3. (1 5 points) In a 1999 survey of teachers by USA Today revealed that 85% of t
ID: 3130750 • Letter: 3
Question
Explanation / Answer
3.The required probability is P(X>=1)
=1-P(X=0)
=1-30C0*(1-0.85)^0*0.85^30
=1-0.007=0.993
4.The mean is np=15*0.28=4.2
The standard deviation is: sq rt of np(1-p)=sq rt of 4.2*(1-0.28)=1.73
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