Find z that has an area
(a) 0.9505 to its left
(b) 0.05 to its left
(c) 0.6915 to its right
(d) 0.1587 to its right
1
Expert's answer
2019-08-29T07:34:35-0400
https://www.sjsu.edu/faculty/gerstman/EpiInfo/z-table.htmIt is table of standard normal distribution (more precisely, in this table "Z(z)=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^z e^{-\\frac{x^2}{2}}dx" )So having this table we can write:a)"Z(z)=0.9505" if "z=1.65" b)Since graph of standard normal distribution is symmetric with respect to line "x=0" , for every "u\\ge 0" we have "\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-u}^0 e^{-\\frac{x^2}{2}}dx=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_0^u e^{-\\frac{x^2}{2}}dx". (1)Let "z" such number that "Z(z)=0,05" . So "0.5=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^0 e^{-\\frac{x^2}{2}}dx=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^z e^{-\\frac{x^2}{2}}dx+\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_z^0 e^{-\\frac{x^2}{2}}dx=""=0.05+\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_z^0 e^{-\\frac{x^2}{2}}dx" . That is "\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_z^0 e^{-\\frac{x^2}{2}}dx=0.45=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_0^{-z} e^{-\\frac{x^2}{2}}dx" by property (1). We have "Z(-z)=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^{-z} e^{-\\frac{x^2}{2}}dx=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^0 e^{-\\frac{x^2}{2}}dx+\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_0^{-z} e^{-\\frac{x^2}{2}}dx=0.95" . From table we can see that "-z\\approx 1.64" , so "z\\approx -1.64" c)We need "z" such "\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_z^{+\\infty} e^{-\\frac{x^2}{2}}dx=0.6915" . Since "1=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^{+\\infty} e^{-\\frac{x^2}{2}}dx=\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_{-\\infty}^z e^{-\\frac{x^2}{2}}dx+\\frac{1}{\\sqrt{2\\pi}}\\int\\limits_z^{+\\infty} e^{-\\frac{x^2}{2}}dx=""=Z(z)+0.6915" , so "Z(z)=1-0.6915=0.3085" Applying the same argument as in b). Since "0.3085=0.5-0.1915" , we have "Z(-z)=0.5+0.1915=0.6915" . So from the table we see that "-z=0.50" and so "z=-0.50" d)Applying the same argument as in c), we have "Z(z)=1-0.1587=0.8413" . From the table we see that "z=1.00"
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