Question 1193053
The mean is np=100(1/6)=16.67 people
variance is np*(1-p)=(100/6)*(5/6)=500/36 people ^2; the 5/6 is 1-p
sd is sqrt (V)=sqrt(500/6)=3.727 people.
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This would be a surprising finding with 50 people's being 8.94 sd above the expected value.

The population distribution is mean=p (0.167) 
The data distribution is mean 0.5, 
The difference (for z testing ) is( phat-p )=z*sqrt (p*(1-p)/n);note the parameter tested against is under the radical.