Normal Distribution Calculations

Worked Examples

Example 1: Heights

Adult male heights are normally distributed with $\mu = 175$ cm, $\sigma = 7$ cm. What proportion are over 190 cm?

Standardise: $z = \frac{190 - 175}{7} = \frac{15}{7} \approx 2.14$

$P(X > 190) = P(Z > 2.14) = 1 - \Phi(2.14)$

From tables: $\Phi(2.14) \approx 0.9838$.

$P(X > 190) = 1 - 0.9838 = 0.0162$

About 1.62% exceed 190 cm.

Example 2: Manufacturing

A machine produces rods with mean length 100 cm and SD 0.5 cm. Specifications require lengths between 99 cm and 101 cm. What proportion are within spec?

$P(99 \leq X \leq 101) = P\left(\frac{99-100}{0.5} \leq Z \leq \frac{101-100}{0.5}\right) = P(-2 \leq Z \leq 2)$

$= \Phi(2) - \Phi(-2) = 0.9772 - 0.0228 = 0.9544$

About 95.44% are within specification.

Example 3: Finding a Value

What height is exceeded by 10% of adult males?

We want $x$ such that $P(X > x) = 0.10$.

$P(X > x) = 0.10 \Rightarrow P(Z > z) = 0.10$

From tables: $P(Z > 1.282) \approx 0.10$ (90th percentile is 1.282).

$z = \frac{x - 175}{7} = 1.282$

$x = 175 + 7 \times 1.282 = 175 + 8.97 = 183.97$

About 184 cm is exceeded by 10% of heights.

Example 4: Two-Sided Interval

Find a symmetric interval around the mean containing 95% of values.

$P(-z \leq Z \leq z) = 0.95$

$\Phi(z) - \Phi(-z) = 0.95$

$\Phi(z) - (1 - \Phi(z)) = 0.95$

$2\Phi(z) - 1 = 0.95$

$\Phi(z) = 0.975$

From tables: $z = 1.96$.

So 95% of values lie within $\mu \pm 1.96\sigma$.

Continuity Correction

When approximating discrete distributions with the normal, use a continuity correction.

For $X \sim \text{Binomial}(n, p)$ approximated by $N(np, np(1-p))$:

$P(X = k) \approx P\left(k - 0.5 < Y < k + 0.5\right)$

$P(X \leq k) \approx P(Y < k + 0.5)$

Example: Coin Flips

Flip a fair coin 100 times. Approximate the probability of 55 or more heads.

$X \sim \text{Bin}(100, 0.5)$. Approximate with $Y \sim N(50, 25)$.

Without correction: $P(X \geq 55) \approx P(Y \geq 55) = P\left(Z \geq \frac{55 - 50}{5}\right) = P(Z \geq 1) = 0.1587$

With continuity correction:

$P(X \geq 55) \approx P(Y \geq 54.5) = P\left(Z \geq \frac{54.5 - 50}{5}\right) = P(Z \geq 0.9) = 0.1841$

Exact: $P(X \geq 55) = \sum_{k=55}^{100} \binom{100}{k} 0.5^{100} \approx 0.184$

The continuity correction improves the approximation significantly.

z-Scores

A z-score (or standard score) measures how many standard deviations a value is from the mean:

$z = \frac{x - \mu}{\sigma}$

Interpretation:

• $z = 0$: value equals the mean
• $z > 0$: value above the mean
• $z < 0$: value below the mean
• $|z| > 2$: unusually far from mean (occurs about 5% of the time)

Example: Comparing Tests

Alice scores 85 on Test A (mean 70, SD 10). Bob scores 78 on Test B (mean 60, SD 8). Who did better relative to their class?

• Alice: $z_A = \frac{85 - 70}{10} = 1.5$
• Bob: $z_B = \frac{78 - 60}{8} = 2.25$

Bob’s z-score is higher, so Bob performed better relative to his class.

Summary

Task	Approach
$P(X > x)$	Standardise, use $1 - \Phi(z)$
$P(X < x)$	Standardise, use $\Phi(z)$
$P(a < X < b)$	$\Phi(z_b) - \Phi(z_a)$
Find $x$ given percentile	$x = \mu + z_p \sigma$
Continuity correction	Add/subtract 0.5 for discrete approx