Question 1051295
On the average, members at a local fitness center work out for 90 minutes with a standard deviation of 15 minutes. The distribution is normal. 

a) What percentage of the fitness club members work out for 45 minutes or less?
z(45) = (45-90)/15 = -45/15 = -3
P(x <= 45) = P(z <= -3) = normalcdf(-100,-3) = 0.0013
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b) What percentage of the fitness club members work out for 2 hours and 15 minutes or more?
z(135) = (135-90)/15 = 45/15 = 3
P(x >= 135 min) = P(z >= 3) = 0.0013
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c) 68% of the fitness club members work out between which two time intervals? 
(Mean - std) and (Mean + std)
(90 - 15) and (90 + 15)
75 and 105 

d) We can say that 99.7% of the fitness club members work out for no more than ____________ minutes. 

(Mean - 3 × std) and (Mean + 3 × std) 
(90 - 3 × 15) and (90 + 3 × 15)
45 and 135 
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Cheers,
Stan H.
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