Question 879699
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Hi
good work on (c). Hope the following is clear to You.
Do recommend stattrek.com as an excellent reference
u = 55.9 and o = 5.7, Normal curve y
y = {{{(1/(alpha*sqrt(2pi)))}}}e^{{{(-(x-mu)^2/2(alpha)^2)}}}
y = {{{(1/5.7sqrt(2pi))}}}e^{{{(-(x-55.9)^2/2(5.7)^2)}}}
You can put this in a graphing calculator or just sketch it...
They all look the same..just different heights and symmetry about the mean:
locate 46.5 and shade to the left (proportion left of x-value = 46.5)
   *[illustration normal_curve_2.jpg].

Do NOT confuse this with the 'standard normal curve' using z-values..
where left of z-value = (46.5-55.9)/5.9 = -1.649 (proportion left of x-value = 46.5)
Note: P(z < -1.649) = .0496
For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
Area under the standard normal curve to the left of the particular z is P(z)
Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50%  to the right
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(1,0,1,exp(-1^2/2)),line(-1,0,-1,exp(-1^2/2))),green(line(2,0,2,exp(-2^2/2)),line(-2,0,-2,exp(-2^2/2))),green(line(3,0,3,exp(-3^2/2)),line(-3,0,-3,exp(-3^2/2))),green(line( 0,0, 0,exp(0^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}