document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1186019>Question 1186019</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2.	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" );
document.write( "\n" );
document.write( "a.	Less than 32 of the television needs repairs\r<BR>\n" );
document.write( "\n" );
document.write( "b.	Between 38 and 42 of the televisions, needs repairs<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849484 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=849484\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this binomial probability problem using the normal approximation:\r<BR>\n" );
document.write( "\n" );
document.write( "**1. Check for Normal Approximation Applicability:**\r<BR>\n" );
document.write( "\n" );
document.write( "*   np = 90 * 0.40 = 36 ≥ 10<BR>\n" );
document.write( "*   n(1-p) = 90 * 0.60 = 54 ≥ 10\r<BR>\n" );
document.write( "\n" );
document.write( "Since both conditions are met, the normal approximation is appropriate.\r<BR>\n" );
document.write( "\n" );
document.write( "**2. Calculate Mean and Standard Deviation:**\r<BR>\n" );
document.write( "\n" );
document.write( "*   Mean (μ) = np = 36<BR>\n" );
document.write( "*   Standard Deviation (σ) = sqrt(np(1-p)) = sqrt(90 * 0.40 * 0.60) = sqrt(21.6) ≈ 4.65\r<BR>\n" );
document.write( "\n" );
document.write( "**3. Apply Continuity Correction:**\r<BR>\n" );
document.write( "\n" );
document.write( "Because we're approximating a discrete distribution (number of TVs) with a continuous one (normal), we adjust the values:\r<BR>\n" );
document.write( "\n" );
document.write( "*   **a. Less than 32:**  Use 31.5 as the upper limit.  P(x < 32) becomes P(x < 31.5)<BR>\n" );
document.write( "*   **b. Between 38 and 42:** Use 37.5 as the lower limit and 42.5 as the upper limit. P(38 ≤ x ≤ 42) becomes P(37.5 < x < 42.5)\r<BR>\n" );
document.write( "\n" );
document.write( "**4. Calculate Z-scores:**\r<BR>\n" );
document.write( "\n" );
document.write( "The z-score formula is: z = (x - μ) / σ\r<BR>\n" );
document.write( "\n" );
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" );
document.write( "\n" );
document.write( "**5. Find Probabilities from Z-table or Calculator:**\r<BR>\n" );
document.write( "\n" );
document.write( "*   **a. P(x < 31.5) = P(z < -0.97) ≈ 0.1660** (This is the area to the left of z = -0.97)<BR>\n" );
document.write( "*   **b. P(37.5 < x < 42.5) = P(0.32 < z < 1.40) = P(z < 1.40) - P(z < 0.32)**<BR>\n" );
document.write( "    *   P(z < 1.40) ≈ 0.9192<BR>\n" );
document.write( "    *   P(z < 0.32) ≈ 0.6255<BR>\n" );
document.write( "    *   P(37.5 < x < 42.5) ≈ 0.9192 - 0.6255 ≈ 0.2937\r<BR>\n" );
document.write( "\n" );
document.write( "**Answers:**\r<BR>\n" );
document.write( "\n" );
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" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );