Notes on Problems for Quiz 4 (10/24/97) from Schaum:

5.27, 5.29, 5.31, 5.32, 5.37, 5.40, 5.42, 5.43, 5.47, 5.48

Problem 5.27:

(1) The six-by-six table below represents the random variable X. The rows and colums are labeled by the outcome of the first and second die respectively. The entry in postion (i,j) of the table represeents X(i,j) = min(i,j). The image(X)={1,2,3,4,5,6}. The distribution function of X is shown on page 102 of the text in Answers to Supplementary Problems. How this distribution function was gotten was discussed in class and should be obvious from the table below.

Problem 5.29:

(1) The basic sample space consists of sixteen points: HHHH (4), HHHT(3) , HHTH(2), HHTT(2), HTHH(2),HTHT(1), HTTH(1), HTTT(1), THHH(3), THHT(2), THTH(1), THTT(1), TTHH(2), TTHT(1),TTTH(1), TTTT(0). The value of Y on each sample point is shown in parentheses. The image(Y) = {0,1,2,3,4} and the distribution is as shown below:

Problem 5.31:

(1) The basic equiprobable sample space is the set

{1 1, 1 2, 1 2, 1 3, 1 2 , 1 2, 1 3, 2 2, 2 3, 2 3}.

In tabular form we have

The four distributions can be this table and are shown on pages 102 and 103 of the text.

Problem 5.32:

(1) The sample space is {H, TH, TTH, TTTH, TTTT} with with probabilities 1/2, 1/4, 1/8, 1/16, 1/16 in the order listed. From this we derive the distribution function

The expected value is 1/2 + 2/4 + 3/8 + 4/8 = 15/8 .

Problem 5.37: The outcomes of the tournament can be listed in lexicographic order:

AA , ABAA, ABABA, ABABB, ABB, BB, BABB, BABAB, BABAA, BAA

or, equvalenlty, the tree diagram can be drawn as was done in class. The distribution of the random variable L (length of the tournament) is as follows:

The E(L) = 1 + 3/4 + 4/8 + 5/8 = 23/8 .

Problem 5.40

(1) There are C(3,3) = 1 way that 3 heads occurs, C(3,2) = 3 ways that 2 heads occurs, C(3,1)=3 ways for 1 head to occur and C(3,0) = 1 way for 0 heads to occur. The probabilities are 1/8 in each case. Thus the expected winnings are ($10 + $15 + $9 + $2)/8 = $36/8 = $4.50.

Problem 5.42:

(1) This is a straightforward calculation using the formulas in the text with the answers given on page103. The correlation is negative, -0.3. You should be able to explain that intuitively (we discussed this in class).

Problem 5.43:

(1) This again is a straightforward computation. The answer given on page 103. The Cov(X,Y) is always zero if X and Y are independent. See 5.19 (iii) on page 96 for a proof.

Problem 5.47:

(1) The cumulative distribution for this distribution funtion is

F(x) = 0, if x<0, F(x) = x^2 /25, if 0<= x <=5, and F(x) = 1, if x>5.

We showed how to get this in class by integrating the distribution function. This can be used to answer (ii). For example, P(2 < X < 4) = F(4) - F(2) = 12/25 .

Problem 5.48:

(1) The graph is shown on page 104.