Hi,

mean of 9.5 minutes and a standard deviation of 2.4 minutes.

a) P(x< 10min)  Yes, z = .2083  P = NORMSDIST(.2083) = .5825

b) P(x> 5min) = 1 - NORMSDIST(-4.5/2.4)

c) P(8 ≤ x ≤ 15) = NORMSDIST(6.5/2.4) - NORMSDIST(-1.5/2.4)



TI function normalcdf(smaller, larger, µ, σ) can also be used 

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