Question

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

Answer 1

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 $z=1.00$