Question

once again I am not getting this.
assume that the population of heights of male college students is normally distibuted with a mean of 69.09 and standard deviation of 4.71. A random sample of 92 heights is obtained. find the mean and standard error of the x distribution.
find P(x>68.5)

Answer 1

standard error is another term for standard deviation

Answer 2

z = (68.5-69.09)(sqrt(92))/4.71 = -1.2

Answer 3

isn;t the error 1/2 of the mean?

Answer 4

i mean the deviation?

Answer 5

no the deviation is given in the problem

Answer 6

so we would do this like the last one, right? z=x-o/n/sqrt

Answer 7

we can yes, but a denom IN the denom is just a multiplier

Answer 8

z=x-u/o~ * sqrt(n)

Answer 9

for a ztable we can use this

http://www.pavementinteractive.org/index.php?title=Normal_Distribution

Answer 10

at -1.2 we have an area of .1151

but we want the area to the right of the z, not the left. 1 - .1151 = .8839

Answer 11

well, .8849