SOLUTION: I'm stuck on this one!
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 probabilit
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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 probabilit
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Question 1050946: I'm stuck on this one!
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.
B) find the probability that the mean for 25 randomly selected distances is greater than 202.80 cm. Found 2 solutions by stanbon, ewatrrr:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 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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You can put this solution on YOUR website! 205 cm and a standard deviation of 8.3 cm.
a.
P(x>217.5) = 1 - P(x <= 217.5)
z = (217.50-205)/8.3 = 1.5
P(z<=1.5) = normalcdf(-10,1.5)= .84
P(x > 217.5) = 1 - .84 = .16 0r 16%
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b. n = 25 , x̄ = 202.8
P(x>202.80) = 1 - P(x <= 202.80)
z = = -2.2/1.66 = -1.3253
P(z <= -1.3253) = .1885
|P(x > 202.8) = 1 - .1885 = .9115 0r 91.15%
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