Hi,
mean of 8 yrs and a standard deviation of 2 years.
(i)Using TI: The syntax is normalcdf(smaller, larger, µ, σ).
Note: The -9999 is used as the smaller value to be at least 5 standard deviations from the mean.
P(x > 10) = 1 - normalcdf(-9999,10, 8,2)
(ii)
P(x < 4) = normalcdf(-9999,4, 8,2)
0r
P(P(x > 10) = 1 - NORMSDIST(1) = 1 - .8413 = .1587
P(x < 4) = = NORMSDIST(-2) = .0028 .28% will have to be replaced