Question 610781
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Hi, 
body temperature of healthy children is normally distributed with mean 98.6 F and standard deviation 0.9 F.
 If one healthy child is selected at random:
P(temp>99.6)= P(z> (99.9-98.6)/.9) = P(z> 1.7778) = .0377
If six healthy children are chosen at random: 
P(6 with temp >99.6) = (.0377)^6