1. Normal approximation to Binomial Only applicable for: Large n P not to small or large ie near 0.5 X~N(np, npq) Find the probability of obtaining 5 heads with 12 tosses of a fair coin X~B(n, p) ie X~B(12,0.5) P(r) = n C r (p) r (q) n-r = 0.1934 P(5) = 12 C 5 (0.5) 5 (0.5) 7

2.  X~N(6,3) Z=5.5 - 6 = -0.2887  3 Z=4.5 - 6 = -0.866  3 Therefore P(x=5) in the Binomial is the same as P(4.5<x<5.5) in the Normal The Binomial Distribution is a DISCRETE Distribution But the Normal Distribution is a CONTINUOUS Distribution Therefore while in the Binomial Distribution x=5 has the width of a “bar” in the Normal it has no width To create width we say that 5 is 4.5-5.5 X~N(np, npq) P(x<5.5)=1-0.6138 = 0.3862 P(x<4.5)=1-0.8067 =0.1933 P(4.5<x<5.5) = 0.3862 - 0.1933 = 0.1929

3. Normal approximation to Binomial Only applicable for: Large n P not to small or large ie near 0.5 X~N(np, npq) Find the probability of obtaining between 4 and 7 heads inclusive with 12 tosses of a fair coin X~B(n, p) ie X~B(12,0.5) F(7) - F(3) = 0.8062 - 0.0730 = 0.7332

4.  X~N(6,3) Z=7.5 - 6 = 0.866  3 Z=3.5 - 6 = -1.4434  3 The Binomial Distribution is a DISCRETE Distribution But the Normal Distribution is a CONTINUOUS Distribution Therefore 4 – 7 inclusive is actually 3.5 – 7.5 X~N(np, npq) P(x<7.5)= 0.8067P(x>3.5)=0.9255 P(4<x<7) = 0.8067 - 0.0745 = 0.7332  P(x<3.5)=0.0745