document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1194474>Question 1194474</A>: <I><SPAN STYLE=\"word-break: break-all;\"> We have been using the normal distribution to approximate situations that are, in fact, binomial events. Create an example of a binomial event that can be approximated by a normal distribution and:<BR>\n" );
document.write( "a)Demonstrate how accurate the approximation is by using both approaches to find the probability of the same event. Hint: Calculate the probability of the event as a binomial (sum of all binomial events) and calculate the probability of the approximated event using a normal distribution, and compare them to see how close the approximation is. <BR>\n" );
document.write( "b)Describe the conditions under which the normal distribution would give a less accurate approximation.<BR>\n" );
document.write( "c)Explain a situation in which the criteria for using the normal approximation would be met, i.e.np≥5 and n(1-p)≥5, and yet you would decide not to use the normal distribution.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848520 by </B><A HREF=/tutors/aboutme.mpl?userid=yurtman><B>yurtman(42)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=yurtman><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848520\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ### Example of a Binomial Event Approximated by a Normal Distribution\r<BR>\n" );
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document.write( "**Scenario**: A factory produces light bulbs, and the probability of a bulb being defective is \( p = 0.02 \). The factory produces \( n = 500 \) bulbs in a day. Let \( X \) represent the number of defective bulbs in a batch of 500.\r<BR>\n" );
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document.write( "We want to find the probability that exactly 15 bulbs are defective (\( P(X = 15) \)).\r<BR>\n" );
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document.write( "### Part A: Calculating the Probability Using Both Approaches\r<BR>\n" );
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document.write( "#### 1. Binomial Approach<BR>\n" );
document.write( "The probability mass function for a binomial distribution is given by:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "For \( n = 500 \), \( p = 0.02 \), and \( k = 15 \), the exact probability is:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "P(X = 15) = \binom{500}{15} (0.02)^{15} (0.98)^{485}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "We'll calculate this value explicitly.\r<BR>\n" );
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document.write( "#### 2. Normal Approximation<BR>\n" );
document.write( "The binomial distribution can be approximated by a normal distribution if \( np \geq 5 \) and \( n(1-p) \geq 5 \). Here:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "np = 500 \times 0.02 = 10, \quad n(1-p) = 500 \times 0.98 = 490<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Both conditions are satisfied.\r<BR>\n" );
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document.write( "The approximating normal distribution has mean and standard deviation:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "\mu = np = 10, \quad \sigma = \sqrt{np(1-p)} = \sqrt{500 \times 0.02 \times 0.98} \approx 3.13<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "To approximate \( P(X = 15) \), we use the continuity correction, finding \( P(14.5 \leq X \leq 15.5) \):\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "Z_1 = \frac{14.5 - 10}{3.13}, \quad Z_2 = \frac{15.5 - 10}{3.13}<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "The corresponding probabilities are found using the standard normal table or software.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### Part B: Conditions for Less Accurate Approximation\r<BR>\n" );
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document.write( "The normal approximation becomes less accurate when:<BR>\n" );
document.write( "1. \( p \) is close to 0 or 1, making the distribution highly skewed.<BR>\n" );
document.write( "2. \( n \) is small, violating the \( np \geq 5 \) and \( n(1-p) \geq 5 \) conditions.<BR>\n" );
document.write( "3. The event involves a discrete probability at the tails of the distribution where the binomial and normal curves differ significantly.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### Part C: When Not to Use the Normal Approximation Despite Meeting the Criteria\r<BR>\n" );
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document.write( "Even if \( np \geq 5 \) and \( n(1-p) \geq 5 \), the normal approximation may not be ideal when:<BR>\n" );
document.write( "1. **Precision is critical**: The binomial distribution provides exact probabilities, whereas the normal approximation involves rounding and continuity corrections.<BR>\n" );
document.write( "2. **Events near the tails**: For probabilities of rare events (e.g., \( P(X = n) \) when \( n \) is far from the mean), the normal approximation may misestimate the probabilities.<BR>\n" );
document.write( "3. **Computational ease**: With modern software, binomial probabilities are easy to compute, making the exact approach preferable.\r<BR>\n" );
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document.write( "For example, if \( n = 20 \) and \( p = 0.25 \), both criteria are satisfied, but the distribution is still not symmetric, making the normal approximation less accurate. Using the exact binomial distribution would yield better results. </SPAN>\n" );
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