Normal Distribution: Z-Scores, Standard Normal, and CLT for Exam P

The Normal Distribution

If X ~ Normal(mu, sigma^2), its PDF is f(x) = (1 / (sigma * sqrt(2*pi))) * exp(-(x - mu)^2 / (2*sigma^2)). The distribution is symmetric about mu, with mean = median = mode = mu and variance sigma^2. About 68% of values fall within one standard deviation of the mean, 95% within two, and 99.7% within three.

The standard normal Z = (X - mu) / sigma has mean 0 and variance 1. All normal probability calculations reduce to looking up values in the standard normal table (or using the function Phi). P(X <= x) = Phi((x - mu) / sigma).

The normal distribution has several important properties beyond symmetry. Its skewness is 0 and its excess kurtosis is 0, making it the reference distribution for measuring asymmetry and tail heaviness. The MGF is M(t) = exp(mu * t + sigma^2 * t^2 / 2). All odd central moments are zero (by symmetry), and the even central moments follow the pattern E[(X - mu)^(2k)] = (2k)! / (2^k * k!) * sigma^(2k).

Z-Score Calculations

For Exam P, you must be comfortable converting between X-values and Z-scores. If claim amounts are Normal(500, 100^2), the probability that a claim exceeds 650 is P(X > 650) = P(Z > 1.5) = 1 - Phi(1.5) = 1 - 0.9332 = 0.0668.

To find a percentile, invert: the 90th percentile of X is mu + z_{0.90} * sigma = 500 + 1.2816 * 100 = 628.16. Exam P provides a normal table, so memorize key values: z_{0.90} = 1.282, z_{0.95} = 1.645, z_{0.975} = 1.960, z_{0.99} = 2.326.

Because the normal distribution is symmetric, Phi(-z) = 1 - Phi(z). This means P(Z < -1.5) = P(Z > 1.5) = 0.0668. Use this symmetry to reduce table lookups. Also, P(-a < Z < a) = 2 * Phi(a) - 1. For example, P(-1.96 < Z < 1.96) = 2 * 0.975 - 1 = 0.95, which is the basis for the 95% confidence interval.

On Exam P, some problems give you Phi(z) for a specific z and ask you to work backwards. For instance, if P(X > c) = 0.10, then Phi((c - mu)/sigma) = 0.90, so (c - mu)/sigma = 1.282, giving c = mu + 1.282 * sigma. Practice both forward (probability given a value) and inverse (value given a probability) calculations.

Linear Combinations

If X1, X2, ..., Xn are independent normal random variables, any linear combination a1*X1 + a2*X2 + ... + an*Xn is also normally distributed with mean = sum(ai*mu_i) and variance = sum(ai^2 * sigma_i^2). This property is unique to the normal distribution and simplifies many Exam P problems involving sums or differences of normal variables.

A common application involves the difference of two normal variables. If X ~ Normal(100, 25) and Y ~ Normal(80, 16) independently, then X - Y ~ Normal(20, 41). To find P(X > Y), compute P(X - Y > 0) = P(Z > (0 - 20)/sqrt(41)) = P(Z > -3.12) = Phi(3.12) = 0.9991.

The sample mean of n iid Normal(mu, sigma^2) variables is Normal(mu, sigma^2/n). This exact result (not just an approximation) holds because of the linear combination property. It is the basis for many statistical inference procedures and appears in Exam P problems involving averages of normal populations.

The Central Limit Theorem Connection

The Central Limit Theorem (CLT) states that the sum or average of many independent, identically distributed random variables is approximately normal, regardless of the original distribution, provided the sample size is large enough (typically n >= 30). This makes the normal distribution the workhorse of probability and statistics.

On Exam P, CLT problems specify a non-normal distribution and a large sample size, then ask for an approximate probability using the normal distribution. The steps are: (1) compute the mean and variance of the individual variable, (2) find the mean and variance of the sum or average, (3) standardize and use the normal table. Practice applying the CLT with and without the continuity correction for discrete underlying distributions.

Example: Claim processing times have mean 15 minutes and standard deviation 6 minutes (distribution unknown). For 100 independent claims, find the probability that the total processing time exceeds 1560 minutes. The total has mean 1500 and standard deviation sqrt(100) * 6 = 60. Z = (1560 - 1500) / 60 = 1.0. P(total > 1560) = 1 - Phi(1.0) = 0.1587.

Common Pitfalls

A major source of errors is confusing sigma with sigma^2. The normal distribution is parameterized by the mean mu and the variance sigma^2, but many formulas use the standard deviation sigma. When a problem says "X is normal with mean 10 and variance 25," the standard deviation is 5, not 25. Always clarify which parameter you are working with before computing.

Another pitfall is forgetting that the normal distribution extends from negative infinity to positive infinity. When modeling quantities that must be positive (like claim amounts), the normal distribution can assign positive probability to negative values. For large mu relative to sigma, this probability is negligible, but it is a theoretical concern that sometimes appears in Exam P as a conceptual question.

Students sometimes try to compute P(X = c) for a continuous normal variable. Remember that P(X = c) = 0 for any specific value c. The normal distribution assigns probability only to intervals, not to individual points.

Study Tips

Commit the key z-values to memory: 1.282 (90th percentile), 1.645 (95th), 1.960 (97.5th), and 2.326 (99th). Practice converting between X and Z in both directions. Work through at least 10 linear combination problems, including sums, differences, and weighted averages. When studying CLT applications, practice both the "sum" version and the "average" version and make sure you apply the correct variance formula for each. On exam day, normal distribution problems should take two to three minutes at most, so build speed through repetition.