Question 970256: Compute P(x) using the binomial probability formula. Then determine whether the normal distribution can be used as an approximation for the binomial distribution. If so, approximate P(x) and compare the result to the exact probability.
N=85 P=.8 x=70
I know that binom pdf= .096972206 using the calculator
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 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.
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