91Ó°ÊÓ

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

(a) The \({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}\)percentile represents the \({\rm{91st}}\) percentile. It signifies that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}\)is \({\rm{0}}{\rm{.91}}\) (or that the region to the right of \({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}\)$ is \({\rm{0}}{\rm{.09}}\)or that:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.09}}}}} \right){\rm{ = 0}}{\rm{.91}}\)

For any \({\rm{z}}\), we examine Appendix Table A.3 to see if \({\rm{f(z)}}\) equals \({\rm{0}}{\rm{.91}}\)

At the point where the \({\rm{1}}{\rm{.3}}\) row and the \({\rm{0}}{\rm{.7}}\) column cross, \({\rm{0}}{\rm{.9099}}\)is the amount there. \({\rm{0}}{\rm{.91}}\) is a pretty close match. As a result, without utilizing interpolation, we may say:

\({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}{\rm{ = 1}}{\rm{.37}}\)

(b) The \({{\rm{z}}_{{\rm{0}}{\rm{.91}}}}\)percentile represents the \({\rm{9th}}\) percentile. It signifies that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.91}}}}\)is \({\rm{0}}{\rm{.09}}\) :

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.91}}}}} \right){\rm{ = 0}}{\rm{.09}}\)

For any \({\rm{z}}\), we examine Appendix Table A.3 to see if \({\rm{f(z)}}\) equals \({\rm{0}}{\rm{.09}}\)

At the point where the \({\rm{ - 1}}{\rm{.3}}\) row and the \({\rm{0}}{\rm{.4}}\) column cross, \({\rm{0}}{\rm{.9091}}\)is the amount there. \({\rm{0}}{\rm{.09}}\) is a pretty close match. As a result, without utilizing interpolation, we may say:

\({{\rm{z}}_{{\rm{0}}{\rm{.91}}}}{\rm{ = 1}}{\rm{.34}}\)

(c) The \({{\rm{z}}_{{\rm{0}}{\rm{.25}}}}\) percentile for the \({\rm{75th}}\)percentile. It means that the region to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.25}}}}\)equals \({\rm{0}}{\rm{.75}}\), or in other words:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}} \right){\rm{ = 0}}{\rm{.75}}\)

For any\({\rm{z}}\), we examine Appendix Table A.3 to see if \({\rm{f(z)}}\)equals \({\rm{0}}{\rm{.75}}\).

\({\rm{0}}{\rm{.7486 and 0}}{\rm{.7517}}\)are the two closest values from the table, with z-values \({\rm{0}}{\rm{.67 and 0}}{\rm{.68,}}\)respectively. As a result of the interpolation, we can write:

\(\begin{array}{*{20}{c}}{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ - 0}}{\rm{.67}}}}{{{\rm{0}}{\rm{.75 - 0}}{\rm{.7486}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.68 - 0}}{\rm{.67}}}}{{{\rm{0}}{\rm{.7517 - 0}}{\rm{.7486}}}}}\\{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ - 0}}{\rm{.67}}}}{{{\rm{0}}{\rm{.0014}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.01}}}}{{{\rm{0}}{\rm{.0031}}}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ - 0}}{\rm{.67 = 0}}{\rm{.0045}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ = 0}}{\rm{.6745}}}\end{array}\)

(d) The \({{\rm{z}}_{{\rm{0}}{\rm{.75}}}}\) percentile for the \({\rm{25th}}\)percentile. It means that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.75}}}}\) equals \({\rm{0}}{\rm{.25}}\), or in other words:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}} \right){\rm{ = 0}}{\rm{.25}}\)

For any \({\rm{z}}\), we examine Appendix Table A. 3 to see if \({\rm{f(z)}}\)equals \({\rm{0}}{\rm{.25}}\)

\({\rm{0}}{\rm{.2514 and 0}}{\rm{.2483}}\)are the two closest values from the table, with z-values \({\rm{ - 0}}{\rm{.67 and - 0}}{\rm{.68}}\)respectively. As a result of the interpolation, we can write:

\(\begin{array}{*{20}{c}}{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}{\rm{ - ( - 0}}{\rm{.67)}}}}{{{\rm{0}}{\rm{.25 - 0}}{\rm{.2514}}}}}&{{\rm{ = }}\frac{{{\rm{ - 0}}{\rm{.68 - ( - 0}}{\rm{.67)}}}}{{{\rm{0}}{\rm{.2483 - 0}}{\rm{.2514}}}}}\\{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}{\rm{ + 0}}{\rm{.67}}}}{{{\rm{ - 0}}{\rm{.0014}}}}}&{{\rm{ = }}\frac{{{\rm{0}}{\rm{.01}}}}{{{\rm{0}}{\rm{.0031}}}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}{\rm{ + 0}}{\rm{.67}}}&{{\rm{ = - 0}}{\rm{.0045}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}}&{{\rm{ = - 0}}{\rm{.6745}}}\end{array}\)

(d) The \({{\rm{z}}_{{\rm{0}}{\rm{.94}}}}\) percentile for the \({\rm{25th}}\)percentile. It means that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.94}}}}\) equals \({\rm{0}}{\rm{.06}}\), or in other words:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}} \right){\rm{ = 0}}{\rm{.06}}\)

For any \({\rm{z}}\), we examine Appendix Table A. 3 to see if \({\rm{f(z)}}\)equals \({\rm{0}}{\rm{.06}}\)

\({\rm{0}}{\rm{.0606 and 0}}{\rm{.0594}}\)are the two closest values from the table, with z-values \({\rm{ - 1}}{\rm{.55 and - 1}}{\rm{.56}}\) respectively. As a result of the interpolation, we can write:

\(\begin{array}{*{20}{c}}{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ - ( - 1}}{\rm{.56)}}}}{{{\rm{0}}{\rm{.06 - 0}}{\rm{.0594}}}}{\rm{ = }}\frac{{{\rm{ - 1}}{\rm{.55 - ( - 1}}{\rm{.56)}}}}{{{\rm{0}}{\rm{.0606 - 0}}{\rm{.0594}}}}}\\{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ + 1}}{\rm{.56}}}}{{{\rm{0}}{\rm{.0006}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.01}}}}{{{\rm{0}}{\rm{.0012}}}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ + 1}}{\rm{.56 = 0}}{\rm{.0050}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ = - 1}}{\rm{.555}}}\end{array}\)