Logout Open In App
Open In App
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

Chapter 8: Q3 E (page 472)

Sketch the p.d.f. of the\({{\bf{\chi }}^{\bf{2}}}\)distribution withmdegrees of freedom for each of the following values ofm. Locate the mean, the median, and the mode on each sketch. (a)m=1;(b)m=2; (c)m=3; (d)m=4.

Short Answer

Expert verified
  1. Graph for the degree of freedom m=1
  2. Graph for the degree of freedom m=2
  3. Graph for the degree of freedom m=3
  4. Graph for the degree of freedom m=4

Step by step solution

01

Given information

Here the p.d.f of the chi-square distribution with m degree of freedom is given

02

The formula for mean, median, and mode

\(\begin{align}Mean &= m\\Median &= m{\left( {1 - \frac{2}{{9m}}} \right)^3}\\Mode &= \max \left\{ {m - 2,0} \right\}\end{align}\)

03

Step 3:: Graph of the chi-square distribution with 1 degree of freedom

\(\begin{align}Form &= 1\\Mean &= 1\\Median &= 0.4705075\\Mode &= 0\end{align}\)

04

Graph of the chi-square distribution with 2 degree of freedom

\(\begin{align}Form &= 2\\Mean &= 2\\Median &= 1.404664\\Mode &= 0\end{align}\)

05

Graph of the chi-square distribution with 3degree of freedom

\(\begin{align}Form &= 3\\Mean &= 3\\Median &= 2.381497\\Mode &= 1\end{align}\)

06

Graph of the chi-square distribution with 4 degree of freedom

\(\begin{align}Form &= 4\\Mean &= 4\\Median &= 3.369684\\Mode &= 2\end{align}\)

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.

Start your free trial

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

Consider the conditions of Exercise 10 again. Suppose also that it is found in a random sample of size n = 10 \(\overline {{{\bf{x}}_{\bf{n}}}} {\bf{ = 1}}\,\,{\bf{and}}\,\,{{\bf{s}}_{\bf{n}}}^{\bf{2}}{\bf{ = 8}}\) . Find the shortest possible interval so that the posterior probability \({\bf{\mu }}\) lies in the interval is 0.95.

Suppose that X1,….., Xn form a random sample from the normal distribution with unknown mean µ and known variance σ2> 0 . Show thatX̄nis an efficient estimator of µ.

Suppose that X1,….., Xn forms a random sample from the Bernoulli distribution with unknown parameter p. Show thatX̄n is an efficient estimator of p.

Find the mode of theχ2 distribution withmdegrees of

freedom(m=1,2, . . .).

Determine whether or not each of the five following matrices is orthogonal:

  1. \(\left( {\begin{align}{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}\end{align}} \right)\)
  2. \(\left( {\begin{align}{{\bf{0}}{\bf{.8}}}&{\bf{0}}&{{\bf{0}}{\bf{.6}}}\\{{\bf{ - 0}}{\bf{.6}}}&{\bf{0}}&{{\bf{0}}{\bf{.8}}}\\{\bf{0}}&{{\bf{ - 1}}}&{\bf{0}}\end{align}} \right)\)
  3. \(\left( {\begin{align}{{\bf{0}}{\bf{.8}}}&{\bf{0}}&{{\bf{0}}{\bf{.6}}}\\{{\bf{ - 0}}{\bf{.6}}}&{\bf{0}}&{{\bf{0}}{\bf{.8}}}\\{\bf{0}}&{{\bf{0}}{\bf{.5}}}&{\bf{0}}\end{align}} \right)\)
  4. \(\left( {\begin{align}{}{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\\{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\\{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\end{align}} \right)\)
  5. \(\left( {\begin{align}{}{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}\end{align}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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