Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider again the joint distribution of heights of husbands

and wives in Example 5.10.6. Find the 0.95 quantileof the conditional distribution of the height of the wife given that the height of the husband is 72 inches.

Short Answer

Expert verified

The 95th quantile is 70.57.\(\)

Step by step solution

01

Given information

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.

02

Denote the random variables

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

Then,

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

\(\)\(\)

03

Denote the conditional pdf

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

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

04

Calculate the probability

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

It is said that a random variable X has the Pareto distribution with parameters\({{\bf{x}}_{\bf{0}}}\,{\bf{and}}\,{\bf{\alpha }}\) if X has a continuous distribution for which the pdf\({\bf{f}}\left( {{\bf{x|}}\,{{\bf{x}}_{\bf{0}}}{\bf{,\alpha }}} \right)\) is as follows

\(\begin{array}{l}{\bf{f}}\left( {{\bf{x|}}\,{{\bf{x}}_{\bf{0}}}{\bf{,\alpha }}} \right){\bf{ = }}\frac{{{\bf{\alpha }}{{\bf{x}}_{\bf{0}}}^{\bf{\alpha }}}}{{{{\bf{x}}^{{\bf{\alpha + 1}}}}}}\,{\bf{,x}} \ge {{\bf{x}}_{\bf{0}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{ = }}\,{\bf{0}}\,\,{\bf{,x < }}{{\bf{x}}_{\bf{0}}}\end{array}\)

Show that if X has this Pareto distribution, then the random variable\({\bf{log}}\left( {{\bf{X|}}\,{{\bf{x}}_{\bf{0}}}} \right)\)has the exponential distribution with parameter ฮฑ.

Suppose that a fair coin is tossed until at least one head and at least one tail has been obtained. Let X denote the number of tosses that are required. Find the p.f. of X

Let \({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}} \right)\) denote the p.d.f. of the bivariate normaldistribution specified by Eq. (5.10.2). Show that the maximumvalue of \({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}} \right)\) is attained at the point at which \({{\bf{x}}_{\bf{1}}} = {{\bf{\mu }}_{\bf{1}}}{\rm{ }}{\bf{and}}{\rm{ }}{{\bf{x}}_{\bf{2}}} = {{\bf{\mu }}_{\bf{2}}}.\)

Suppose that seven balls are selected at random withoutreplacement from a box containing five red balls and ten blue balls. If \(\overline X \) denotes the proportion of red balls in the sample, what are the mean and the variance of \(\overline X \) ?

Prove Theorem 5.3.3. Hint:Prove that\[\mathop {lim}\limits_{n \to \infty } {c_n}log\left( {1 + {a_n}} \right) - {a_n}{c_n} = 0\]by applying Taylorโ€™s theorem with remainder (see Exercise 13 in Sec. 4.2) to the functionf (x)=log(1+x)aroundx=0.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free