Question 857016
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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 &#8804; x &#8804; 15) = NORMSDIST(6.5/2.4) - NORMSDIST(-1.5/2.4)

TI function normalcdf(smaller, larger, µ, &#963;) can also be used