91Ó°ÊÓ

A married couple is selected at randomfrom a certain population of married couples and the joint distribution of theheight of the wife and the height of her husband is a bivariate normal distribution.

The heights of the wives have a mean of 66.8 inches and a standard deviation of 2 inches, the heights of the husbands have a mean of 70 inches and a standard deviation of 2 inches, and the correlation between these two heights is 0.68.

Let X denote the height of the wife, and let Y denote the height of her husband.

Then,

\(\begin{array}{*{20}{l}}{{\mu _x} = 66.8\;}\\{{\sigma _x} = 2}\\{{\mu _y} = 70}\\{{\sigma _y} = 2}\\{p = 0.68}\end{array}\)

\(\)\(\)

For a\(BVN{\rm{ }}\left( {66.8,70,2,2,0.680} \right)\)

\(\begin{aligned}{c}X|Y& = y \sim N\left( {{\mu _x} + p\frac{{{\sigma _x}}}{{{\sigma _y}}}\left( {y - {\mu _y}} \right),{\sigma _x}^2\left( {1 - {p^2}} \right)} \right)\\ &= N\left( {66.8 + 0.68\frac{2}{2}\left( {y - 70} \right),\,\,{2^2}\left( {1 - {{0.68}^2}} \right)} \right)\\ &= N\left( {66.8 + 0.68\left( {y - 70} \right),\,\,2.1504} \right)\end{aligned}\)

It is given that the height of the husband is 72 inches. Therefore Y=72.

The above equation reduces to,

\(\begin{array}{l} = N\left( {66.8 + 0.68\left( {72 - 70} \right),\,\,2.1504} \right)\\ = N\left( {66.8 + 0.68 \times 2,\,\,2.1504} \right)\\ = N\left( {68.16,\,\,2.1504} \right)\end{array}\)

Therefore, the pdf is,

\(f\left( {x,\mu ,\sigma } \right) = \frac{1}{{\sqrt {2\pi } \times 2.1504}}{e^{ - \frac{1}{2}}}{\left( {\frac{{x - 68.16}}{{2.1504}}} \right)^2}\)

Therefore,the 0.95 quantile

\(\int\limits_{ - \infty }^{0.95} {\frac{1}{{\sqrt {2\pi } \times 2.1504}}{e^{ - \frac{1}{2}}}{{\left( {\frac{{x - 68.16}}{{2.1504}}} \right)}^2}dx} = 70.57\).

From the normal tables the value is 70.57.