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Math  /  Data & Statistics

Question22. An instructor has given a short quiz consisting of two parts. For a randomly selected student, let X=X= the number of points eamed on the first part and Y=Y= the number of points eamed on the second part. Suppose that the joint pmf of XX and YY is given in the accompanying table. \begin{tabular}{|c|c|c|c|c|c|} \hline \multicolumn{2}{|l|}{\multirow[b]{2}{*}{p(x,y)p(x, y)}} & \multicolumn{4}{|c|}{yy} \\ \hline & & 0 & 5 & 10 & 15 \\ \hline & 0 & . 02 & . 06 & . 02 & . 10 \\ \hline xx & 5 & . 04 & . 15 & . 20 & . 10 \\ \hline & 10 & .01 & .15 & . 14 & . 01 \\ \hline \end{tabular} a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score E(X+Y)E(X+Y) ? b. If the maximum of the two scores is recorded, what is the expected recorded score?

Studdy Solution

STEP 1

1. X X and Y Y are discrete random variables representing points earned on two parts of a quiz.
2. The joint probability mass function (pmf) p(x,y) p(x, y) is given in the table.
3. The total score is X+Y X + Y .
4. The maximum score is max⁡(X,Y) \max(X, Y) .

STEP 2

1. Calculate E(X+Y) E(X+Y) .
2. Calculate E(max⁡(X,Y)) E(\max(X, Y)) .

STEP 3

Calculate the expected value E(X+Y) E(X+Y) .
E(X+Y)=∑x∑y(x+y)⋅p(x,y) E(X+Y) = \sum_{x} \sum_{y} (x + y) \cdot p(x, y)

STEP 4

Compute each term (x+y)⋅p(x,y) (x + y) \cdot p(x, y) and sum them:
\begin{align*} 0 + 0: & \quad (0 + 0) \cdot 0.02 = 0 \\ 0 + 5: & \quad (0 + 5) \cdot 0.06 = 0.30 \\ 0 + 10: & \quad (0 + 10) \cdot 0.02 = 0.20 \\ 0 + 15: & \quad (0 + 15) \cdot 0.10 = 1.50 \\ 5 + 0: & \quad (5 + 0) \cdot 0.04 = 0.20 \\ 5 + 5: & \quad (5 + 5) \cdot 0.15 = 1.50 \\ 5 + 10: & \quad (5 + 10) \cdot 0.20 = 3.00 \\ 5 + 15: & \quad (5 + 15) \cdot 0.10 = 2.00 \\ 10 + 0: & \quad (10 + 0) \cdot 0.01 = 0.10 \\ 10 + 5: & \quad (10 + 5) \cdot 0.15 = 2.25 \\ 10 + 10: & \quad (10 + 10) \cdot 0.14 = 2.80 \\ 10 + 15: & \quad (10 + 15) \cdot 0.01 = 0.25 \\ \end{align*}

STEP 5

Sum all the computed terms:
E(X+Y)=0+0.30+0.20+1.50+0.20+1.50+3.00+2.00+0.10+2.25+2.80+0.25=14.10E(X+Y) = 0 + 0.30 + 0.20 + 1.50 + 0.20 + 1.50 + 3.00 + 2.00 + 0.10 + 2.25 + 2.80 + 0.25 = 14.10

STEP 6

Calculate the expected value E(max⁡(X,Y)) E(\max(X, Y)) .
E(max⁡(X,Y))=∑x∑ymax⁡(x,y)⋅p(x,y) E(\max(X, Y)) = \sum_{x} \sum_{y} \max(x, y) \cdot p(x, y)

STEP 7

Compute each term max⁡(x,y)⋅p(x,y) \max(x, y) \cdot p(x, y) and sum them:
\begin{align*} \max(0, 0): & \quad 0 \cdot 0.02 = 0 \\ \max(0, 5): & \quad 5 \cdot 0.06 = 0.30 \\ \max(0, 10): & \quad 10 \cdot 0.02 = 0.20 \\ \max(0, 15): & \quad 15 \cdot 0.10 = 1.50 \\ \max(5, 0): & \quad 5 \cdot 0.04 = 0.20 \\ \max(5, 5): & \quad 5 \cdot 0.15 = 0.75 \\ \max(5, 10): & \quad 10 \cdot 0.20 = 2.00 \\ \max(5, 15): & \quad 15 \cdot 0.10 = 1.50 \\ \max(10, 0): & \quad 10 \cdot 0.01 = 0.10 \\ \max(10, 5): & \quad 10 \cdot 0.15 = 1.50 \\ \max(10, 10): & \quad 10 \cdot 0.14 = 1.40 \\ \max(10, 15): & \quad 15 \cdot 0.01 = 0.15 \\ \end{align*}

STEP 8

Sum all the computed terms:
E(max⁡(X,Y))=0+0.30+0.20+1.50+0.20+0.75+2.00+1.50+0.10+1.50+1.40+0.15=9.90E(\max(X, Y)) = 0 + 0.30 + 0.20 + 1.50 + 0.20 + 0.75 + 2.00 + 1.50 + 0.10 + 1.50 + 1.40 + 0.15 = 9.90
The expected recorded score for the total is:
E(X+Y)=14.10 E(X+Y) = 14.10
The expected recorded score for the maximum is:
E(max⁡(X,Y))=9.90 E(\max(X, Y)) = 9.90

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