Question

Let X be normal with mean 80 and variance 9. Find P(X > 83), P(X < 81), P(X < 80), and P(78 < X < 82). I found the answer for P(X>83) and P(X<80), but I can't seem to get the correct values of the other two. If you could, please show me how to get the other two, and go step by step Let X be normal with mean 80 and variance 9. Find P(X > 83), P(X < 81), P(X < 80), and P(78 < X < 82). I found the answer for P(X>83) and P(X<80), but I can't seem to get the correct values of the other two. If you could, please show me how to get the other two, and go step by step @Mathematics

Answer 1

since variance = 9, standard deviation is the sqrt of variance; sd = 3

Answer 2

the values can either be found with a ztable, or a stats calculator like a TI83

Answer 3

with a TI83, you hit 2nd, vars, normalCDF() and enter:

normalCDF(lower bound, upper bound, mean, sd)

if the lower bound goes to infinity just enter in an arbitrarily small number like -99999

if the upper bound goes to infinity just enter in an arbitrarily large number like 99999

Answer 4

for example:

P(X < 81) would be, normalCDF(-99999,81,80,3)

Answer 5

yeah, I understand that, but i am trying to do it by hand with the formula
z=x-mean/variance but i keep getting the wrong values for it. I am thinking for P(78<=x<=82) it should be an integral but I am not sure though

Answer 6

its not "/variance".

\[z=\frac{x-\mu}{\sigma}\]
\[\sigma = \sqrt{variance}\]

Answer 7

if you were to take the equation of the normal curve and integrate it, then yes; but that is complicated so that others have done it and created ztables to refer to instead.

Answer 8

since the interval has endpoints that are equal distance from the mean; then you only need to do one of them and then double it

Answer 9

depending on the author of the ztable, the measure that you obtain is either in the tail or its a measure from the mean. If its a measure from the mean then you are good to double it from z= .66