Using the z-table, find the corresponding area between π§ = 0
and each of the following:
1. z = 0.92
2. z = 1.29
3. z = 2.73
4. z = β0.50
5. z = β2.98
As we know from calculus, the area under the graph of a positive function "f(x)" between two values "x_1 <x_2" is equal to the integral "\\int\\limits_{x_1}^{x_2}f(x)dx =F(x_2)-F(x_1)" , where F(x) is any antidevative of the function "f(x)". If we take "f(x)=\\frac{1}{\\sqrt{2\\pi}}e^{-x^2\/2}" then its antiderivative is "\\Phi(x)=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^{x}e^{-t^2\/2}dt", which is a cumulative function of distribution for standard normal distribution.
Using the table of CFD for standard normal distribution, we have
"\\begin{matrix}\n z& -0.50 & 0.00 & 0.92& 1.29 & 2.73 & 2.98 \\\\\n \\Phi(z)& 0.3085 & 0.5000 & 0.8212 & 0.9015 & 0.9968 & 0.9986\n\\end{matrix}"
The area under the graph of density of standard normal distribution between π§ = 0 and "z=a" is:
if "a=-0.50" then "S=\\Phi(0.00)-\\Phi(-0.50)=0.5000-0.3085=0.1915"
if "a=0.92" then "S=\\Phi(0.92)-\\Phi(0.00)=0.8212-0.5000=0.3212"
if "a=1.29" then "S=\\Phi(1.29)-\\Phi(0.00)=0.9015-0.5000=0.4015"
if "a=2.73" then "S=\\Phi(2.73)-\\Phi(0.00)=0.9968-0.5000=0.4968"
if "a=2.98" then "S=\\Phi(2.98)-\\Phi(0.00)=0.9986-0.5000=0.4986"
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