document.write( "Question 133641This question is from textbook Statistics
\n" ); document.write( ": I want to see if I did this problem right:
\n" ); document.write( "The average afe of CEO's is 56. Assume the cariable is normally distrubited. If the standard deviation is 4 years, find the probability that the age of a randomly selected CEO will be in the following range:
\n" ); document.write( "Between 53 and 59 years old.\r
\n" ); document.write( "\n" ); document.write( "This is how I did it:
\n" ); document.write( "53-56/4 = -.75 then I looked at my chart and found that the average was .2580\r
\n" ); document.write( "\n" ); document.write( "59-56/4 = .75 then I looked at the chart and found the average was .2580 \r
\n" ); document.write( "\n" ); document.write( "then I added the 2 averages together: .2580 + .2580 = .516 = 51.6%
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Algebra.Com's Answer #98717 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
The average afe of CEO's is 56. Assume the cariable is normally distributed. If the standard deviation is 4 years, find the probability that the age of a randomly selected CEO will be in the following range:
\n" ); document.write( "Between 53 and 59 years old.
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\n" ); document.write( "Find the z-score for 53 and for 59:
\n" ); document.write( "z(53) = (53-56)/4 = -0.75
\n" ); document.write( "z(59) = (59-56)/4 = 3/3 = +0.75
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\n" ); document.write( "P(53 < x < 59) = P(-.075 < z < 0.75) = 0.5467
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\n" ); document.write( "Let me know if you do not understand all of this.
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\n" ); document.write( "If you have a TI calculator use normalcdf(53,59,56,4) = 0.5467
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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