Solution :
Given that ,
mean = = 14.2
standard deviation = =0.90
a) P(x >16.8 ) = 1 - p( x< 16.8 )
=1- p [(x - ) / < (16.8 - 14.2) /0.90 ]
=1- P(z < 2.89 )
= 1 - 0.9981 = 0.0019
probability = 0.0019
b)
n = 115
= = 14.2
= / n = 0.90 / 115 = 0.0839
P( > 16.8 ) = 1 - P( <16.8 )
= 1 - P[( - ) / < (16.8 -14.2) /0.0839 ]
= 1 - P(z < 30.99)
= 1 -1 = 0
Probability = 0
c) The result from part (a) should be considered only average individuals should be considered .
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