Difficulty: Intermediate | Prerequisites: normal distribution basics, z-scores, confidence intervals for means (Doc 1 of this set).
Once you know how a single normal random variable behaves, the next step is combining them: what happens when you subtract one normal from another, or average several? This is how you compare two options (which train is faster?) or two groups (which candidate is ahead?). This material also introduces confidence intervals for proportions, which work slightly differently from CIs for means because the variance of a proportion depends on the proportion itself. If you are comfortable with z-scores, the normal PDF, and basic confidence intervals, you are ready for this.
The difference of two independent normals is itself normal, with mean = difference of means and variance = sum of variances. This lets you compute the probability one variable exceeds another. For proportions, the confidence interval uses p-hat * (1 - p-hat) / n as the estimated variance.
Linear combination of normal random variables
If X ~ N(mu_X, sigma_X^2) and Y ~ N(mu_Y, sigma_Y^2) are independent, then aX + bY ~ N(amu_X + bmu_Y, a^2sigma_X^2 + b^2sigma_Y^2).
In simple terms, adding or subtracting independent normal variables gives you another normal variable, and you can work out its mean and variance using simple rules.
Difference of two independent normals
A special case of the above: X - Y ~ N(mu_X - mu_Y, sigma_X^2 + sigma_Y^2). Note the variances add even when you subtract.
Think of it as: uncertainty from both sources piles up regardless of the direction.
Sample proportion (p-hat)
The fraction of successes in a sample: p-hat = X / n, where X is the number of successes. For large n, p-hat is approximately normal.
In simple terms, this is the percentage you observe in your sample, used to estimate the true population percentage.
Standard error of a proportion
SE(p-hat) = sqrt(p-hat * (1 - p-hat) / n). Unlike the mean's SE, this depends on the estimated proportion itself.
Think of it as: the spread of your estimate is largest when p-hat is near 0.5 and shrinks as p-hat approaches 0 or 1.
Confidence interval for a proportion
An interval p-hat +/- z* * SE(p-hat) that, in repeated sampling, would contain the true proportion a stated percentage of the time.
In simple terms, the same idea as a confidence interval for a mean, but with a different formula for the spread.
Setup: the subway problem (PS4 Q3)
R train: mean 30 min, SD 6 min. So R ~ N(30, 36).
6 train: mean 32 min, SD 5 min. So T6 ~ N(32, 25).
The two are independent.
Probability the 6 train is faster on a single day (Q3a)
Want P(T6 < R), equivalently P(T6 - R < 0).
D = T6 - R ~ N(32 - 30, 25 + 36) = N(2, 61).
SD of D = sqrt(61) ≈ 7.81.
Standardise: Z = (0 - 2) / 7.81 = -0.256.
P(Z < -0.256) ≈ 0.399, so roughly 40% of the time the 6 train is faster.
Comparing sample means over 10 trips (Q3b)
Average of 10 R-train trips: R-bar ~ N(30, 36/10) = N(30, 3.6).
Average of 10 6-train trips: T6-bar ~ N(32, 25/10) = N(32, 2.5).
D-bar = T6-bar - R-bar ~ N(2, 3.6 + 2.5) = N(2, 6.1).
SD of D-bar = sqrt(6.1) ≈ 2.47.
Z = (0 - 2) / 2.47 = -0.81.
P(Z < -0.81) ≈ 0.209, so about a 21% chance the 6 train beats the R on average over two weeks.
Key point: averaging over more trips makes the 6 train's disadvantage (higher mean) harder to overcome.
Which train to catch a 45-minute deadline (Q3c)
R train: P(R <= 45) = P(Z <= (45 - 30)/6) = P(Z <= 2.5) = 0.9938.
6 train: P(T6 <= 45) = P(Z <= (45 - 32)/5) = P(Z <= 2.6) = 0.9953.
The 6 train gives a slightly higher probability of arriving on time (99.53% vs 99.38%), despite its higher mean, because its lower variance keeps it more tightly clustered.
Takeaway: when you care about the tail (will I be late?), variance matters as much as the mean.
Setup: the polling problem (PS4 Q4)
Population of 300 million. Poll of n = 1000. Democratic candidate: 510 (p-hat_D = 0.51). Republican: 490 (p-hat_R = 0.49).
95% CI for each candidate's support (Q4a)
SE for Democratic candidate: sqrt(0.51 * 0.49 / 1000) = sqrt(0.0002499) ≈ 0.01581.
95% CI: 0.51 +/- 1.96 * 0.01581 = (0.479, 0.541).
SE for Republican candidate: sqrt(0.49 * 0.51 / 1000) ≈ 0.01581 (same value, since p-hat * (1 - p-hat) is the same).
95% CI: 0.49 +/- 1.96 * 0.01581 = (0.459, 0.521).
Both intervals overlap and both contain 0.5, so you cannot conclude either candidate is ahead at 95% confidence.
Probability of observing 51%+ for the Democrat if p = 0.5 (Q4b)
Under p = 0.5: SE = sqrt(0.5 * 0.5 / 1000) = sqrt(0.00025) ≈ 0.01581.
Z = (0.51 - 0.50) / 0.01581 = 0.632.
P(p-hat >= 0.51) = P(Z >= 0.632) ≈ 0.264.
There is roughly a 26% chance of seeing 51% or more support for the Democrat in a sample of 1000 even when the population is perfectly split.
This illustrates why small leads in polls are often not statistically significant.
D = X - Y \sim N(\mu_X - \mu_Y,\; \sigma_X^2 + \sigma_Y^2) \quad \text{(independent normals)}\text{SE}(\hat{p}) = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\text{CI for } p = \hat{p} \pm z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}Students often subtract the variances when computing the variance of X - Y. The variances always add for independent variables, regardless of the sign.
Students sometimes use the population proportion p in the SE formula when constructing a CI. In practice, you use p-hat because p is unknown. (You use p under the null hypothesis only for hypothesis tests.)
A common error is assuming the candidate with the higher sample proportion is "ahead." If the confidence intervals overlap and both contain 0.5, the lead is not statistically significant.
Students sometimes forget that the normal approximation for proportions requires np-hat and n(1 - p-hat) both to be at least 5 (some texts say 10). Always check this before using the formula.
⚠️ "Variance of the difference" questions almost always test whether you add or subtract the variances. The answer is always add (for independent variables).
⚠️ Comparing two options under a deadline ("which maximises P(X <= threshold)?") is a common exam format. You need to compute two separate probabilities and compare.
⚠️ Proportion CI questions may ask you to interpret overlapping intervals. Overlapping does not automatically mean no difference, but it strongly suggests the difference is not significant.
⚠️ Be ready to compute the probability of a sample result under a null hypothesis (as in Q4b). This is the conceptual bridge to p-values and hypothesis testing.
True or false: Var(X - Y) = Var(X) - Var(Y) when X and Y are independent. ___
Fill in the blank: If X ~ N(10, 4) and Y ~ N(7, 9) independently, then X - Y ~ N(___, ___).
True or false: The standard error of a sample proportion is largest when p-hat = 0.5. ___
Fill in the blank: A 95% CI for a proportion with p-hat = 0.51 and n = 1000 has a margin of error of approximately ___.
True or false: If two 95% CIs for candidate support both contain 0.5, the election is a statistical tie. ___
Answers: 1. False (variances add). 2. N(3, 13). 3. True. 4. About 0.031 (1.96 * 0.0158). 5. True in the sense that neither lead is statistically significant, though the point estimates differ.
Q: The R train has mean 30 min (SD 6) and the 6 train has mean 32 min (SD 5). What is the probability the 6 train is faster on a given day?
A: D = T6 - R ~ N(2, 61). P(D < 0) = P(Z < -2/sqrt(61)) = P(Z < -0.256) ≈ 0.399, or about 40%.
Q: Over 10 trips on each line, what is the probability the 6 train has a lower average time?
A: D-bar ~ N(2, 6.1). P(D-bar < 0) = P(Z < -2/sqrt(6.1)) = P(Z < -0.81) ≈ 0.209, or about 21%.
Q: You need to arrive in 45 minutes. Which train gives you the best chance?
A: R train: P(R <= 45) = P(Z <= 2.5) = 0.9938. 6 train: P(T6 <= 45) = P(Z <= 2.6) = 0.9953. The 6 train is slightly better because its lower variance more than compensates for its higher mean at this threshold.
Q: In a poll of 1000, 510 support the Democrat. Construct a 95% CI for the Democrat's true support.
A: SE = sqrt(0.51 * 0.49 / 1000) ≈ 0.0158. CI = 0.51 +/- 1.96 * 0.0158 = (0.479, 0.541).
Q: If the true split is 50/50, what is the probability of seeing 51%+ for the Democrat in a sample of 1000?
A: SE = sqrt(0.25/1000) ≈ 0.0158. Z = (0.51 - 0.50)/0.0158 = 0.632. P(Z >= 0.632) ≈ 0.264, or about 26%.
The difference-of-normals technique is the basis for two-sample t-tests and z-tests, which you will encounter when comparing treatment and control groups in experiments.
Confidence intervals for proportions connect directly to hypothesis testing for proportions (the one-sample z-test for p). The computation in Q4b is, in effect, a p-value calculation.
The subway problem (choosing between mean and variance) foreshadows decision theory and risk analysis, where you trade off expected value against tail risk.
Difference of normals, sum of normals, linear combination, independent random variables, variance of a difference, confidence interval for proportion, sample proportion, p-hat, standard error of proportion, polling, margin of error, normal approximation to binomial, z-test for proportions, two-sample comparison, tail probability, Analytical Statistics, ASTATS, NYU, PS4, subway problem, election polling