Mean=18×⅔=12; SD=√(12(1-⅔))=2.
We use these values to apply to normal distribution. We are using a continuous distribution (normal) to approximate to a discreet distribution (binomial).
(1) For exact value X=7, we use the interval [6.5,7.5] to compute two Z scores, Z₁ and Z₂.
Z₁=(6.5-12)/2=-2.75; Z₂=(7.5-12)/2=-2.25.
p(-2.75)=0.0030; p(-2.25)=0.0122.
Difference is 0.009 or 0.9% approx. This is the normal approximation to the binomial distribution.
(2) Z₁=(8.5-12)/2=-1.75, Z₂=(13.5-12)/2=0.75. p(-1.75)=0.0401; p(0.75)=0.7734.
Difference is 0.7333, or 73.3% approx.
(3) Since 12 is the mean, it splits the distribution in half, so 50% will be 12 or above. But we need to work out what’s above, so we start at 12.5. Z=(12.5-12)/2=0.25. This corresponds to a probability of 59.9% approx.
