document.write( "Question 1070493: When a California household is randomly selected, the number of televisions and the corresponding probabilities are: 0(.03);1(.15);2(.29);3(.26);4(.16); 5(.11)
\n" ); document.write( "Verify that this qualifies as a probability distribution. Then find the average number of TV's per household and the standard deviation. Is it unusual for a household in California to not have a television?
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Algebra.Com's Answer #685605 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
If it's a distribution,
\n" ); document.write( "\"0%3C=P%28x%29%3C=1\"
\n" ); document.write( "\"sum%28P%29=1\"
\n" ); document.write( "First one, check.
\n" ); document.write( "Second one,
\n" ); document.write( "\".03%2B.15%2B.29%2B.26%2B.16%2B.11=1\"
\n" ); document.write( "Check.
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\n" ); document.write( ".\r
\n" ); document.write( "\n" ); document.write( "\"mu=sum%28x%2AP%28x%29%29=0%2A.03%2B1%2A.15%2B2%2A.29%2B3%2A.26%2B4%2A.16%2B5%2A.11=2.7\"
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\n" ); document.write( "\"sigma=1.29\"
\n" ); document.write( "So then a value of \"X=0\" would be
\n" ); document.write( "\"2.7-N%2A1.29=0\"
\n" ); document.write( "\"N=2.1\"
\n" ); document.write( "This value is more than two standard deviations away from the mean.
\n" ); document.write( "Yes, it's unusual.
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