Why Standardize Normal Values?
A normal model describes a variable in its original units, such as millimetres, minutes, or dollars. Standardizing converts a value to a dimensionless score that measures how far it is from the mean in standard deviations. Once the value is standardized, probabilities can be read from the same standard normal distribution regardless of the original units or parameters.
This is useful whenever a question asks for the probability that a normal observation is below a threshold, above a threshold, or between two bounds. The central task is to preserve the direction and meaning of the event while converting each original-scale value to its standard-normal score.
The subtraction of \(\mu\) measures the value's distance from the mean. Division by \(\sigma\) expresses that distance in standard deviation units. A positive score is above the mean, a negative score is below the mean, and a score of zero is exactly at the mean. A score is not itself a probability.
Why the Standardization Formula Works
For a normal variable \(X\sim N(\mu,\sigma^2)\), its density is \[ f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right). \] Set \(z=(x-\mu)/\sigma\), so \(x=\mu+\sigma z\) and \(dx=\sigma\,dz\). Under this change of variable, the density for \(Z\) is \[ f_Z(z)=f_X(\mu+\sigma z)\left|\frac{dx}{dz}\right| =\frac{1}{\sigma\sqrt{2\pi}}e^{-z^2/2}\sigma =\frac{1}{\sqrt{2\pi}}e^{-z^2/2}. \] This is the standard normal density. The factor \(\sigma\) from the change in scale cancels the \(1/\sigma\) in the original density, which is why all normal models can use the same standard normal cumulative probabilities.
Because \(\sigma>0\), standardizing is an increasing transformation: if \(a<b\), then \((a-\mu)/\sigma<(b-\mu)/\sigma\). Consequently, interval endpoints keep their order. The resulting probability rules are:
The normal distribution is continuous, so \(P(X=a)=0\) for any single value \(a\). Therefore, changing a strict inequality to a non-strict inequality at an endpoint does not change the probability. This fact does not mean that endpoints can be reversed or that a right-tail probability can be read directly as a left-tail probability.
A Reliable Calculation Process
A standard normal table or calculator may report cumulative left-tail values \(\Phi(z)\). Check the table convention before using it: a table of areas between zero and \(z\) does not directly give \(\Phi(z)\). When using a cumulative table, a right-tail probability is one minus the left-tail cumulative probability, and a between-value probability is the larger cumulative probability minus the smaller one.
Identify \(\mu\) and \(\sigma\) from \(X\sim N(\mu,\sigma^2)\). Remember that the second parameter is the variance, so the standard deviation is its square root.
For each original-scale value \(x\), calculate \(z=(x-\mu)/\sigma\), retaining the sign.
Use \(\Phi(z)\) for a left tail, \(1-\Phi(z)\) for a right tail, and a difference of cumulative probabilities for an interval.
Confirm the probability is between zero and one and that its size is plausible for the event. Use full-precision cumulative values for subtraction, then round the final probability to four decimal places.
Worked Applications
Worked Example: Left and Right Tails of a Measurement Model
A hypothetical laboratory sensor reading \(X\), in units, is modeled as \(X\sim N(72,8^2)\). Find the probabilities that a reading is below 60 and above 84. This is a measurement model, not an insurance cash-flow calculation; no interest rate or mortality assumption applies. The distribution and parameters are \(N(72,64)\), with mean \(\mu=72\) and standard deviation \(\sigma=8\).
Method. Standardize each threshold, then use a left-tail cumulative probability for “below” and its complement for “above.”
For 60, the score is \[ z_{60}=\frac{60-72}{8}=\frac{-12}{8}=-1.5. \] Therefore, \[ P(X<60)=\Phi(-1.5)=0.0668072013\ldots\approx 0.0668. \]
For 84, the score is \[ z_{84}=\frac{84-72}{8}=\frac{12}{8}=1.5. \] The event is a right tail, so \[ P(X>84)=1-\Phi(1.5)=1-0.9331927987\ldots =0.0668072013\ldots\approx 0.0668. \]
Both results are small, as expected: each threshold is 1.5 standard deviations from the mean in its respective direction. Common mistake avoided: using \(\Phi(1.5)\) as the probability above 84 would give 0.9332, the probability below 84, not the requested right-tail probability. The complement gives the correct answer.
Worked Example: Probability Between Two Bounds
A hypothetical battery test records operating time \(X\), in hours, with model \(X\sim N(100,12^2)\). Find the probability that a battery lasts between 82 and 124 hours, and the probability that it lasts outside that range. This is a measurement model; no interest rate or mortality assumption applies. The distribution and parameters are \(N(100,144)\), with \(\mu=100\) and \(\sigma=12\).
Method. Standardize both endpoints and subtract the lower cumulative probability from the upper one.
The lower endpoint gives \[ z_{82}=\frac{82-100}{12}=\frac{-18}{12}=-1.5, \] and the upper endpoint gives \[ z_{124}=\frac{124-100}{12}=\frac{24}{12}=2. \] Thus, \[ \begin{aligned} P(82<X<124) &=\Phi(2)-\Phi(-1.5)\\ &=0.9772498681\ldots-0.0668072013\ldots\\ &=0.9104426668\ldots\\ &\approx 0.9104. \end{aligned} \] The probability outside the interval is the complement: \[ P(X<82\text{ or }X>124)=1-0.9104426668\ldots =0.0895573332\ldots\approx 0.0896. \] The two probabilities sum to \(0.9104+0.0896=1.0000\) after rounding. Common mistake avoided: the lower score is negative, but its cumulative probability is \(\Phi(-1.5)\), not \(-\Phi(1.5)\). A probability cannot be negative; use the cumulative value at the signed score.
Worked Example: Combining Two Tail Events
A hypothetical quality-control device measures the length \(X\) of a component in millimetres. The model is \(X\sim N(50,2^2)\). Find the probability that a component is either shorter than 48 mm or longer than 54 mm. This is a measurement model; no interest rate or mortality assumption applies. The distribution and parameters are \(N(50,4)\), with \(\mu=50\) and \(\sigma=2\).
Method. Standardize both cutoffs, calculate the two disjoint tail probabilities, and add them. The events are disjoint because a component cannot be both shorter than 48 mm and longer than 54 mm.
For the lower cutoff, \[ z_{48}=\frac{48-50}{2}=-1, \qquad P(X<48)=\Phi(-1)=0.1586552539\ldots. \] For the upper cutoff, \[ z_{54}=\frac{54-50}{2}=2, \qquad P(X>54)=1-\Phi(2) =1-0.9772498681\ldots =0.0227501319\ldots. \] Therefore, \[ \begin{aligned} P(X<48\text{ or }X>54) &=0.1586552539\ldots+0.0227501319\ldots\\ &=0.1814053858\ldots\\ &\approx 0.1814. \end{aligned} \] As a check, the probability of being between 48 and 54 mm is \(1-0.1814053858\ldots=0.8185946142\ldots\), which rounds to 0.8186. The rounded probabilities add to 1.0000. Common mistake avoided: adding two standardized scores, or adding a right-tail cutoff’s cumulative value \(\Phi(2)\), does not calculate the probability of the two tail events. Calculate each event’s probability first, then add.
Precision and Common Exam Traps
A standard normal table usually prints only a limited number of digits. When finding an interval probability, subtracting two displayed, rounded table entries can differ slightly from subtracting their underlying values. For example, the exact cumulative values for the interval from \(-1\) to \(2\) give \[ \Phi(2)-\Phi(-1) =0.9772498681\ldots-0.1586552539\ldots =0.8185946142\ldots, \] which rounds to 0.8186. If a four-decimal table is used first, \(0.9772-0.1587=0.8185\); that is the subtraction of rounded table entries, not the exact probability rounded to four decimals. Use a calculator with adequate precision when possible, or retain extra table digits before the final rounding.
The most frequent errors are conceptual rather than arithmetic. A variance is not the denominator in the score formula; use the standard deviation. A right-tail event requires a complement. A between-value probability requires the upper cumulative value minus the lower one, even when one bound is negative. For a union of two disjoint tails, calculate both tail probabilities in the correct directions before adding. Finally, carry full precision until the final result: early rounding can alter the last reported digit.
Check Your Understanding
For each problem, standardize the relevant value or values before calculating. Round final probabilities to four decimal places.
- A hypothetical calibration reading \(X\sim N(36,4^2)\). What is the probability that a reading is below 30? (Answer: 0.0668)
- A hypothetical service duration \(X\sim N(18,3^2)\) minutes. What is the probability the duration exceeds 24 minutes? (Answer: 0.0228)
- A hypothetical instrument output \(X\sim N(500,25^2)\). Find the probability that the output is between 450 and 550. (Answer: 0.9545)
- A hypothetical component dimension \(X\sim N(75,10^2)\). Find the probability that the dimension is below 60 or above 90. (Answer: 0.1336)
- A hypothetical test duration \(X\sim N(240,20^2)\) seconds. Find the probability it is between 210 and 270 seconds, and the probability it is outside those bounds. (Answer: between 0.8664; outside 0.1336)