document.write( "Question 245191: X has a normal distribution with a mean of 75.0 and a standard deviation of 3.5. Find the following probabilities:
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document.write( "(A) P(x < 73.0)
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document.write( "(B) P(72.0 < x < 78.0)
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document.write( "(C) P(x>79.0)
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Algebra.Com's Answer #179207 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! X has a normal distribution with a mean of 75.0 and a standard deviation of 3.5. Find the following probabilities:\r \n" ); document.write( "\n" ); document.write( "(A) P(x < 73.0) \n" ); document.write( "z(73) = (73-75)/(3.5) = -2/3.5 = -0.571 \n" ); document.write( "P(x < 73) = P(z < -0.571) = 0.284 \n" ); document.write( "=============================================== \n" ); document.write( "(B) P(72.0 < x < 78.0) \n" ); document.write( "z(72) = (72-75)/3.5 = -0.857 \n" ); document.write( "z(78) = (78-75)/3.5 = +0.857 \n" ); document.write( "--- \n" ); document.write( "P(72 < x < 78) = P(-857 < z < 0.857) = 0.609 \n" ); document.write( "=============================================== \n" ); document.write( "(C) P(x>79.0) \n" ); document.write( "z(79) = (79-75)/3.5 = 1.143 \n" ); document.write( "P(x > 79) = P(z > 1.143) = 0.127 \n" ); document.write( "=============================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |