Question 1050946
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.3 cm.
A) Find the probability that an individual distance is greater than 217.50 cm.
z(217.5) = (217.5-205)/8.3 = 12.5/8.3 = 1.5060
P(x > 217.5) = P(z > 1.5060 = normalcdf(1.5060,100) = 0.0660
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B) find the probability that the mean for 25 randomly selected distances is greater than 202.80 cm.
Note:: std of means of samples of size 25 = 8.3/sqrt(25) = 8.3/5 = 1.66
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z(202.80) = (202.8-205)/1.66 = -2.2/1.66 = -1.3253 
P(x-bar > 202.8) = P(z > -1.3253) = normalcdf(-1.3253,100) = 0.9075
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Cheers,
Stan H.
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