document.write( "Question 1195864: About 20% of all professional foot ball players are injured during a given season a team has four star players a) what is the probability that at list two of the star players gets injured?
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Algebra.Com's Answer #828454 by ikleyn(52803)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "                        Part  (a)\r
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document.write( "This problem is a binomial distribution with the parameters\r\n" );
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document.write( "    n = 4    (the number of trials);\r\n" );
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document.write( "    k >= 2   (the number of success);\r\n" );
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document.write( "    p = 0.2  (the probability of individual success).\r\n" );
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document.write( "The probability is  \r\n" );
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document.write( "    P = P(2) + P(3) + P(4) = P(n=4, k>=2, p=0.2) = \"sum%28C%5B4%5D%5Ek%2Ap%5Ek%2A%281-p%29%5E%28n-k%29%2C+k=2%2C4%29\" = \"sum%28C%5B4%5D%5Ek%2A0.2%5Ek%2A0.8%5E%284-k%29%2C+k=2%2C4%29\".\r\n" );
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document.write( "     To facilitate my calculations, I used online calculator at this site  https://stattrek.com/online-calculator/binomial.aspx\r\n" );
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document.write( "     It provides nice instructions  and  a convenient input and output for all relevant options/cases.\r\n" );
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document.write( "          The resulting number is P = 0.1808  (rounded).    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "                        Part  (b)\r
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document.write( "The binomial distribution has the following properties:\r\n" );
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document.write( "    The mean of the distribution (μx) is equal to n * P  = 4*0.2 = 0.8.\r\n" );
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document.write( "    The variance (σ2x) is n * P * ( 1 - P ) = 4*0.2*0.8 = 0.64.\r\n" );
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document.write( "    The standard deviation (σx) is sqrt[ n * P * ( 1 - P ) ] = \"sqrt%280.64%29\" = 0.8.\r\n" );
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document.write( "For the reference,  see this web-site\r\n" );
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document.write( "https://stattrek.com/probability-distributions/binomial#:~:text=The%20binomial%20distribution%20has%20the,(%201%20%2D%20P%20)%20%5D.\r\n" );
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document.write( "or your textbook.\r\n" );
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