document.write( "Question 1208645: An integrated circuit manufacturer produces wafers that contain
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document.write( "20 chips. Each chip has a probability of 0.085 of not placed correctly on the wafer.
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document.write( "A) Find the probability that a wafer contains at least 3
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document.write( "incorrectly place chips. (Ans = 0.2390)
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document.write( "b) What is the probability that a wafer no more than one
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document.write( "incorrectly place chips on a wafer? (Ans = 0.4836)
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document.write( "c) What is the expected number of incorrectly placed chips on a
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document.write( "wafer? (Ans = 1.7) \n" );
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Algebra.Com's Answer #847029 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Please do not type in all upper case. \n" ); document.write( "I have edited your question to a more readable format.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We can use the Binomial probability distribution because:
\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Part A\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n = 20 = sample size = number of chips \n" ); document.write( "p = 0.085 = probability of a chip being incorrectly placed \n" ); document.write( "x = number of chips incorrectly placed \n" ); document.write( "x is a whole number that ranges from x = 0 to x = 20, i.e. it's an integer from the set {0,1,2,...,19,20}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "B(x) = binomial probability of exactly x chips being incorrectly placed \n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x) \n" ); document.write( "The nCx refers to the nCr combination formula. These values are found in Pascal's Triangle. A quick way to calculate the nCr values is to use the Combin function in a spreadsheet. Or you can use a TI Calculator. \n" ); document.write( "There are many options to calculate nCr.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's find the probability of exactly x = 0 chips that are incorrectly placed. \n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x) \n" ); document.write( "B(x) = (20Cx)*(0.085^x)*(1-0.085)^(20-x) \n" ); document.write( "B(0) = (20C0)*(0.085^0)*(1-0.085)^(20-0) \n" ); document.write( "B(0) = (1)*(0.085^0)*(1-0.085)^(20-0) \n" ); document.write( "B(0) = 0.16920839 approximately\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now let's find the probability of exactly x = 1 chip being incorrectly placed. \n" ); document.write( "B(x) = (20Cx)*(0.085^x)*(1-0.085)^(20-x) \n" ); document.write( "B(1) = (20C1)*(0.085^1)*(1-0.085)^(20-1) \n" ); document.write( "B(1) = (20)*(0.085^1)*(1-0.085)^(20-1) \n" ); document.write( "B(1) = 0.31437624 approximately\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Repeat similar steps to find that B(2) = 0.27744132 approximately which is the probability of having exactly 2 chips incorrectly placed.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then we could say, \n" ); document.write( "( B(0)+B(1)+B(2) ) + ( B(3)+B(4)+...+B(19)+B(20) ) = 1 \n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 1 - ( B(0)+B(1)+B(2) ) \n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 1 - ( 0.16920839+0.31437624+0.27744132 ) \n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 1 - 0.76102595 \n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 0.23897405 \n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 0.2390 which is the answer that your textbook mentioned. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If you want to use technology to quickly find the answer, then you can use a TI83 or similar. \n" ); document.write( "The command you'll use is called BinomCDF \n" ); document.write( "The order of inputs is: n, p, k n = 20\r\n" ); document.write( "p = 0.085\r\n" ); document.write( "k = 2Type in 1-BinomCDF(20,0.085,2) to get the approximate result 0.2390 when rounding to 4 decimal places. \n" ); document.write( "You can round manually or you can use the Round command in the TI83. \n" ); document.write( "Note that the BinomCDF(20,0.085,2) portion computes the sum B(0)+B(1)+B(2)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Here are some alternative technology options.
\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "------------------------------------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Part B\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Refer to the previous part. \n" ); document.write( "We calculated these approximate values \n" ); document.write( "B(0) = 0.16920839 \n" ); document.write( "B(1) = 0.31437624 \n" ); document.write( "They add up to 0.48358463 which rounds to 0.4836 \n" ); document.write( "This is the approximate probability of having at most 1 chip being incorrectly placed (i.e. 1 or fewer).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If you're using technology, then...
\n" ); document.write( "------------------------------------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Part C\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n = 20 \n" ); document.write( "p = 0.085 \n" ); document.write( "mean = expected value = n*p = 20*0.085 = 1.7 \n" ); document.write( "This result is exact and hasn't been rounded.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "More practice with Binomial distributions is found here \n" ); document.write( " \n" ); document.write( " |