Normal Distribution (Further)

This lesson extends your understanding of the normal distribution to cover linear combinations of normal random variables, the distribution of sample means, and the foundations for inference. These results are essential for hypothesis testing and confidence intervals.

---

Recap: The Normal Distribution

If $X \sim N(\mu, \sigma^2)$X∼N(μ,σ2):

$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}}$f(x)=σ2π​1​e−2σ2(x−μ)2​

Standardisation: $Z = \frac{X - \mu}{\sigma} \sim N(0, 1)$Z=σX−μ​∼N(0,1).

---

Linear Combinations of Normal Variables

If $X \sim N(\mu_X, \sigma_X^2)$X∼N(μX​,σX2​) and $Y \sim N(\mu_Y, \sigma_Y^2)$Y∼N(μY​,σY2​) are independent:

Combination	Distribution
$aX + b$aX+b	$N(a\mu_X + b, \; a^2\sigma_X^2)$N(aμX​+b,a2σX2​)
$X + Y$X+Y	$N(\mu_X + \mu_Y, \; \sigma_X^2 + \sigma_Y^2)$N(μX​+μY​,σX2​+σY2​)
$X - Y$X−Y	$N(\mu_X - \mu_Y, \; \sigma_X^2 + \sigma_Y^2)$N(μX​−μY​,σX2​+σY2​)
$aX + bY$aX+bY	$N(a\mu_X + b\mu_Y, \; a^2\sigma_X^2 + b^2\sigma_Y^2)$N(aμX​+bμY​,a2σX2​+b2σY2​)

Critical point: For $X - Y$X−Y, the variances are added, not subtracted. This is the most common error in this topic.

Worked Example 1

The weight of a large egg is $X \sim N(68, 16)$X∼N(68,16) grams and a small egg is $Y \sim N(50, 9)$Y∼N(50,9) grams, independently.

Find the probability that a large egg weighs more than 15 grams more than a small egg.

We need $P(X - Y > 15)$P(X−Y>15). Now $X - Y \sim N(68 - 50, 16 + 9) = N(18, 25)$X−Y∼N(68−50,16+9)=N(18,25).

$P(X - Y > 15) = P\left(Z > \frac{15 - 18}{5}\right) = P(Z > -0.6) = \Phi(0.6) = 0.7257$P(X−Y>15)=P(Z>515−18​)=P(Z>−0.6)=Φ(0.6)=0.7257

Worked Example 2

Packets of flour weigh $X \sim N(1005, 100)$X∼N(1005,100) grams. A box holds 12 packets. Find the probability that the total weight exceeds 12100 grams.

$T = X_1 + X_2 + \cdots + X_{12} \sim N(12 \times 1005, 12 \times 100) = N(12060, 1200)$T=X1​+X2​+⋯+X12​∼N(12×1005,12×100)=N(12060,1200).

$P(T > 12100) = P\left(Z > \frac{12100 - 12060}{\sqrt{1200}}\right) = P\left(Z > \frac{40}{34.64}\right) = P(Z > 1.155)$P(T>12100)=P(Z>1200​12100−12060​)=P(Z>34.6440​)=P(Z>1.155)

$= 1 - \Phi(1.155) \approx 1 - 0.876 = 0.124$=1−Φ(1.155)≈1−0.876=0.124

Exam Tip: When adding $n$n independent identical normal variables, the mean is multiplied by $n$n and the variance is multiplied by $n$n (not $n^2$n2). This is different from scaling a single variable by $n$n, which multiplies the variance by $n^2$n2.

---

Sum vs Scalar Multiple

Be very careful to distinguish: