Question 139732: Having a lot of trouble with this. Please help and show me how so I can actually learn from all this. Thanks!
PS- I am allowed to use Megastat on this so any direction there would be welcomed as well.
The bi-monthly starting salaries of recent statistician graduates follow the normal distribution with a mean of $2,625 and a standard deviation of $350. (shos all your work please)
A. What is the z-value for a salary of $2,200?
B. What is the approximate percent of statisticians making between $2,625 & $2,975?
C. What is the approximate percent of statisticians making between $2,275 & $2,625?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The bi-monthly starting salaries of recent statistician graduates follow the normal distribution with a mean of $2,625 and a standard deviation of $350. (shos all your work please)
A. What is the z-value for a salary of $2,200?
z(2200) = (2200-2625)/350 = -1.21428...
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B. What is the approximate percent of statisticians making between $2,625 & $2,975?
Since the mean is 2625 the z-value of 2625 is zero
z(2975) = (2975-2625)/350 = 1
P(2625 < x < 2975) = P(0 < z < 1) = 0.3413..
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C. What is the approximate percent of statisticians making between $2,275 & $2,625?
P( 2275 < x < 2625) = P(-1 < z < 0) = 0.3413
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Cheers,
Stan H.
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