Why Normal Quantiles Matter
A normal probability calculation starts with a specified distribution and asks for the chance of an event. An inverse probability problem reverses the direction: it specifies a probability and asks for the value that marks its boundary. That value is a normal quantile. Quantiles are used to set reference limits, describe percentiles, and work backward from a probability statement to an unknown model parameter.
The key is to distinguish a cumulative probability, \(P(X\leq x)\), from an upper-tail probability, \(P(X>x)\). A standard-normal table or calculator typically returns a quantile for a cumulative probability. Convert an upper-tail probability to its complement before looking up that quantile.
Finding a Cutoff From a Cumulative Probability
For \(X\sim N(\mu,\sigma^2)\), standardization gives \(Z=(X-\mu)/\sigma\sim N(0,1)\). Thus \(P(X\leq x)=p\) is equivalent to \(P(Z\leq (x-\mu)/\sigma)=p\). The standardized cutoff must be \(z_p\), so the original-scale cutoff is \(x_p=\mu+\sigma z_p\). In this notation, the second parameter of \(N(\mu,\sigma^2)\) is the variance; the standard deviation used in the calculation is \(\sigma\).
The function \(\Phi\) is strictly increasing, so each cumulative probability strictly between 0 and 1 determines one finite standard-normal quantile. Common values include \(z_{0.90}\approx1.2816\), \(z_{0.95}\approx1.6449\), and \(z_{0.975}\approx1.9600\). For a lower-tail probability, the quantile can be negative: \(z_{0.10}\approx-1.2816\).
Worked Example: Finding the Central 80% of a Normal Measurement
A hypothetical instrument’s repeated calibration error, in a suitable measurement scale, is modeled as \(X\sim N(72,10^2)\). Find the cutoffs for the central 80% of readings and state the probability that a reading falls between them. The model is assumed appropriate, and \(\mu=72\) and \(\sigma=10\) are known.
Method. A central 80% interval leaves \(0.20\) outside the interval, split equally between the two tails. Each tail therefore has probability \(0.10\). The lower cutoff is the 10th percentile and the upper cutoff is the 90th percentile.
From the standard-normal quantiles, \(z_{0.10}\approx-1.2816\) and \(z_{0.90}\approx1.2816\). Convert each to the measurement scale:
The cutoffs are approximately 59.18 and 84.82. The probability between the exact 10th and 90th percentiles is \(0.90-0.10=0.80\), or 80%. (Rounding each cutoff to two decimal places does not change the stated percentile interpretation.) Check: the cutoffs are equally far from the mean, as expected for a central interval under a symmetric normal model. The common mistake avoided here is using the 80th percentile as the upper cutoff; a central 80% interval instead uses the 10th and 90th percentiles.
Solving for an Unknown Mean
If \(\sigma\), a cutoff \(c\), and a cumulative probability \(p\) are given, standardize the event \(X\leq c\). Its equation is \((c-\mu)/\sigma=z_p\), which can be rearranged to find the mean:
For an upper-tail statement \(P(X>c)=r\), use \(P(X\leq c)=1-r\), since a continuous normal variable has probability zero of equaling exactly \(c\). Then use \(z_{1-r}\), not \(z_r\). The sign of the quantile conveys whether the cutoff is below or above the mean.
Worked Example: Inferring a Mean From a Cumulative Probability
A hypothetical processing-time measure is modeled as \(X\sim N(\mu,6^2)\), with known standard deviation 6 but unknown mean. Suppose \(P(X\leq41)=0.6915\). Estimate \(\mu\), then find the probability that \(X>41\). Treating 0.6915 as a table probability rounded to four decimals, use \(z_{0.6915}\approx0.5000\).
Method. The probability is cumulative, so standardize the cutoff directly:
Multiply both sides by 6: \(41-\mu=3\). Therefore \(\mu=41-3=38\). The estimated mean is 38 in the same units as the processing-time measure.
The requested upper-tail probability is the complement of the given cumulative probability:
Check. The cutoff 41 is three units above the inferred mean, or \(3/6=0.5\) standard deviations above it, consistent with a cumulative probability greater than 0.5. The answer uses the table’s rounded probability, so the inferred mean is an estimate. The common mistake avoided is treating \(0.6915\) as a right-tail probability, which would put the cutoff on the wrong side of the mean.
Solving for an Unknown Standard Deviation
If \(\mu\), a cutoff \(c\), and a probability are known but \(\sigma\) is unknown, standardization produces an equation for \(\sigma\). For a cumulative probability \(p\),
This division is valid only when \(z_p\neq0\). Since a standard deviation must be positive, the numerator and denominator must have the same sign for a finite solution. If \(c=\mu\), the cumulative probability is 0.5 regardless of the positive value of \(\sigma\), so that probability statement cannot identify \(\sigma\). If the probability is not 0.5 but the cutoff equals the mean, the stated model and probability are inconsistent.
For an upper-tail probability, first convert to the cumulative probability. This also makes the sign check transparent: if \(c>\mu\), then \(P(X>c)<0.5\), and the corresponding cumulative quantile is positive.
Worked Example: Inferring a Standard Deviation From an Upper Tail
A hypothetical measurement \(X\) is modeled as normal with known mean 100 and unknown standard deviation \(\sigma>0\). Suppose \(P(X>113)=0.1000\). Estimate \(\sigma\), report it to two decimal places, and verify the direction of the tail.
Method. Convert the upper-tail probability to a cumulative probability:
The standard-normal quantile for cumulative probability 0.9000 is \(z_{0.90}\approx1.2816\). Standardize the threshold and solve:
Rounded only at the final step to two decimal places, \(\sigma\approx10.14\). Check. The threshold 113 is above the mean, and a positive standard deviation gives a positive standardized value, \(13/10.1435705\ldots=1.2816\). Its upper-tail probability is \(1-\Phi(1.2816)\approx0.1000\), as specified.
A common wrong calculation uses \(z_{0.10}=-1.2816\) directly and obtains \(13/(-1.2816)<0\), an impossible standard deviation. The correction is to recognize that 0.1000 is a right-tail probability and convert it to the left-tail probability 0.9000 before selecting the quantile.
Why the Inverse Formula Works
The quantile equation is not a separate rule to memorize: it follows directly from the standardized normal probability. Suppose \(X\sim N(\mu,\sigma^2)\) with \(\sigma>0\). For any proposed cutoff \(x\),
The inequality direction remains unchanged because \(\sigma\) is positive. If \(x\) is the \(p\)-quantile, the left side equals \(p\); hence \(\Phi((x-\mu)/\sigma)=p\). Applying the inverse function \(\Phi^{-1}\) gives \((x-\mu)/\sigma=z_p\), and multiplying by \(\sigma\) and adding \(\mu\) gives \(x=\mu+\sigma z_p\). The same equation can then be rearranged for \(\mu\) or \(\sigma\), subject to the positivity and identifiability conditions already noted.
Choosing the Correct Tail and Checking the Answer
Before consulting a table or calculator, rewrite the probability in cumulative form. For a continuous normal variable, \(P(X>c)=r\) is equivalent to \(P(X\leq c)=1-r\). For a lower-tail event, the given probability is already cumulative. For a central interval with probability \(1-\alpha\), each tail has probability \(\alpha/2\), so its endpoints use quantiles \(\alpha/2\) and \(1-\alpha/2\).
After calculating, check whether the result has the right direction. A quantile above 0.5 must be above the mean; a quantile below 0.5 must be below it. For an inferred standard deviation, check positivity and substitute it back into the standardized equation. Keep full calculator precision during intermediate steps and round only as requested. This topic is aligned with the SOA VEE Mathematical Statistics syllabus.
Check Your Understanding
For each problem, state the relevant cumulative probability before using a quantile. Unless specified otherwise, report cutoffs and parameters to two decimal places.
- A hypothetical sensor reading has distribution \(X\sim N(40,5^2)\). Find its 80th percentile. (Answer: \(44.21\))
- A normal variable has distribution \(X\sim N(\mu,4^2)\), and \(P(X\leq27)=0.1587\). Estimate \(\mu\), using \(z_{0.1587}\approx-1.0000\). (Answer: \(\mu\approx31.00\))
- A hypothetical instrument output has distribution \(X\sim N(120,\sigma^2)\), with \(\sigma>0\), and \(P(X>132)=0.0228\). Estimate \(\sigma\), using \(z_{0.9772}\approx2.0000\). (Answer: \(\sigma\approx6.00\))
- Find the central 90% interval for \(X\sim N(60,7^2)\), using \(z_{0.95}=1.6449\). (Answer: approximately \((48.49,71.51)\))
- A hypothetical duration has distribution \(X\sim N(\mu,5^2)\), and \(P(X>35)=0.9332\). Estimate \(\mu\), using \(z_{0.0668}\approx-1.5000\). (Answer: \(\mu\approx42.50\))