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Statistics / Mathematical Statistics with Applications 7 / Chapter 4 / Problem 10E

Mathematical Statistics with Applications | 7th Edition | ISBN: 9780495110811 | Authors: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer

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Refer to the density function given in Exercise 4.8.a Find

Chapter 4, Problem 10E

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QUESTION:

Refer to the density function given in Exercise $4.8$ .
a. Find the .95-quantile, \(\phi_{.95}\), such that \(P\left(Y \leq \phi_{.95}\right)=.95\).
b. Find a value \(y_{0}\) so that \(P\left(Y<y_{0}\right)=.95\).
c. Compare the values for \(\phi_{.95}\) and \(y_{0}\) that you obtained in parts (a) and (b). Explain the relationship between these two values.

Equation Transcription:

𝜙.95

$P(Y\leq{}{\phi{}}_{.95})=.95$

${y}_{0}$

$P(Y<{y}_{0})=.95$

𝜙.95

${y}_{0}$

Text Transcription:

phi_.95

P(Y</=phi_.95)=.95

y_0

P(Y<y_0)=.95

phi_.95

y_0

Questions & Answers

QUESTION:

Refer to the density function given in Exercise $4.8$ .
a. Find the .95-quantile, \(\phi_{.95}\), such that \(P\left(Y \leq \phi_{.95}\right)=.95\).
b. Find a value \(y_{0}\) so that \(P\left(Y<y_{0}\right)=.95\).
c. Compare the values for \(\phi_{.95}\) and \(y_{0}\) that you obtained in parts (a) and (b). Explain the relationship between these two values.

Equation Transcription:

𝜙.95

$P(Y\leq{}{\phi{}}_{.95})=.95$

${y}_{0}$

$P(Y<{y}_{0})=.95$

𝜙.95

${y}_{0}$

Text Transcription:

phi_.95

P(Y</=phi_.95)=.95

y_0

P(Y<y_0)=.95

phi_.95

y_0

ANSWER:

Answer:

Step 1 of 3:

(a)

By referring to the density function given in exercise 4.8.

We need to find the $0.95-quantile,\ {\phi{}}_{0.95},$ such that $P(Y\leq{}\ {{\phi{}}_{0.95})\ =0.95.}_{\ }$

$Y$ has a probability density function,

$f(y)\ =ky(1\ -y),\ \ \ \ \ \ \ 0\leq{}y\leq{}1,$

$\ =0\ \ \ \ \ \ \ \ \ \ \ \ elsewhere\ \ \$

If $f(y)\$ is a density function for a continuous random variable, then

$\int\limits_{-\infty{}}^{\infty{}}f(y)dy\ =1$ ……….(1)

Hence we can calculate the value of $k$ using equation (1),

$\int\limits_{0}^{1}ky(1\ -y)dy\ =1$

$k\int\limits_{0}^{1}y(1\ -y)dy\ =1$

$\int\limits_{0}^{1}y(1\ -y)dy\ =\frac{1}{k}$

$\int\limits_{0}^{1}y(1\ -y)dy\ =\left({\frac{{y}^{2}}{2}\ -\frac{{y}^{3}}{3}}\right){|}_{0}^{1}\ =\frac{1}{6}$

$k\ =6$

Let $Y$ denote a random variable, then the cumulative distribution function of $Y,$ denoted by,

$F(y)\ =P(Y\leq{}y)\ =\int\limits_{-\infty{}}^{y}f(t)dt,\ \ \ \ \ \ \ \ \ \ \ -\infty{}<y<\infty{}$

………….(2)

We have given $P(Y\leq{}\ {{\phi{}}_{0.95})\ =0.95.}_{\ }$

Hence, we can write, using equation (2),

$F({\varphi{}}_{0.95})\ =P(Y\leq{}\ {{\phi{}}_{0.95})\ =\int\limits_{0}^{{\varphi{}}_{0.95}}6t(1\ -t)dt\ =0.95}_{\ }$

$\int\limits_{0}^{{\varphi{}}_{0.95}}6t(1\ -t)dt\ =0.95$

6 $\times{}$ $\left({\frac{{t}^{2}}{2}\ -\frac{{t}^{3}}{3}}\right){|}_{0}^{{\varphi{}}_{0.95}}\ =0.95$

$\left({3{{\varphi{}}^{2}}_{0.95}\ -2{{\varphi{}}^{3}}_{0.95}\ -0.95}\right)\ =0$

$\left({\ -2{{\varphi{}}^{3}}_{0.95}\ +\ 3{{\varphi{}}^{2}}_{0.95}\ -0.95}\right)\ =0$

On solving the equation, we get,

${{\varphi{}}^{\ }}_{0.95}\ =0.865$ ………(3)

Hence the $0.95-quantile,\ {\phi{}}_{0.95},$ is $0.865.$

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