Standard Normal Probabilities and Quantiles

From Normal Values to Standard Normal Areas

Normal probability questions often look different on the surface: a cutoff may be stated in measurement units, a question may ask for a tail area, or an interval may be described as containing a given percentage of the distribution. The calculations become systematic once the cutoff is expressed in standard normal units and the requested area is matched to the correct tail convention.

The standard normal distribution has mean zero and variance one. As established in the earlier material on normal location-scale parameterization, if \(X\sim N(\mu,\sigma^2)\) with \(\sigma>0\), then \(Z=(X-\mu)/\sigma\) has the standard normal distribution. The remaining task is to evaluate its probabilities or invert them to find a cutoff.

Definition: The standard normal density is \( \phi(z)=(2\pi)^{-1/2}\exp(-z^2/2) \), for \(z\in\mathbb{R}\). The standard normal CDF is \( \Phi(z)=P(Z\le z)=\int_{-\infty}^{z}\phi(t)\,dt \), where \(Z\sim N(0,1)\). For \(0<p<1\), the standard normal \(p\)-quantile is the unique number \(z_p\) satisfying \( \Phi(z_p)=p \).

Because the standard normal density is positive everywhere, \(\Phi\) increases strictly: moving the cutoff to the right adds positive probability. Its limits are zero and one at the two ends of the real line. Consequently every probability \(p\) strictly between zero and one has exactly one quantile \(z_p\). In practice, CDF values and quantiles are obtained from a table or numerical software; reported decimal values are usually rounded.

Symmetry and Tail Conventions

The standard normal density is symmetric about zero: \(\phi(-z)=\phi(z)\). This means the area to the left of \(-z\) equals the area to the right of \(z\). It is important to distinguish a lower-tail probability, \(P(Z\le z)=\Phi(z)\), from an upper-tail probability, \(P(Z>z)=1-\Phi(z)\). The inverse CDF takes a lower-tail probability as its input.

Theorem: For every real \(z\), \( \Phi(-z)=1-\Phi(z) \). Thus \(P(Z>z)=\Phi(-z)\). For \(0<p<1\), the quantiles satisfy \(z_{1-p}=-z_p\).

To prove the CDF identity, substitute \(u=-t\) in the integral defining \(\Phi(-z)\). Since \(\phi(-u)=\phi(u)\), this gives

\Phi(-z) =\int_{-\infty}^{-z}\phi(t)\,dt =\int_z^\infty\phi(u)\,du =1-\Phi(z).

The last equality holds because the density integrates to one and a single point has probability zero. Therefore \(P(Z>z)=1-\Phi(z)=\Phi(-z)\). For the quantile identity, \(\Phi(z_p)=p\) implies \(\Phi(-z_p)=1-p\), so uniqueness of the quantile gives \(z_{1-p}=-z_p\). \( \square \)

For an interval, subtract the CDF values at its endpoints. Since the normal distribution is continuous, the choice of strict or non-strict inequalities at a finite endpoint does not change the probability.

P(a<Z\le b)=\Phi(b)-\Phi(a),\qquad a\le b.

In particular, a two-sided probability outside symmetric cutoffs is not the same as a single upper-tail probability: for \(c>0\), each tail contributes \(1-\Phi(c)\), so the total is \(2[1-\Phi(c)]\).

Worked Example: A Lower Tail, an Upper Tail, and a Symmetric Pair

Let \(Z\sim N(0,1)\). Calculate \(P(Z\le -1.25)\), \(P(Z>1.25)\), and \(P(|Z|>1.25)\).

A standard normal table or numerical CDF evaluation gives \(\Phi(1.25)\approx0.8944\). By symmetry, \( \Phi(-1.25)=1-\Phi(1.25)\approx1-0.8944=0.1056 \). Thus \(P(Z\le-1.25)\approx0.1056\).

The upper-tail probability at \(1.25\) is also \(1-\Phi(1.25)\approx0.1056\). The event \(|Z|>1.25\) is the disjoint union of \(Z<-1.25\) and \(Z>1.25\), so \(P(|Z|>1.25)\approx0.1056+0.1056=0.2112\).

As a check, the probability between the two cutoffs is \(1-0.2112=0.7888\), which also equals \(\Phi(1.25)-\Phi(-1.25)\approx0.8944-0.1056=0.7888\).

Standardizing Probabilities for a General Normal Variable

If \(X\sim N(\mu,\sigma^2)\), with \(\sigma>0\), any cutoff \(x\) can be standardized as \(z=(x-\mu)/\sigma\). The same transformation applies to both endpoints of an interval. In particular, upper-tail questions use a complement, while two-sided questions require accounting for both tails unless the event is an interval.

Takeaway: For \(X\sim N(\mu,\sigma^2)\), write \(Z=(X-\mu)/\sigma\). Then \(P(X\le x)=\Phi((x-\mu)/\sigma)\), \(P(X>x)=1-\Phi((x-\mu)/\sigma)\), and \(P(a<X\le b)=\Phi((b-\mu)/\sigma)-\Phi((a-\mu)/\sigma)\).

Worked Example: An Upper-Tail Probability in Measurement Units

Suppose \(X\sim N(70,10^2)\), where measurements are in the same units as the mean. Find the probability that \(X\) exceeds 85.

First standardize the cutoff: \( z=(85-70)/10=1.5 \). Therefore \( P(X>85)=P(Z>1.5)=1-\Phi(1.5) \). Using \(\Phi(1.5)\approx0.9332\), we obtain \(P(X>85)\approx1-0.9332=0.0668\).

The units matter in the standardization: 85 is 15 units above the mean, or \(15/10=1.5\) standard deviations above it. As a related check, symmetry gives \(P(X<55)=P(Z<-1.5)\approx0.0668\). The probability of being more than 15 units from the mean in either direction is twice this amount, approximately \(0.1336\).

Quantiles, One-Sided Cutoffs, and Central Intervals

A quantile answers the reverse question: given a lower-tail probability \(p\), find the cutoff with probability \(p\) to its left. For a standard normal variable this cutoff is \(z_p=\Phi^{-1}(p)\). For \(X\sim N(\mu,\sigma^2)\), the corresponding quantile is obtained by reversing the standardization.

Quantile rule: For \(0<p<1\), the \(p\)-quantile of \(X\sim N(\mu,\sigma^2)\), with \(\sigma>0\), is \(x_p=\mu+\sigma z_p\), where \(\Phi(z_p)=p\). Thus \(P(X\le x_p)=p\) and \(P(X>x_p)=1-p\).

A central interval containing probability \(1-\alpha\), with equal probability in its two tails, leaves \(\alpha/2\) in each tail. Its standard normal endpoints are therefore \(z_{\alpha/2}\) and \(z_{1-\alpha/2}\). By symmetry they are \(-z_{1-\alpha/2}\) and \(z_{1-\alpha/2}\). This equal-tail convention is common, but not the only possible way to specify an interval with a given probability.

Theorem: For \(0<\alpha<1\), the standard normal equal-tail central interval with probability \(1-\alpha\) is \( [-z_{1-\alpha/2},\,z_{1-\alpha/2}] \). Its probability is exactly \(1-\alpha\).

To verify the claim, the two endpoint probabilities are equal by the quantile symmetry already proved. The probability in the interval is

\Phi(z_{1-\alpha/2})-\Phi(z_{\alpha/2}) =(1-\alpha/2)-\alpha/2 =1-\alpha.

Since \(z_{\alpha/2}=-z_{1-\alpha/2}\), these are precisely the stated symmetric endpoints. For \(X\sim N(\mu,\sigma^2)\), transforming both endpoints gives the central interval \([\mu-\sigma z_{1-\alpha/2},\,\mu+\sigma z_{1-\alpha/2}]\). \( \square \)

Worked Example: A Central 95% Interval

Let \(X\sim N(70,10^2)\). Find the symmetric interval about the mean containing 95% of the probability, leaving equal tails.

Here \(1-\alpha=0.95\), so \(\alpha=0.05\) and each tail has probability \(\alpha/2=0.025\). The positive endpoint in standard normal units is the quantile with lower-tail probability \(0.975\): \(z_{0.975}\approx1.96\). The symmetric standard normal interval is \([-1.96,1.96]\).

Transforming back to the measurement scale gives \(70\pm10(1.96)=70\pm19.6\), or approximately \([50.4,89.6]\). The probability in that interval is \(\Phi(1.96)-\Phi(-1.96)\approx0.975-0.025=0.95\).

For comparison, a one-sided upper cutoff with 95% below it uses \(z_{0.95}\approx1.645\), not 1.96. Its measurement-scale value is \(70+10(1.645)=86.45\), and \(P(X\le86.45)\approx0.95\). The two cutoffs differ because the central interval allocates probability to both tails, whereas this one-sided cutoff allocates all 5% above a single endpoint.

Worked Example: Finding a 90th Percentile

Suppose \(X\sim N(12,3^2)\). Find its 90th percentile and verify the tail areas.

The 90th percentile is the value \(x_{0.90}\) with 0.90 probability below it. The standard normal quantile is \(z_{0.90}\approx1.2816\), since \(\Phi(1.2816)\approx0.9000\). Hence \(x_{0.90}=12+3(1.2816)=15.8448\), or about \(15.84\).

Standardizing this value gives \((15.8448-12)/3=1.2816\), so \(P(X\le15.8448)\approx0.9000\), and the probability above it is approximately \(1-0.9000=0.1000\). The calculation verifies that the requested percentile is a lower-tail quantile, even though one may also describe it as leaving 10% above the cutoff.

Common Mistake: Using the Wrong Tail Probability

A frequent error is to treat every “95%” request as a call for \(z_{0.95}\). That is correct for a cutoff with 95% below it, but it is not correct for a central 95% interval. For the latter, 2.5% lies below the interval and 2.5% above it, so the positive endpoint is \(z_{0.975}\). Conversely, a one-sided upper cutoff leaving 5% above it has 95% below it and uses \(z_{0.95}\).

Another error is to double a tail probability when the question asks for the probability between two cutoffs. For example, \(P(-c<Z<c)=\Phi(c)-\Phi(-c)\), whereas \(P(|Z|>c)=2[1-\Phi(c)]\). The events are complements, not interchangeable descriptions of the same probability. Before using a table or inverse CDF, identify the event and say explicitly which area is being supplied or requested.

Key takeaway: Standardize first, then match the probability statement to the CDF convention. The inverse CDF takes a lower-tail probability; symmetry supplies the corresponding upper tail and converts equal-tail central intervals into symmetric cutoffs.

Check Your Understanding

For each question, identify the tail or interval convention before calculating.

If \(\Phi(1.10)\approx0.8643\), what is \(P(Z<-1.10)\), and what is \(P(|Z|>1.10)\)?
For \(X\sim N(20,4^2)\), express \(P(X>26)\) using \(\Phi\), and evaluate it using \(\Phi(1.5)\approx0.9332\).
Which standard normal quantile gives a one-sided cutoff with 97.5% below it? Which positive quantile gives the upper endpoint of a central 95% interval?
If \(z_{0.90}\approx1.2816\), what is \(z_{0.10}\), and why?
For \(X\sim N(12,3^2)\), write the formula for its 90th percentile and explain which probability is to the right of that value.