document.write( "Question 624015: I am having trouble understanding \"at most\" and \"at least\" \r
\n" ); document.write( "\n" ); document.write( "If, for example, the problem was \" ... Their times are normally distributed with a mean of 184 minutes and a standard deviation of 55 minutes. For a randomly selected person from the control group, find the probability that the time is at least 210 minutes.\" \r
\n" ); document.write( "\n" ); document.write( "I used the following function on my TI-84:\r
\n" ); document.write( "\n" ); document.write( "2ND, VARS, normalcdf(209.5,1E99,184,55)
\n" ); document.write( "=0.321454
\n" ); document.write( "=32.14% \r
\n" ); document.write( "\n" ); document.write( "is this correct?\r
\n" ); document.write( "\n" ); document.write( "Thanks for your time. \n" ); document.write( "
Algebra.Com's Answer #392496 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( "for at least 210, using 209.5 is the correct spproach
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document.write( " z = (209.5 - 184)/25 = .4636 which results in  .6786  or 67.86% to the left
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document.write( "Note: we know as 210 >184 that this P has to be greater than 50% for 
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document.write( "Also as to Your P(x≥210) = \"highlight%28.3214%29\" Note (1-.3214) = .6786
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document.write( "Important to Understand z -values as they relate to the Standard Normal curve:
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document.write( "Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
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document.write( "Note: z = 0 (x value the mean) 50% of the area under the curve is to the left and %50 to the right
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