document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1186410>Question 1186410</A>: <I><SPAN STYLE=\"word-break: break-all;\"> It has been estimated that 40% of the televisions that the manufacturer makes need repairs in the first three years operation. A new hotel buys 90 televisions form the company. Find the probability that in the first three years operation:\r<BR>\n" );
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document.write( "a.	Less than 32 of the television needs repairs\r<BR>\n" );
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document.write( "b.	Between 38 and 42 of the televisions,  needs repairs<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849444 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=849444\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this problem using the normal approximation to the binomial distribution:\r<BR>\n" );
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document.write( "**1. Check if the normal approximation is appropriate:**\r<BR>\n" );
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document.write( "*   n * p = 90 * 0.40 = 36 >= 10<BR>\n" );
document.write( "*   n * (1-p) = 90 * 0.60 = 54 >= 10\r<BR>\n" );
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document.write( "Since both conditions are met, the normal approximation is reasonable.\r<BR>\n" );
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document.write( "**2. Calculate the mean (μ) and standard deviation (σ) of the binomial distribution:**\r<BR>\n" );
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document.write( "*   μ = n * p = 90 * 0.40 = 36<BR>\n" );
document.write( "*   σ = sqrt(n * p * (1-p)) = sqrt(90 * 0.40 * 0.60) = sqrt(21.6) ≈ 4.65\r<BR>\n" );
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document.write( "**3. Apply the continuity correction:**\r<BR>\n" );
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document.write( "Since we are dealing with a discrete variable (number of televisions) and approximating it with a continuous distribution (normal), we need a continuity correction.\r<BR>\n" );
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document.write( "*   **a. Less than 32 televisions:**  We want P(x < 32).  With the continuity correction, we use P(x < 31.5).<BR>\n" );
document.write( "*   **b. Between 38 and 42 televisions:** We want P(38 <= x <= 42). With the continuity correction, we use P(37.5 < x < 42.5).\r<BR>\n" );
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document.write( "**4. Calculate the z-scores:**\r<BR>\n" );
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document.write( "The z-score formula is: z = (x - μ) / σ\r<BR>\n" );
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document.write( "*   **a. For x < 31.5:** z = (31.5 - 36) / 4.65 ≈ -0.97<BR>\n" );
document.write( "*   **b. For x < 37.5:** z = (37.5 - 36) / 4.65 ≈ 0.32<BR>\n" );
document.write( "*   **c. For x < 42.5:** z = (42.5 - 36) / 4.65 ≈ 1.40\r<BR>\n" );
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document.write( "**5. Find the probabilities using the z-table or calculator:**\r<BR>\n" );
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document.write( "*   **a. P(x < 31.5):** Look up the probability corresponding to z = -0.97.  P(z < -0.97) ≈ 0.1660.<BR>\n" );
document.write( "*   **b. P(37.5 < x < 42.5):** Look up the probabilities for z = 1.40 and z = 0.32. P(z < 1.40) ≈ 0.9192 and P(z < 0.32) ≈ 0.6255. Then subtract the smaller probability from the larger: P(0.32 < z < 1.40) = P(z < 1.40) - P(z < 0.32) ≈ 0.9192 - 0.6255 ≈ 0.2937.\r<BR>\n" );
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document.write( "**Answers:**\r<BR>\n" );
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document.write( "*   a. The probability that less than 32 televisions need repairs is approximately **0.1660**.<BR>\n" );
document.write( "*   b. The probability that between 38 and 42 televisions need repairs is approximately **0.2937**.<BR>\n" );
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