Question

6. X is binomial with n = 500 and p =.38. Use the standard normal distribution to approximate: a. P(172 SX S216) Ansba b. P(x > 220) Ans 6b What allows us to use the standard normal distributi approximate binomial probabilities when n is large Ans 6c

Answer 1

6. Here mean=np=500*0.38=190 and standard deviation is $\sigma=\sqrt{npq}=10.8536$

a. We need to find $P(172\leq X\leq 216)$

As we are asked to use normal approximation, we can convert x to z

$P(172\leq X\leq 216)=P(\frac{172-190}{10.8536}<z<\frac{216-190}{10.8536})=P(-1.66<z<2.40)=0.9433$

b. $P(X>220)=P(z>\frac{220-190}{10.8536})=P(z>2.76)=0.0029$

c, As np>5 and nq>5 so normal approximation is satisfied.