Question 970256
Formula (85 choose 70)* 0.8^70*0.2^15=0.097, as you obtained.



np=mean  np(1-p)=V

E(x)=68; 85 * 0.8
V(x)=85*0.8*0.2=13.6
SD (x)=sqrt (13.6)=3.69

Normal distribution: 
z=(70-68)/3.69 =2/3.69 or 0.542, The probability >= is 0.2939.

The problem I have here is that the normal distribution would be the z-value for everything >=70, not just the point 70, which actually mathematically doesn't exist.  We are comparing a density function with a non-density function.  

If you do the binomial probabilities from x=70-x=85, you should get a reasonable approximation.  It turns out that the cumulative probability is approaching 0.29 (the probabilities of getting more than 80 are small, and most of the probability comes with x=70-75.  

The normal distribution can be approximated in this instance if np>5 (or 10, depending upon the source).  Here, it is, but it does not count with one value.