document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1199567>Question 1199567</A>: <I><SPAN STYLE=\"word-break: break-all;\"> . In the School of ICT, one person in 80, on average, has blood of type O. If 200 student blood donors are <BR>\n" );
document.write( "taken at random, find an approximation to the probability that they include at least five persons having <BR>\n" );
document.write( "blood of type O.<BR>\n" );
document.write( "How many student donors must be taken at random in order that the probability of including at least one <BR>\n" );
document.write( "student donor of type O shall be 0.9 or more </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848275 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.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=848275\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve this, we approach each part of the problem systematically using probability theory and approximations.\r<BR>\n" );
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document.write( "### **Part 1: Approximation for at least 5 persons with blood type O**\r<BR>\n" );
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document.write( "1. **Given data**:<BR>\n" );
document.write( "   - Probability of having type O blood: \( p = \frac{1}{80} = 0.0125 \)<BR>\n" );
document.write( "   - Sample size: \( n = 200 \)<BR>\n" );
document.write( "   - Random variable: Let \( X \) be the number of students with blood type O. \( X \) follows a **binomial distribution**:<BR>\n" );
document.write( "     \[<BR>\n" );
document.write( "     X \sim \text{Binomial}(n = 200, p = 0.0125)<BR>\n" );
document.write( "     \]\r<BR>\n" );
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document.write( "2. **Normal approximation**:<BR>\n" );
document.write( "   Since \( n \) is large and \( p \) is small, we approximate \( X \) using a normal distribution with:<BR>\n" );
document.write( "   - Mean: \( \mu = n \cdot p = 200 \cdot 0.0125 = 2.5 \)<BR>\n" );
document.write( "   - Standard deviation: \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{200 \cdot 0.0125 \cdot 0.9875} \approx 1.567 \)<BR>\n" );
document.write( "   - Approximation: \( X \sim N(\mu = 2.5, \sigma = 1.567) \)\r<BR>\n" );
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document.write( "3. **Find \( P(X \geq 5) \):**<BR>\n" );
document.write( "   Using the normal approximation, apply continuity correction:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   P(X \geq 5) \approx P\left(Z \geq \frac{5 - 0.5 - \mu}{\sigma}\right)<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   Substitute values:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   Z = \frac{5 - 0.5 - 2.5}{1.567} = \frac{2}{1.567} \approx 1.276<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   From standard normal tables or a calculator:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   P(Z \geq 1.276) = 1 - P(Z \leq 1.276) \approx 1 - 0.8980 = 0.1020<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   Thus:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   P(X \geq 5) \approx 0.1020<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "### **Part 2: Number of donors for \( P(\text{at least one type O}) \geq 0.9 \)**\r<BR>\n" );
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document.write( "1. **Given data**:<BR>\n" );
document.write( "   - Probability of having type O blood: \( p = 0.0125 \)<BR>\n" );
document.write( "   - Required: Find \( n \) such that \( P(\text{at least one type O}) \geq 0.9 \).\r<BR>\n" );
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document.write( "2. **Complement rule**:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   P(\text{at least one type O}) = 1 - P(\text{none with type O})<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   For \( P(\text{none with type O}) \), the probability is:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   P(\text{none with type O}) = (1 - p)^n<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   Thus:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   1 - (1 - p)^n \geq 0.9 \quad \Rightarrow \quad (1 - p)^n \leq 0.1<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "3. **Solve for \( n \):**<BR>\n" );
document.write( "   Taking the natural logarithm:<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   \ln((1 - p)^n) \leq \ln(0.1) \quad \Rightarrow \quad n \cdot \ln(1 - p) \leq \ln(0.1)<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   Substitute \( p = 0.0125 \):<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   \ln(1 - 0.0125) \approx \ln(0.9875) \approx -0.0126<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   n \cdot (-0.0126) \leq \ln(0.1) \approx -2.3026<BR>\n" );
document.write( "   \]<BR>\n" );
document.write( "   \[<BR>\n" );
document.write( "   n \geq \frac{-2.3026}{-0.0126} \approx 183<BR>\n" );
document.write( "   \]\r<BR>\n" );
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document.write( "Thus, at least **183 donors** are required to ensure the probability of including at least one donor with type O blood is \( \geq 0.9 \).\r<BR>\n" );
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document.write( "### **Final Answers**:<BR>\n" );
document.write( "1. Probability of at least 5 type O donors: **0.1020**.<BR>\n" );
document.write( "2. Number of donors needed for \( P(\text{at least one type O}) \geq 0.9 \): **183**. </SPAN>\n" );
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