Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
Question
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Chapter 5.8, Problem 104E

a.

To determine

Calculate the values of E(Y1+Y2) and V(Y1+Y2) by using the probability distribution of Y1+Y2.

a.

Expert Solution
Check Mark

Answer to Problem 104E

The value of E(Y1+Y2) by using the probability distribution of Y1+Y2 is 73_.

The value of V(Y1+Y2) by using the probability distribution of Y1+Y2 is 718_.

Explanation of Solution

Calculation:

The possible values of Y1+Y2 are 1, 2, 3. Let Y=Y1+Y2. Using the fact that each of Y1 and Y2 can take values 0, 1, 2, 3, the possible values of Y, along with the corresponding probabilities using the joint probability distribution, are calculated below:

Y=Y1+Y2Y1=y1Y2=y2p(y1,y2)=(4y1)(3y2)(23y1y2)(93)
101(40)(31)(      2301)(93)=384
110(41)(30)(      2310)(93)=484
202(40)(32)(      2302)(93)=684
211(41)(31)(      2311)(93)=2484
220(42)(30)(      2320)(93)=1284
303(40)(33)(      2303)(93)=184
312(41)(32)(      2312)(93)=1284
321(42)(31)(      2321)(93)=1884
330(43)(30)(      2330)(93)=484
Total1

Extracting the unique values of Y from the above table, and adding up all the probabilities corresponding to that value gives the probability distribution of Y=Y1+Y2, as follows:

Y=Y1+Y2p(y)
1784(384+484)
24284(684+2484+1284)
33584(184+1284+1884+484)
Total1

Thus, E(Y), that is, E(Y1+Y2) is calculated below:

E(Y1+Y2)=E(Y)=yyp(y)=(1×784)+(2×4284)+(3×3584)=73.

Thus, the value of E(Y1+Y2) is 73_.

Again, V(Y1+Y2) is calculated below:

V(Y1+Y2)=V(Y)=[(173)2×784]+[(273)2×4284]+[(373)2×4284]=(169×784)+(19×4284)+(49×3584)=718.

Thus, the value of V(Y1+Y2) is 718_.

b.

To determine

Calculate the values of E(Y1+Y2) and V(Y1+Y2) by using Theorem 5.12.

b.

Expert Solution
Check Mark

Answer to Problem 104E

The value of E(Y1+Y2) by using Theorem 5.12 is 73_.

The value of V(Y1+Y2) by using Theorem 5.12 is 718_.

Explanation of Solution

Calculation:

Hypergeometric distribution:

A discrete real valued random variable Y is said to follow hypergeometric distribution if the probability mass function of Y is,

p(y)=(ry)(Nrny)(Nn),

Where integer y takes value from 0, 1, 2,…, n with the conditions yr and nyNr.

Consider that Y1 and Y2 are two discrete real valued random variables with joint probability mass function of p(y1,y2).

Then, the marginal probability functions of Y1 and Y2 are defined as,

p1(y1)=ally2p(y1,y2) and p2(y2)=ally1p(y1,y2).

The joint probability distribution of Y1 and Y2 is,

p(y1,y2)=(4y1)(3y2)(23y1y2)(93), where y1 and y2 are integers, 0y13,0y23and1y1+y23.

Thus, the marginal probability distribution of Y1 is obtained below:

p1(y1)=1y13y1p(y1,y2)=1y13y1(4y1)(3y2)(23y1y2)(93)=(4y1)(93)1y13y1(3y2)(23y1y2)=(4y1)(93)[(31y1)(23y11+y1)+....+(33y1)(23y13+y1)]=(4y1)(93)[(31y1)(22)+....+(33y1)(20)]=(4y1)(93)(53y1)=(4y1)(943y1)(93)

Thus, the marginal probability distribution of Y1 is Hypergeometric distribution with N=9, n=3 and r=4.

Now, the marginal probability distribution of Y2 is obtained below:

p2(y2)=1y23y2p(y1,y2)=1y23y2(4y1)(3y2)(23y1y2)(93)=(3y2)(93)1y23y2(4y1)(23y1y2)=(3y2)(93)[(41y2)(231+y2y2)+....+(43y2)(233+y2y2)]=(3y2)(63y2)(93)

Thus, marginal probability distribution of Y2 is Hypergeometric distribution with N=9, n=3 and r=3.

For a random variable, Y1, with a hypergeometric distribution and parameters n, r and N, the expectation is, E(Y1)=nrN, and the variance is, V(Y1)=nrN(NrN)(NnN1).

Thus, the value of E(Y1) is 43(=3×49); the value of E(Y2) is 1(=3×39).

Thus, E(Y1+Y2) is calculated below:

E(Y1+Y2)=E(Y1)+E(Y2)=43+1=73.

Thus, the value of E(Y1+Y2) is 73_.

Again, the value of V(Y1) is 59(=43(949)(9391)); the value of E(Y2) is 12(=1(939)(9391)).

From Exercise 5.90, the value of Cov(Y1,Y2) is 13.

Thus, V(Y1+Y2) is calculated below:

V(Y1+Y2)=V(Y1)+V(Y2)+2Cov(Y1,Y2)=59+12(2×13)=718.

Thus, the value of V(Y1+Y2) is 718_.

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Chapter 5 Solutions

Mathematical Statistics with Applications

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