Question 1048255
mean μ = 10.2 kg and standard deviation σ = 0.8 kg. distributed. X ~ N(10.2, 0.8)
*Note: {{{z = blue(x - mu)/blue(sigma)}}}
a. 11 kg,  z = .8/.8 = 1      (84%  weigh less)
b. 7.9 kg  z = -2.3/.8 = -28/8 = -3.5   (Far... left of average(z=0) 98.98% weigh more)
c. 12.2 kg z = 2/.8 = 20/8 = 2.5   (Far right on curve 99.38% weigh less)
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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))}}}