Unit 10, lesson 65

The Normal Distribution

What a normal distribution is, how its mean and standard deviation shape the normal density, how standardization leads to the standard normal, and how to find central and tail probabilities.

  • Video 65
  • Normal Distributions

Repeated readings of a measurement often scatter around a central value, a little high or a little low, with large departures rare. The normal distribution is a precise model for this kind of symmetric variation, fixed by a mean and a standard deviation. This lesson reads its density, shows how standardization turns every normal calculation into one about a single reference curve, and computes central and tail probabilities. It also marks what the model assumes, including that every real value is possible.

Key terms

Normal distribution

Definition Let μ\mu be a real number and σ>0\sigma > 0. A random variable XX has the normal distribution with mean μ\mu and variance σ2\sigma^{2}, written X∼N(μ,σ2)X \sim N(\mu, \sigma^{2}), if it has the density

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

for every real xx. Then E[X]=μE[X] = \mu, Var⁡(X)=σ2\operatorname{Var}(X) = \sigma^{2} and SD⁡(X)=σ\operatorname{SD}(X) = \sigma. In this course the second entry of N(μ,σ2)N(\mu, \sigma^{2}) is always the variance. The support is the whole real line, since every interval of positive length has positive probability. The density is symmetric about μ\mu and highest there, so the mean, the median and the unique mode all equal μ\mu.

In plain words A normal model has a bell-shaped density curve fixed by two numbers. The mean says where the center is, and the standard deviation sets the horizontal spread. A roughly bell-shaped histogram is a reason to consider the model, not proof that a quantity is normal. State whether the model is assumed exactly or used as an approximation, especially for a quantity that cannot be negative or has a hard limit, because a normal law gives positive probability beyond every bound.

Example A constructed measurement model has mean 4040 units and standard deviation 33 units, so X∼N(40,9)X \sim N(40, 9) with variance 99 units squared. Writing N(40,3)N(40, 3) would describe a different model, with variance 33 and standard deviation 3≈1.732\sqrt{3} \approx 1.732. Values above 5252, four standard deviations above the mean, are possible under the model, but their probability is only about 0.000030.00003.

Normal density

Definition The normal density with parameters μ\mu and σ>0\sigma > 0 is the function ff of the previous card. It is positive for every real xx and symmetric about μ\mu, so f(μ+d)=f(μ−d)f(\mu + d) = f(\mu - d) for every real dd. It increases up to x=μx = \mu, where it takes its largest value 1/(σ2π)1/(\sigma\sqrt{2\pi}), and decreases after that, with inflection points at μ−σ\mu - \sigma and μ+σ\mu + \sigma. Its total area is 11, which rests on the Gaussian integral

∫−∞∞e−z2/2 dz=2π,\int_{-\infty}^{\infty} e^{-z^{2}/2}\,dz = \sqrt{2\pi},

a calculus result used here without proof. Probabilities are areas under ff, so P(a≤X≤b)P(a \le X \le b) is the integral of ff from aa to bb for a≤ba \le b, and P(X=x)=0P(X = x) = 0 for every real xx.

In plain words Read the exponent from the inside out. The distance from the mean is squared, which makes the curve symmetric, and divided by 2σ22\sigma^{2}, which measures it against the spread. The minus sign makes the density fall as the distance grows, and the factor in front makes the total area 11. Changing μ\mu slides the curve without changing its shape. Doubling σ\sigma doubles the width and halves the peak, so the area stays 11. The height of the curve is a density, not a probability, and the peak is larger than 11 whenever σ<1/2π≈0.399\sigma < 1/\sqrt{2\pi} \approx 0.399.

Example For X∼N(40,9)X \sim N(40, 9) the peak height is f(40)=1/(32π)≈0.1330f(40) = 1/(3\sqrt{2\pi}) \approx 0.1330 per unit, yet P(X=40)=0P(X = 40) = 0. One standard deviation away from the mean on either side,

f(37)=f(43)=f(40) e−1/2≈0.0807.\begin{aligned} f(37) = f(43) &= f(40)\,e^{-1/2} \\[0.4em] &\approx 0.0807. \end{aligned}

A normal model with σ=0.1\sigma = 0.1 has peak height about 3.9893.989, a valid density value larger than 11.

Standard normal distribution

Definition The standard normal distribution is N(0,1)N(0, 1), with mean 00 and standard deviation 11. A variable with this law is usually called ZZ. Its CDF is

Φ(z)=P(Z≤z)=∫−∞ze−u2/22π du.\begin{aligned} \Phi(z) &= P(Z \le z) \\[0.4em] &= \int_{-\infty}^{z} \frac{e^{-u^{2}/2}}{\sqrt{2\pi}}\,du. \end{aligned}

The function Φ\Phi is continuous and strictly increasing, with Φ(0)=0.5\Phi(0) = 0.5 by symmetry. It cannot be written with elementary functions, so its values come from tables or numerical software and are reported as approximations.

In plain words The standard normal is the one reference curve to which every normal calculation is reduced. Its horizontal axis counts standard deviations from the mean. A table of Φ\Phi lists areas to the left of each value, and other areas are found by subtraction.

Example Approximate values are Φ(1)≈0.8413\Phi(1) \approx 0.8413, Φ(−1)≈0.1587\Phi(-1) \approx 0.1587 and Φ(2)≈0.9772\Phi(2) \approx 0.9772. So P(−1<Z<1)=Φ(1)−Φ(−1)≈0.6827P(-1 < Z < 1) = \Phi(1) - \Phi(-1) \approx 0.6827, and P(Z>2)=1−Φ(2)≈0.0228P(Z > 2) = 1 - \Phi(2) \approx 0.0228.

Standardization

Definition For a random variable XX with mean μ\mu and standard deviation σ>0\sigma > 0, standardization is the change of variable

Z=X−μσ,Z = \frac{X - \mu}{\sigma},

which has mean 00 and standard deviation 11. If X∼N(μ,σ2)X \sim N(\mu, \sigma^{2}), then Z∼N(0,1)Z \sim N(0, 1), and conversely μ+σZ∼N(μ,σ2)\mu + \sigma Z \sim N(\mu, \sigma^{2}) whenever Z∼N(0,1)Z \sim N(0, 1). So for a normal XX and any a<ba < b,

P(a<X<b)=Φ(b−μσ)−Φ(a−μσ).\begin{aligned} &P(a < X < b) \\[0.4em] &\quad = \Phi\Bigl(\frac{b - \mu}{\sigma}\Bigr) \\[0.4em] &\qquad - \Phi\Bigl(\frac{a - \mu}{\sigma}\Bigr). \end{aligned}

Standardization does not change the shape of a distribution, so the standardized variable is normal exactly when XX is normal.

In plain words The standardized value says how many standard deviations a value lies above the mean, and a negative value means it lies below the mean. It has no units. Shifting and rescaling relabel the axis, but a skewed distribution stays skewed and a flat one stays flat. So standardizing does not make a variable normal, and reading probabilities from Φ\Phi rests on the assumption that XX itself is normal.

Example For X∼N(40,9)X \sim N(40, 9) the value 4343 standardizes to (43−40)/3=1(43 - 40)/3 = 1, the value 3737 to −1-1 and the value 3434 to −2-2. So P(37<X<43)=P(−1<Z<1)≈0.6827P(37 < X < 43) = P(-1 < Z < 1) \approx 0.6827. In contrast, a Uniform⁡(0,1)\operatorname{Uniform}(0, 1) variable has mean 0.50.5 and standard deviation 1/121/\sqrt{12}, and its standardized version is uniform on (−3,3)(-\sqrt{3}, \sqrt{3}), roughly (−1.732,1.732)(-1.732, 1.732). Its density is still flat, and none of its values lies farther than 3\sqrt{3} from 00.

68-95-99.7 rule

Definition For X∼N(μ,σ2)X \sim N(\mu, \sigma^{2}) and any k>0k > 0,

P(∣X−μ∣<kσ)=P(∣Z∣<k)=2Φ(k)−1.\begin{aligned} &P(\lvert X - \mu \rvert < k\sigma) \\[0.4em] &\quad = P(\lvert Z \rvert < k) \\[0.4em] &\quad = 2\Phi(k) - 1. \end{aligned}

For k=1,2,3k = 1, 2, 3 this gives approximately 0.68270.6827, 0.95450.9545 and 0.99730.9973, often rounded to 6868, 9595 and 99.799.7 percent. These values hold for every normal distribution, whatever μ\mu and σ\sigma. They are not general facts about other distributions. For any distribution with finite positive variance, Chebyshev’s inequality guarantees only that at least 1−1/k21 - 1/k^{2} of the probability lies within kk standard deviations of the mean, for example at least 0.750.75 for k=2k = 2.

In plain words For a normal model, about two thirds of the probability lies within one standard deviation of the mean, about 9595 percent within two and almost all of it within three. Almost all is not all. The rule describes a normal model, not every data set or every probability law, and its percentages are rounded.

Example For X∼N(40,9)X \sim N(40, 9), the intervals (37,43)(37, 43), (34,46)(34, 46) and (31,49)(31, 49) have probabilities about 0.68270.6827, 0.95450.9545 and 0.99730.9973. A Uniform⁡(0,1)\operatorname{Uniform}(0, 1) variable puts only 1/3≈0.57741/\sqrt{3} \approx 0.5774 of its probability within one standard deviation of its mean, and all of it within two.

Normal tail probability

Definition For X∼N(μ,σ2)X \sim N(\mu, \sigma^{2}) and a threshold cc with standardized value z=(c−μ)/σz = (c - \mu)/\sigma, the upper and lower tail probabilities are

P(X>c)=1−Φ(z),P(X<c)=Φ(z).\begin{gathered} P(X > c) = 1 - \Phi(z), \\[0.4em] P(X < c) = \Phi(z). \end{gathered}

For k≥0k \ge 0, the probability of lying more than kk standard deviations from the mean on either side is P(∣X−μ∣>kσ)=2(1−Φ(k))P(\lvert X - \mu \rvert > k\sigma) = 2\bigl(1 - \Phi(k)\bigr), twice one tail. Every tail probability is positive, however far the threshold lies from μ\mu.

In plain words A tail probability is the chance of landing beyond a threshold on one side. A two-sided question asks for both tails together, so keep one tail and two tails apart. The tails shrink very quickly but never reach 00, and drawing the curve only out to three standard deviations is a choice of picture, not a limit of the model.

Example For X∼N(40,9)X \sim N(40, 9) the threshold 4646 has standardized value 22, so P(X>46)=1−Φ(2)≈0.0228P(X > 46) = 1 - \Phi(2) \approx 0.0228. By symmetry P(X<34)≈0.0228P(X < 34) \approx 0.0228 as well, and the probability of lying outside [34,46][34, 46] is about 0.04550.0455. Further out the tails get small fast, with P(Z>3)≈0.00135P(Z > 3) \approx 0.00135 and P(Z>4)≈0.00003P(Z > 4) \approx 0.00003.

Check yourself

  1. A constructed normal model has mean 100100 and standard deviation 88. State its variance and standardize the values 8888 and 116116.
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The variance is 82=648^{2} = 64, so the model is N(100,64)N(100, 64). The standardized values are

88−1008=−1.5,116−1008=2.\begin{gathered} \frac{88 - 100}{8} = -1.5, \\[0.4em] \frac{116 - 100}{8} = 2. \end{gathered}

The value 8888 lies one and a half standard deviations below the mean. A negative standardized value is an ordinary result, not an error.

  1. In the same model, find the probability of lying outside the interval [84,116][84, 116] and the probability of lying above 116116.
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The endpoints are 100±2⋅8100 \pm 2 \cdot 8, two standard deviations from the mean, so the interval has probability 2Φ(2)−1≈0.95452\Phi(2) - 1 \approx 0.9545. The probability outside is about 1−0.9545=0.04551 - 0.9545 = 0.0455. By symmetry it splits equally between the two tails, so P(X>116)≈0.0228P(X > 116) \approx 0.0228. Reporting 0.04550.0455 for the upper tail alone would double the correct value.

  1. A constructed model for a waiting time in minutes is written N(5,16)N(5, 16). Find its standard deviation and the probability it gives to a negative waiting time. What does the result say about the model?
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The second entry is the variance, so the standard deviation is 16=4\sqrt{16} = 4 minutes. The value 00 standardizes to (0−5)/4=−1.25(0 - 5)/4 = -1.25, so P(X<0)=Φ(−1.25)≈0.1056P(X < 0) = \Phi(-1.25) \approx 0.1056. A waiting time cannot be negative, yet the model puts more than 1010 percent of its probability there, so it is a poor model for this quantity. Reading 1616 as the standard deviation would be a separate mistake and would give Φ(−0.3125)≈0.3773\Phi(-0.3125) \approx 0.3773.

How this lesson connects

It also uses using a standard normal table or calculator.