SOLUTION: Can someone show me how to work these type of math questions? Please, check the ones I tried to do. Thank you for your help.
On the average, members at a local fitness center
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Question 1051295: Can someone show me how to work these type of math questions? Please, check the ones I tried to do. Thank you for your help.
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?
b) What percentage of the fitness club members work out for 2 hours and 15 minutes or more?
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
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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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