Question 815385
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Hi, Re reply: P(z &#8804; 1.7)= NORMSDIST(1.7)(Excel)  
0rP(z &#8804; 1.7)=  normalcdf(-10,1.7)(TI Calculator) See Graph Below
 z = (117-100)/10 = 1.7.  P(z>1.7)  = 1 - P(z &#8804; 1.7) = 1 - .9554 = .0446 0r 4.46% 
B) find the probability that of 49 randomly selected people, their average IQ is less than 105.
 z = {{{(105-100)/(10/sqrt(49)) = 35/10 = 3.5}}}
P(z &#8804; 3.5) = .9554  0r  95.54%

z = 1.7
Note: P(z &#8804; 1.7) is the Area under the Normal Curve to the 'left' of Blue Line
P(z > 1.7)is the Area under the Normal Curve to the 'right' of Blue Line

{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, 0,exp(-x^2/2)), blue(line( 1.7,0, 1.7,exp(-1.7^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}