....
c) select 4
P(all four play > 5hr) = (.2119)^4