document.write( "Question 1050446: The mean incubation time of fertilized eggs is 19 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day.\r
\n" ); document.write( "\n" ); document.write( "​(a) Determine the 16th percentile for incubation times.
\n" ); document.write( "​(b) Determine the incubation times that make up the middle 97% of fertilized eggs.\r
\n" ); document.write( "\n" ); document.write( "How can I solve this (preferably) using a TI calculator?
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Algebra.Com's Answer #666030 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
a) z = invNorm(.16)-.9945 Use Your TI
\n" ); document.write( "\"z+=blue+%28x+-+mu%29%2Fblue%28sigma%29\"
\n" ); document.write( "0r
\n" ); document.write( " \"blue%28sigma%29%2Az+%2B+mu=blue+%28x%29\"
\n" ); document.write( " \"1%2A-.9945+%2B+19+=+18.0054\" = 18days
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\n" ); document.write( "b middle 97%: from .015 to .985 \"%28100-.97%29%2F2+=+.015\"
\n" ); document.write( "z = invNorm( 98.5)= 2.17 and invNorm(.015)= -2.17
\n" ); document.write( " P(-2.17 ≤ z ≤ 2.17).
\n" ); document.write( "TI syntax is normalcdf(smaller z, larger z)
\n" ); document.write( "P(-2.17 ≤ z ≤ 2.17) = normalcdf(-2.17, 2.17) Use Your TI\r
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