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 

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