The Central Limit Theorem and the Continuity Correction for Normal Approximations: Practice Problems and Solutions

Some problems in this section will involve the Central Limit Theorem, whose statement is provided by Larsen and Marx:

"Let W₁, W₂,... be an infinite sequence of independent random variables, each with the same distribution. Suppose that the mean μ and the variance σ² of f_W(w) are both finite. Then for any numbers a and b, lim_n→∞Pr(a ≤ (W₁ + .... + W_n - nμ)/(√(n)*σ) ≤ b) = Φ(b) - Φ(a)." (302)

Here, Φ(x) = Pr(Z ≤ x), where Z is the standard normal random variable.

You can find Φ(x) in MS Excel by using the function "=NormSDist(x)".

In other words, when you have a sum of n identically distributed independent random variables, each with mean μ and variance σ², you can use the Central Limit Theorem to approximate the probability that this sum is between any two values by using the probabilities associated with the standard normal distribution. Let S_n be the sum (W₁ + .... + W_n). Then

Pr(a ≤ (S_n - nμ)/(√(n)*σ) ≤ b) ≈ Φ(b) - Φ(a).

However, when making approximations using the normal distribution, it is often necessary to use a continuity correction. This needs to happen when you use a continuous distribution such as the normal distribution to approximate a distribution where only discrete outcomes (such as integer outcomes) are possible. For instance, using the normal distribution in approximating a binomial distribution with a large number of possible discrete outcomes requires the continuity correction.

When you have an integer g, then, by the continuity correction,

Pr(More than g) = Pr(At least g + 1) = Pr(g + 1 or more) = 1 - Φ((g + 0.5 - μ)/σ).

When you have an integer g, then, by the continuity correction,

Pr(Less than g) = Pr(At most g - 1) = Pr(g - 1 or less) = Φ((g - 0.5 - μ)/σ).

Some of the problems in this section were designed to be similar to problems from past versions of the Casualty Actuarial Society's Exam 3L and the Society of Actuaries' Exam MLC. They use original exam questions as their inspiration - and the specific inspiration for each problem is cited so as to give students a chance to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

Source: Broverman, Sam. Actuarial Exam Solutions - CAS Exam 3L - Spring 2008.

Larsen, Richard J. and Morris L. Marx. An Introduction to Mathematical Statistics and Its Applications. Fourth Edition. Pearson Prentice Hall: 2006. p. 302.

Mahler, Howard. "Sample Pages for Stochastic Models (CAS 3L / SOA MLC)."

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L59-1. You are examining a binomially distributed random variable G, with mean 67 and variance 54. Find the probability that G ≤ 55. Use a normal approximation with a continuity correction.

Solution S3L59-1. We want to find the probability that G is at most 55, which, by the formula

Pr(At most g - 1) = Φ((g - 0.5 - μ)/σ) (for g = 56), is = Φ((56 - 0.5 - 67)/√(54)) = Φ(-1.56495178), which we find in MS Excel using the function "=NormSDist(-1.56495178)" = about 0.058797116.

Problem S3L59-2. You are examining a binomially distributed random variable G, with mean 127 and variance 23. Find the probability that G ≥ 129. Use a normal approximation with a continuity correction.

Solution S3L59-2. We want to find the probability that G is at least 129, which, by the formula

Pr(At least g + 1) = 1 - Φ((g + 0.5 - μ)/σ) (for g = 128) is 1 - Φ((128 + 0.5 - 127)/√(23)) =

1 - Φ(0.3127716211), which we find in MS Excel using the function "=1 - NormSDist(0.3127716211)" = about 0.377227154.

Problem S3L59-3. You are examining a binomially distributed random variable G, with mean 25 and variance 10. Find the probability that G < 28. Use a normal approximation with a continuity correction.

Solution S3L59-3. We want to find the probability that G is less than 28, which, by the formula

Pr(Less than g) = Φ((g - 0.5 - μ)/σ) (for g = 28) = Φ((28 - 0.5 - 25)/√(10)) = Φ(0.790569415), which we find in MS Excel using the function "=NormSDist(0.790569415)" = about 0.785402416.

Problem S3L59-4. The variable S is a sum of 45 independent, identically distributed, and continuous random variables for each of which the mean is 930 and the variance is 3451. Find the approximate probability that 41650 ≤ S ≤ 42150 using the Central Limit Theorem.

Solution S3L59-4. Since the random variables being added are each continuous, no continuity correction needs to be applied when using the normal approximation. We use the formula

Pr(a ≤ (S_n - nμ)/(√(n)*σ) ≤ b) ≈ Φ(b) - Φ(a). We need to find a and b.

We find nμ = 45*930 = 41850. We also find √(n)*σ = √(45*3451) = 399.1804103.

Thus, b = (42150 - 41850)/399.1804103 = about 0.7515398858 and

a = (41650 - 41850)/399.1804103 = about -0.5010265905.

Thus, we want to find Φ(0.7515398858) - Φ(-0.5010265905), which we do in MS Excel using the function "=NormSDist(0.7515398858) - NormSDist(-0.5010265905)" = about 0.465659972.

Problem S3L59-5. Similar to Question 1 from the Casualty Actuarial Society's Spring 2008 Exam 3L. Nrocinu is a missionary for the World Church of the Pink Unicorn. Nrocinu's job is to hand out pamphlets to people on the street, convincing them to believe in the Pink Unicorn. His probability of convincing any given person on the street whom he approaches is 0.03. Nrocinu will be given a promotion to Grand Unicorn if he manages to convince at least 340 people to believe in the Pink Unicorn during the course of 10 days. During each of these days, Nrocinu will hand out pamphlets to 1000 people. What is Nrocinu's probability of getting the promotion?

Solution S3L59-5. Because each person whom Nrocinu approaches has a 0.03 probability of being convinced, and Nrocinu can only convince whole numbers of people, the number of people he convinces (N) follows a binomial distribution with n = 10*1000 = 10000 and p = 0.03. Thus, μ = np = 10000*0.03 = μ = 300, and σ² = np(1-p) = 300*0.97 = σ² = 291. We want to find Pr(N ≥ 340). In approximating this probability using the normal distribution, we apply the continuity correction: Pr(At least 339 + 1) = 1 - Φ((339 + 0.5 - μ)/σ) = 1 - Φ((339 + 0.5 - 300)/√(291)) = 1 - Φ(2.315531008) ), which we find in MS Excel using the function "=1 - NormSDist(2.315531008)" = about 0.010291919.

See other sections of The Actuary's Free Study Guide for Exam 3L.