Question 909240
 <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi
{{{mu}}} = 1.005, {{{sigma}}}= .10
*Note: {{{z = blue(x - mu)/blue(sigma)}}}
The idea (<u>the x-value</u> - the mean) and then divide by sd to find z
a) P(X< 1) = P(z < (1 - 1.005)/.10) = P(z < -.05)  = 48.1%
b) P(.95 < x < 1) = P(x<1) - P(x < .95) = P(z <-.05) - P(z < -.55) =  48.1% - 29.12%= 19.26%
19.26% the Area Under the Standard Normal curve between the Green Lines
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(-.05,0,-.05,exp(-.05^2/2)),line(-.55,0,-.55,exp(-.55^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}
c) P(1 < x < 1.05) = P(x <1.05) - P(x <1 ) = P(z<.45) - p(z <-.05)  = <u>67.36</u> - 48.1 = 
19.26% the Area Under the Standard Normal curve between the Green Lines
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(-.05,0,-.05,exp(-.05^2/2)),line(.45,0,.45,exp(-.45^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}
d) P(x<.95 0r x>1.05) = P(z<-.55) + [1 - P(z < .45)] = 29.12 + (1-<u>.6736</u>) = .2912 + .3264  = ..6176 0r 61.76%
61.76% the Area Under the Standard Normal curve <u>Outside</u> the Green Lines
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(-.55,0,-.55,exp(-.55^2/2)),line(.45,0,.45,exp(-.45^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}