The normal distribution
Introduced| English | Meaning |
|---|---|
| normal distribution/ˈnɔːml ˌdɪstrɪˈbjuːʃn/ | A continuous, symmetric bell-shaped probability distribution specified by its mean and standard deviation. |
| continuous variable/kənˈtɪnjuːəs ˈveərɪəbl/ | A variable that can take any value within a specified interval, rather than only separate values. |
| standardize/ˈstændədaɪz/ | standardize |
| approximate/əˈprɒksɪmət/ | approximate |
| continuity correction/kɒntɪˈnjuːɪti kəˈrekʃn/ | An adjustment of discrete interval boundaries by half a unit when using a continuous approximation to unit-spaced counts. |
The curve that rules the world
- Heights, IQ scores, measurement errors, exam results — all follow the same symmetric bell-shaped curve.
- The normal distribution 正态分布 is the most important distribution in statistics. Master it and you can model almost any continuous variable 连续变量.
The normal distribution
- The normal distribution models a continuous variable with a symmetric bell shape: $X \sim N(\mu, \sigma^2)$.
- The mean $\mu$ determines the centre; the standard deviation $\sigma$ determines the spread.
The 68-95-99.7 rule. About 68% of values lie within 1 standard deviation of the mean, 95% within 2, and 99.7% within 3.
The normal curve is symmetric about the mean; about 68% of the area lies within one standard deviation.
The normal curve
P(Z < z)
The area under the bell up to z is the probability P(Z < z) — slide z to read it off.
The normal distribution curve is:
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It is the symmetric bell-shaped curve centred on the mean μ.
Approximately what percentage of values lie within 1 standard deviation of the mean?
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The 68-95-99.7 rule: ~68% within 1σ, ~95% within 2σ, ~99.7% within 3σ.
Complete the standardising formula: Z = (X − μ) / ______.
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Z = (X − μ)/σ converts any normal variable to the standard normal N(0, 1).
Standardising
- To use the tables, standardize 标准化 to $Z \sim N(0, 1)$:
- Then $P(X < x) = P\!\left(Z < \dfrac{x - \mu}{\sigma}\right)$, read from the $\Phi$ table.
A Galton board: balls fall through rows of pegs and pile up into the bell-shaped normal distribution
X ~ N(50, 4) means μ = 50, σ = 2. Standardize x = 56: what is Z = (x − μ)/σ?
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Z = (56 − 50)/2 = 6/2 = 3.
X ~ N(100, 25) means μ = 100, σ = 5. Standardize x = 90: Z = (90 − 100)/5. Find Z.
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Z = (90 − 100)/5 = −10/5 = −2.
Worked example
- $X \sim N(50, 16)$ (so $\mu = 50$, $\sigma = 4$). Find $P(X < 58)$.
- $Z = \dfrac{58 - 50}{4} = 2$.
- $P(Z < 2) = \Phi(2) = 0.9772$.
Standardise before using tables. The $\Phi$ table only works for $Z \sim N(0, 1)$. You must convert $X$ to $Z$ first using $Z = \dfrac{X - \mu}{\sigma}$.
A Galton board produces the bell-shaped normal distribution
If Z = 2, Φ(2) = 0.9772. What is P(Z < 2) (to 4 dp)?
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P(Z < 2) = Φ(2) = 0.9772 from the standard normal table.
Approximating 近似 the binomial
- For large $n$, the normal distribution approximates the binomial: $B(n, p) \approx N(np, np(1-p))$.
- Because you replace a discrete variable by a continuous one, apply a continuity correction 连续性校正 (adjust by 0.5).
For large n the binomial bars closely follow the smooth normal curve N(np, np(1−p))
The binomial distribution B(10, 0.3); its mean is np = 3
- The normal models a continuous random variable; use standardisation $Z=(X-\mu)/\sigma$, and as an approximation to the binomial.
When the normal distribution approximates the binomial, a continuity correction of 0.5 is applied.
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Replacing a discrete variable with a continuous one needs a ±0.5 continuity correction.
You've got it
- $X \sim N(\mu, \sigma^2)$; standardize with $Z = \dfrac{X - \mu}{\sigma}$
- read probabilities as areas from the $\Phi$ table for $Z \sim N(0,1)$
- the normal approximates the binomial for large $n$ (use a continuity correction)
Write from a z-score of 4
- A value 62 comes from a normal distribution with mean 50 and standard deviation 3.
- Find z.
A normal distribution has mean 50 and standard deviation 3. What is the z-score of 62?
12 divided by 3 is 4.
What is the z-score?
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62 minus 50 is 12, and 12 divided by 3 is 4.