document.write( "Question 1031822: I am a little stuck from this point\r
\n" ); document.write( "\n" ); document.write( "A survey found that people keep their television sets an average of 4.8 years. The standard deviation is 0.89 year, and assume that this variable (length of time) is normally distributed. If a person decides to by a new TV set, find the probability that he or she has owned the old set for the following amount of time:Z=X-µ/б
\n" ); document.write( " a. Less than 2.5 years
\n" ); document.write( "-P(X<2.5) Z=2.5-4.8/0.89
\n" ); document.write( " b. Between 3 and 4 years
\n" ); document.write( "-P(3\n" ); document.write( " \r
\n" ); document.write( "\n" ); document.write( "c. More than 4.2 years
\n" ); document.write( "-P(X>4.2)
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Algebra.Com's Answer #646550 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi,
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document.write( "Using Excel function NORMSDIST
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document.write( "P(X <2.5)= P( z = (2.5-4.8)/0.89) = P(z = -2.584)) = NORMSDIST(-2.5841) = .00493
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document.write( "Let You finish these up ...
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document.write( "P(3 ≤ x ≤ 4) = P(x< 4) - P(x<3)= P( z = (4-4.8)/0.89) - P( z = (3-4.8)/0.89)
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document.write( "P(X > 4.2) = 1 - P(x < 4.2)) = 1 - P( z = (4.2-4.8)/0.89 ) =  
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document.write( "1 - NORMSDIST(-.8989) = 1 - .184 = .816 Or 81.6% \n" );
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