document.write( "Question 1201704: WidgCo is a company that manufactures widgets. It is known that 1 out of every 50 widgets that the company produces is defective. A batch of 150 widgets is produced.\r
\n" ); document.write( "\n" ); document.write( "Find the probability that more than 5 widgets in the batch are defective.\r
\n" ); document.write( "\n" ); document.write( "Round your answer to 4 decimal places.\r
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Algebra.Com's Answer #836194 by ikleyn(52781) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "WidgCo is a company that manufactures widgets. It is known that 1 out of every 50 widgets
\n" ); document.write( "that the company produces is defective. A batch of 150 widgets is produced.
\n" ); document.write( "Find the probability that more than 5 widgets in the batch are defective.
\n" ); document.write( "Round your answer to 4 decimal places.
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document.write( "It is a binomial distribution probability problem.\r\n" );
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document.write( "The number of trials is n= 150. The number of successful trials is \"more than 5\",\r\n" );
document.write( "or k >= 6 (here the success is finding a defective widget). The probability of an individual success\r\n" );
document.write( "is p = 1/50 = 0.02.\r\n" );
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document.write( "Since the number of trials is very big, the theory recommends to use a normal\r\n" );
document.write( "approximation to binomial distribution.\r\n" );
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document.write( "The parameters of the normal approximation are \r\n" );
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document.write( "    - the mean =  =  =  = 1.732051 (rounded);\r\n" );
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document.write( "    - the standard deviation SD =  =  = 1.71464 (rounded).\r\n" );
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document.write( "So, we should find the area under the normal curve with the mean 1.732051 and standard deviation 1.71464282\r\n" );
document.write( "on the right of the raw score 5.5 (using the continuity correction). Use the calculator normalcdf function\r\n" );
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document.write( "                z1    z2     mean     SD       <<<---=== the formatting pattern\r\n" );
document.write( "    P = normcdf(5.5, 9999, 1.732051, 1.71464)\r\n" );
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document.write( "The answer is P = 0.014.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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\n" ); document.write( "\n" ); document.write( "Instead of using a regular calculator, you can use free of charge online calculator\r
\n" ); document.write( "\n" ); document.write( "https://onlinestatbook.com/2/calculators/normal_dist.html\r
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A post-solution note

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\n" ); document.write( "\n" ); document.write( "        As a Math problem, it is not very well posed.\r
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\n" ); document.write( "\n" ); document.write( "        From one side, the number of trials is very big (150). Thus the \"good style\" does not leave us \r
\n" ); document.write( "\n" ); document.write( "        the possibility to solve the problem as a binomial distribution and FORCES us to use \r
\n" ); document.write( "\n" ); document.write( "        a normal distribution as an APPROXIMATION.\r
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\n" ); document.write( "\n" ); document.write( "        From the other side, formal requirements for applicability of the normal distribution approximation \r
\n" ); document.write( "\n" ); document.write( "        (that are n*p >= 5) are not satisfied, so we can not expect that this approximation will be good.\r
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\n" ); document.write( "\n" ); document.write( "        It only allows to give an estimation of the order of a value, but does not allow to evaluate a value itself\r
\n" ); document.write( "\n" ); document.write( "        with a perfect precision.\r
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