# Question

In Chapter 6, you will see that the normal distribution can sometimes be used to approximate the binomial distribution. This exercise is designed to illustrate how good the normal approximation to the binomial can be.

a. Consider a binomial distribution with n = 200 and p = 0.5.Calculate P(x ≤ 100).

b. What is the mean and standard deviation for the distribution you used in part a?

c. Using a normal distribution with the same mean and standard deviation you calculated in part b, calculate P(x ≤ 100).

d. The probabilities you calculated in parts a and c should be similar. Why do you think this is so?

e. Now calculate P(x ≤ 100.5) for the normal distribution. The extra 0.5 is added to provide a “continuity correction factor.”It compensates for the fact that the normal distribution is continuous (so P(x ≤ 100) = P(x < 100), for example), and the binomial distribution is not (P(x ≤ 100) ≠ P(x < 100)).How close are the binomial and normal probabilities now?

a. Consider a binomial distribution with n = 200 and p = 0.5.Calculate P(x ≤ 100).

b. What is the mean and standard deviation for the distribution you used in part a?

c. Using a normal distribution with the same mean and standard deviation you calculated in part b, calculate P(x ≤ 100).

d. The probabilities you calculated in parts a and c should be similar. Why do you think this is so?

e. Now calculate P(x ≤ 100.5) for the normal distribution. The extra 0.5 is added to provide a “continuity correction factor.”It compensates for the fact that the normal distribution is continuous (so P(x ≤ 100) = P(x < 100), for example), and the binomial distribution is not (P(x ≤ 100) ≠ P(x < 100)).How close are the binomial and normal probabilities now?

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