Step 1
Cracking
\(\displaystyle\Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right.}\)

\(\displaystyle{a}=-{25.49}\)

\(\displaystyle{b}={1.15}\)

\(\displaystyle{c}={4.45}\)

\(\displaystyle{d}=-{1.78}\)

\(\displaystyle{e}={0.171}\) Determine the corresponding probability using the normal probability table. \(\displaystyle{t}={5}\Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{25.49}+{1.15}{\left({5}\right)}}}{{{\left({4.45}+{\left(-{1.78}\right)}{\left({5}\right)}+{0.171}{\left({5}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{19.74}}}{{{\left(-{0.175}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\) =Underfined \(\displaystyle{t}={10}\Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{25.49}+{1.15}{\left({10}\right)}}}{{{\left({4.45}+{\left(-{1.78}\right)}{\left({10}\right)}+{0.171}{\left({10}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\)

\(\displaystyle=\Phi{\left(-{7.22}\right)}\)

\(\displaystyle\approx{0}\) Note that the probability is undefined when \(\displaystyle{t}={5}\), because the expression under the square root (1/2) is negative and the square root of a negative number doesn't exist. Step 2 Cracking \(\displaystyle\Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle{a}=-{21.27}\)

\(\displaystyle{b}={0.0325}\)

\(\displaystyle{c}={0.972}\)

\(\displaystyle{d}=-{0.00028}\)

\(\displaystyle{e}={0.00022}\) Determine the corresponding probability using the normal probability y table. \(\displaystyle\Phi{\left({x}\right)}\) is approximately when x is a value smaller than all z-scores in the table \(\displaystyle{t}={5}\ \Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{21.27}+{0.0325}{\left({5}\right)}}}{{{\left({0.972}+{\left(-{0.00028}\right)}{\left({5}\right)}+{0.00022}{\left({5}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right\rbrace}\)

\(\displaystyle=\Phi{\left(-{21.36}\right)}\)

\(\displaystyle\approx{0}\)

\(\displaystyle{t}={10}\ \Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{21.27}+{0.0325}{\left({10}\right)}}}{{{\left({0.972}+{\left(-{0.00028}\right)}{\left({10}\right)}+{0.00022}{\left({10}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\)

\(\displaystyle\Phi{\left(-{21.04}\right)}\)

\(\displaystyle\approx{0}\)

\(\displaystyle{a}=-{25.49}\)

\(\displaystyle{b}={1.15}\)

\(\displaystyle{c}={4.45}\)

\(\displaystyle{d}=-{1.78}\)

\(\displaystyle{e}={0.171}\) Determine the corresponding probability using the normal probability table. \(\displaystyle{t}={5}\Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{25.49}+{1.15}{\left({5}\right)}}}{{{\left({4.45}+{\left(-{1.78}\right)}{\left({5}\right)}+{0.171}{\left({5}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{19.74}}}{{{\left(-{0.175}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\) =Underfined \(\displaystyle{t}={10}\Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{25.49}+{1.15}{\left({10}\right)}}}{{{\left({4.45}+{\left(-{1.78}\right)}{\left({10}\right)}+{0.171}{\left({10}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\)

\(\displaystyle=\Phi{\left(-{7.22}\right)}\)

\(\displaystyle\approx{0}\) Note that the probability is undefined when \(\displaystyle{t}={5}\), because the expression under the square root (1/2) is negative and the square root of a negative number doesn't exist. Step 2 Cracking \(\displaystyle\Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle{a}=-{21.27}\)

\(\displaystyle{b}={0.0325}\)

\(\displaystyle{c}={0.972}\)

\(\displaystyle{d}=-{0.00028}\)

\(\displaystyle{e}={0.00022}\) Determine the corresponding probability using the normal probability y table. \(\displaystyle\Phi{\left({x}\right)}\) is approximately when x is a value smaller than all z-scores in the table \(\displaystyle{t}={5}\ \Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{21.27}+{0.0325}{\left({5}\right)}}}{{{\left({0.972}+{\left(-{0.00028}\right)}{\left({5}\right)}+{0.00022}{\left({5}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right\rbrace}\)

\(\displaystyle=\Phi{\left(-{21.36}\right)}\)

\(\displaystyle\approx{0}\)

\(\displaystyle{t}={10}\ \Phi{\left(\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right)}\)

\(\displaystyle=\Phi{\left({\frac{{-{21.27}+{0.0325}{\left({10}\right)}}}{{{\left({0.972}+{\left(-{0.00028}\right)}{\left({10}\right)}+{0.00022}{\left({10}\right)}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}}}\right)}\)

\(\displaystyle\Phi{\left(-{21.04}\right)}\)

\(\displaystyle\approx{0}\)