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The normal distribution

The normal distribution · Diagrams · Complete reading

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Location and spread

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Six normal curves. With variance 1, means 0, 1 and −2 shift the location. With mean 0, variances 1, 0.5 and 2 give peaks of approximately 0.399, 0.564 and 0.282 respectively. Smaller variance gives a narrower, taller curve. The source spread figures were corrected to use the printed variance convention.
Six normal curves. With variance 1, means 0, 1 and −2 shift the location. With mean 0, variances 1, 0.5 and 2 give peaks of approximately 0.399, 0.564 and 0.282 respectively. Smaller variance gives a narrower, taller curve. The source spread figures were corrected to use the printed variance convention.
Six normal curves. With variance 1, means 0, 1 and −2 shift the location. With mean 0, variances 1, 0.5 and 2 give peaks of approximately 0.399, 0.564 and 0.282 respectively. Smaller variance gives a narrower, taller curve. The source spread figures were corrected to use the printed variance convention.
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Areas within standard deviations

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Areas between μ−σ and μ+σ approximately 68.3%; μ−2σ and μ+2σ approximately 95.4%; μ−3σ and μ+3σ approximately 99.7%. All three shaded regions are centred at the mean.
Areas between μ−σ and μ+σ approximately 68.3%; μ−2σ and μ+2σ approximately 95.4%; μ−3σ and μ+3σ approximately 99.7%. All three shaded regions are centred at the mean.
Areas between μ−σ and μ+σ approximately 68.3%; μ−2σ and μ+2σ approximately 95.4%; μ−3σ and μ+3σ approximately 99.7%. All three shaded regions are centred at the mean.
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Method map: Translate a normal event before calculating

Identify the model and its spread, keep the event when standardising, then match the calculation to the shaded area. An interval probability and a density height are different quantities.

  1. Read the spread before entering a calculator value

    In the variance convention N(μ, σ²), the second parameter is the variance. Standardisation requires a positive standard deviation.

    X∼N(μ,σ2)X\sim N(\mu,\sigma^2)

    σ>0\sigma>0

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    Variance v is givenUse σ = √v

    For a model N(70,25)N(70,25), enter standard deviation 5. Do not enter 25 as the standard deviation.

    Standard deviation is givenUse it directly

    For N(100,62)N(100,6^2), the standard deviation is 6 and the variance is 36. Label the parameter rather than guessing from the size of the number.

    Check that the normal model is stated or justified in context. A zero spread does not give a nondegenerate standard normal variable and cannot be used as the divisor here.

  2. Transform every cutoff and preserve the event

    Define Z from the same variable X. Subtract the mean and divide every bound by the positive standard deviation; the inequality directions stay unchanged.

    Z=X−μσ∼N(0,1)Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)

    Left or right tailKeep the shaded side

    The event X<cX<c becomes Z<(c−μ)/σZ<(c-\mu)/\sigma; the event X>cX>c becomes Z>(c−μ)/σZ>(c-\mu)/\sigma. A right-tail event does not turn into a left-tail event merely because the variable is standardised.

    Between two boundsTransform both endpoints

    For a<X<ba<X<b, set za=(a−μ)/σz_a=(a-\mu)/\sigma and zb=(b−μ)/σz_b=(b-\mu)/\sigma. The event is za<Z<zbz_a<Z<z_b, with the same probability. With a nondegenerate normal model, including an individual endpoint does not change that probability.

    The source density stages have areas AA, AA, A/σA/\sigma, then AA. Horizontal compression of a drawn density alone changes its area; the compensating vertical stretch restores a normalised density. Compare the four stages and their labelled areas in the source figure.

  3. Choose an area calculation or an inverse cutoff

    Decide whether the unknown is a probability or a cutoff. Sketch the shaded region and name the calculator's area convention before using it.

    Φ(z)=P(Z≤z)\Phi(z)=P(Z\le z)

    Known cutoffsCalculate the probability of the event

    For a standardised cutoff z, a left area is Φ(z)\Phi(z) and a right area is 1−Φ(z)1-\Phi(z). For the standardised interval za<Z<zbz_a<Z<z_b, use Φ(zb)−Φ(za)\Phi(z_b)-\Phi(z_a). For a direct normalcdf calculation, use the original bounds with μ and σ, or the standardised bounds with mean 0 and standard deviation 1. Do not mix those two parameter sets.

    Known one-sided probability pMatch the inverse to the tail

    For 0<p<10<p<1, a lower-tail-only inverse uses p for a stated left area and 1−p for a stated right area. When a standard-normal inverse returns z, recover the original cutoff with x=μ+σzx=\mu+\sigma z. An inverse supplied with the original μ and σ already returns the original cutoff. An interface with an explicit RIGHT selector may instead use the right area directly; do not complement it twice.

    Known central probability pUse symmetry only when justified

    For a central interval symmetric about the mean and 0<p<10<p<1, the lower-tail endpoint areas are (1−p)/2(1-p)/2 and (1+p)/2(1+p)/2. CENTER returns endpoints, not the half-width. An arbitrary interval is not determined by its area alone.

    Check whether a cutoff should be above or below the mean and substitute back into the original event. Keep full calculator precision until the final requested rounding. A density height is not a probability.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

A large right area gives a cutoff below the mean

The source gives X∼N(100,62)X\sim N(100,6^2) and P(X>b)=0.770P(X>b)=0.770. Before calculating, predict which side of 100 contains b.

Show reasoning
  1. 1. Identify the spread and shade the stated area

    The standard deviation is 6, not the variance 36. More than half the curve lies to the right of b, so b is below the mean.

  2. 2. Complement only for a lower-tail interface

    A RIGHT-tail interface can use the given 0.770 directly. A lower-tail-only inverse uses the complementary area 0.230. Both describe the same cutoff.

    P(X<b)=1−0.770=0.230.\begin{aligned}P(X<b)&=1-0.770\\&=0.230.\end{aligned}
  3. 3. Recover the cutoff and check its direction

    Keep calculator precision until the source’s final rounding to three significant figures. The result is below 100 as predicted.

    b=100+6Φ−1(0.230)≈95.56691888≈95.6(3 s.f.).\begin{aligned}b&=100+6\Phi^{-1}(0.230)\\&\approx95.56691888\\&\approx95.6\quad(3\text{ s.f.}).\end{aligned}
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A central interval has two endpoints and one half-width

The source gives X∼N(70,25)X\sim N(70,25) and P(∣X−70∣<a)=0.8P(|X-70|<a)=0.8. Find the central interval, then distinguish its endpoints from the half-width a.

Show reasoning
  1. 1. Expand the event and split the remaining area

    The standard deviation is the positive square root of 25, namely 5. The interval has probability 0.8 and its bounds are symmetric about 70, so the remaining 0.2 splits into tails of 0.1.

    70−a<X<70+a.70-a<X<70+a.
  2. 2. Use lower-tail areas 0.1 and 0.9

    A CENTER interface returns the pair of endpoints; the two lower-tail inverse calculations return the same pair. Neither returned endpoint is a itself.

    L=70+5Φ−1(0.1)≈63.59224217,U=70+5Φ−1(0.9)≈76.40775783.\begin{aligned}L&=70+5\Phi^{-1}(0.1)\\&\approx63.59224217,\\U&=70+5\Phi^{-1}(0.9)\\&\approx76.40775783.\end{aligned}
  3. 3. Find the half-width separately

    Either endpoint gives the same half-width. The source reports a = 6.41 and interval limits 63.6 and 76.4, all to three significant figures. Retain the full endpoint values until rounding.

    a=U−70=70−L≈6.40775783≈6.41.\begin{aligned}a&=U-70=70-L\\&\approx6.40775783\\&\approx6.41.\end{aligned}
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Compare with the density transformation

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Four stages. Start with X normal mean 5 standard deviation 2, interval 6 to 8 and areaA. Translation left 5 gives interval 1 to 3 with areaA. Horizontal compression by 1/2 gives interval.5 to 1.5 but areaA/2, so this intermediate graph is not a normalised density. Vertical stretch 2 restores areaA and the standard normal density. In general the successive areas are A,A,A/σ,A.
Four stages. Start with X normal mean 5 standard deviation 2, interval 6 to 8 and areaA. Translation left 5 gives interval 1 to 3 with areaA. Horizontal compression by 1/2 gives interval.5 to 1.5 but areaA/2, so this intermediate graph is not a normalised density. Vertical stretch 2 restores areaA and the standard normal density. In general the successive areas are A,A,A/σ,A.
Four stages. Start with X normal mean 5 standard deviation 2, interval 6 to 8 and areaA. Translation left 5 gives interval 1 to 3 with areaA. Horizontal compression by 1/2 gives interval.5 to 1.5 but areaA/2, so this intermediate graph is not a normalised density. Vertical stretch 2 restores areaA and the standard normal density. In general the successive areas are A,A,A/σ,A.
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Six source binomial-to-normal comparisons

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Six reconstructed panels compare exact binomial probability bars with the normal density having the same mean and variance: np=2 or 5 and n=10,50,100. Keeping small np fixed while increasing n leaves skewness; the source heuristic requires both np and n(1−p) strictly greater than 5.
Six reconstructed panels compare exact binomial probability bars with the normal density having the same mean and variance: np=2 or 5 and n=10,50,100. Keeping small np fixed while increasing n leaves skewness; the source heuristic requires both np and n(1−p) strictly greater than 5.
Six reconstructed panels compare exact binomial probability bars with the normal density having the same mean and variance: np=2 or 5 and n=10,50,100. Keeping small np fixed while increasing n leaves skewness; the source heuristic requires both np and n(1−p) strictly greater than 5.
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