document.write( "Question 1197456: Really Big Medical Center (RBMC) is testing patients who have spent time in grassy areas this summer and fall for Lyme disease. Suppose that the probability of someone with exposure to tall grass having Lyme is .3 (or 30 percent). In a sample of 15 patients, what is the standard deviation of the number of patients with Lyme disease? Round your answer to 1 decimal place. \n" ); document.write( "
Algebra.Com's Answer #830740 by ikleyn(52848)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "You are given a Binomial distribution with n= 15 trials and p= 0.3 of the
\n" ); document.write( "success probability for each of 15 independent trials.\r
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\n" ); document.write( "\n" ); document.write( "To find the standard deviation of a binomial distribution \r
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\n" ); document.write( "\n" ); document.write( "        Compute the variance as n × p × (1-p) , where n is the number of trials
\n" ); document.write( "        and p is the probability of success.
\n" ); document.write( "        Take the square root of the number obtained in Step 1. \r
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\n" ); document.write( "\n" ); document.write( "Thus for the given problem the standard deviation is \r
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\n" ); document.write( "\n" ); document.write( "            \"sqrt%2815%2A0.3%2A%281-0.3%29%29\" = \"sqrt%283.15%29\" = 1.8     (rounded).         ANSWER
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