document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1162804>Question 1162804</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A multiple choice test has 25 questions, each with 5 possible choices, exactly one of which is correct.<BR>\n" );
document.write( "The test is marked by giving four points to each correct choice, and subtracting one point for each<BR>\n" );
document.write( "incorrect choice. No points are either given or subtracted for questions for which no choice is selected.<BR>\n" );
document.write( "Mickey and Bianca each know the answer to 12 of the 25 questions, but are unsure about the rest.<BR>\n" );
document.write( "Mickey leaves the other 13 unanswered, whereas Bianca answers them randomly, picking one choice<BR>\n" );
document.write( "for each question with equal probabilities, independently between questions.<BR>\n" );
document.write( "(a) Find the expectation and variance of the number of questions that Bianca gets correct.<BR>\n" );
document.write( "(b) Find the expectation and variance of the mark that Bianca gets.<BR>\n" );
document.write( "(c) Find the expectation and variance of the mark that Mickey gets. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854395 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2264)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=854395\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve this problem, we treat Bianca's random guessing as a binomial distribution and Mickey's strategy as a constant.\r<BR>\n" );
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document.write( "### (a) Expectation and variance of Bianca's correct questions<BR>\n" );
document.write( "For the 13 questions Bianca guesses, each has a probability $p = \frac{1}{5} = 0.2$ of being correct. Let $X$ be the number of questions she gets correct out of these $n = 13$ guesses. This follows a binomial distribution: $X \sim \text{Bin}(13, 0.2)$.\r<BR>\n" );
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document.write( "**Expectation:**<BR>\n" );
document.write( "$$E[X] = n \cdot p = 13 \cdot 0.2 = 2.6$$\r<BR>\n" );
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document.write( "**Variance:**<BR>\n" );
document.write( "$$\text{Var}(X) = n \cdot p \cdot (1 - p) = 13 \cdot 0.2 \cdot 0.8 = 2.08$$\r<BR>\n" );
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document.write( "Bianca also knows 12 questions are correct for certain. Let $C_B$ be her total correct answers: $C_B = 12 + X$.<BR>\n" );
document.write( "* **$E[C_B] = 12 + 2.6 = 14.6$**<BR>\n" );
document.write( "* **$\text{Var}(C_B) = \text{Var}(X) = 2.08$** (since the 12 known questions have 0 variance).\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### (b) Expectation and variance of Bianca's mark<BR>\n" );
document.write( "Let $M_B$ be Bianca's mark. For the 12 questions she knows, she earns $12 \times 4 = 48$ points. For the 13 she guesses, let $G_i$ be the mark for a single guess:<BR>\n" );
document.write( "* $P(G_i = 4) = 0.2$<BR>\n" );
document.write( "* $P(G_i = -1) = 0.8$\r<BR>\n" );
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document.write( "**Expectation of one guess ($E[G_i]$):**<BR>\n" );
document.write( "$$E[G_i] = (4 \times 0.2) + (-1 \times 0.8) = 0.8 - 0.8 = 0$$<BR>\n" );
document.write( "Since the expectation of a single guess is 0, the total expected mark for the 13 guesses is $13 \times 0 = 0$.<BR>\n" );
document.write( "* **$E[M_B] = 48 + 0 = 48$**\r<BR>\n" );
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document.write( "**Variance of one guess ($\text{Var}(G_i)$):**<BR>\n" );
document.write( "First, find $E[G_i^2]$:<BR>\n" );
document.write( "$$E[G_i^2] = (4^2 \times 0.2) + ((-1)^2 \times 0.8) = (16 \times 0.2) + (1 \times 0.8) = 3.2 + 0.8 = 4$$<BR>\n" );
document.write( "$$\text{Var}(G_i) = E[G_i^2] - (E[G_i])^2 = 4 - 0^2 = 4$$<BR>\n" );
document.write( "Since the 13 guesses are independent, we sum the variances:<BR>\n" );
document.write( "* **$\text{Var}(M_B) = 13 \times 4 = 52$**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### (c) Expectation and variance of Mickey's mark<BR>\n" );
document.write( "Mickey only answers the 12 questions he knows. He leaves the other 13 blank, earning 0 points for them. His mark $M_M$ is a constant.\r<BR>\n" );
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document.write( "**Expectation:**<BR>\n" );
document.write( "He gets exactly $12 \times 4$ points.<BR>\n" );
document.write( "* **$E[M_M] = 48$**\r<BR>\n" );
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document.write( "**Variance:**<BR>\n" );
document.write( "Since Mickey's mark is not subject to randomness (he knows the 12 are correct and the rest are definitely 0), there is no deviation from the mean.<BR>\n" );
document.write( "* **$\text{Var}(M_M) = 0$**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### Summary Table<BR>\n" );
document.write( "| Statistic | Bianca | Mickey |<BR>\n" );
document.write( "| :--- | :--- | :--- |<BR>\n" );
document.write( "| **Expected Mark** | 48 | 48 |<BR>\n" );
document.write( "| **Variance of Mark** | 52 | 0 |\r<BR>\n" );
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document.write( "**Insight:** Both students have the same expected score because the penalty for guessing ($1/4$ of the points gained) perfectly offsets the $1/5$ chance of being right. However, Bianca's strategy is much \"riskier\" due to the high variance. </SPAN>\n" );
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