Normal Probabilities Across Disjoint Intervals

Why Separate Intervals Need Careful Addition

A question about a normal variable may ask for the probability of falling in several separated ranges rather than in one continuous interval. For example, a quality-control rule might flag measurements that are unusually low, moderately high, or unusually high. Each range has its own area under the normal curve. The total probability is the sum of those areas when the ranges are disjoint.

The essential steps are to translate each cutoff to the standard normal scale, find the area for each interval, and combine the results. The main safeguard is to check that no value is counted in more than one interval. Adding areas is justified by disjointness, not merely by the fact that several ranges appear in the question.

Definition: Events are disjoint if they cannot occur at the same time. For disjoint events $A_1,\ldots,A_k$, the addition rule is $$ P(A_1\text{ or }\cdots\text{ or }A_k)=\sum_{i=1}^{k}P(A_i). $$ A union of disjoint intervals is a set of values consisting of two or more separated ranges, with no value belonging to more than one range.

Standardizing Each Cutoff

Suppose $X\sim N(\mu,\sigma^2)$, where $\mu$ is the population mean and $\sigma>0$ is the population standard deviation. The standardized variable is $Z=(X-\mu)/\sigma$, and it follows the standard normal distribution $N(0,1)$. Standardizing a cutoff $c$ gives $z=(c-\mu)/\sigma$. Thus a probability about $X$ can be evaluated as a probability about $Z$.

Write $\Phi(z)=P(Z\le z)$ for the cumulative standard normal probability to the left of $z$. A single interval between cutoffs $a$ and $b$, with $a<b$, has probability

P(a<X<b) =\Phi\left(\frac{b-\mu}{\sigma}\right) -\Phi\left(\frac{a-\mu}{\sigma}\right).

For several disjoint intervals, apply this difference to each interval and add the results. A left-tail interval uses $\Phi(z)$; a right-tail interval uses $1-\Phi(z)$. These are the same cumulative-area and complement rules used for single normal probabilities, now applied interval by interval.

List the intervals.
Write each range clearly and check that the ranges do not overlap.

Standardize the endpoints.
For every cutoff $c$, calculate $z=(c-\mu)/\sigma$, retaining enough precision for the lookup.

Find each area.
Use cumulative probabilities, subtracting the lower cumulative area from the upper one for a bounded interval.

Add the disjoint areas.
Combine the component probabilities using their unrounded values when possible, then round the total.

Worked Example: Two Symmetric Tail Intervals

Suppose a hypothetical measurement $X$ follows $N(100,15^2)$, with mean 100 units and standard deviation 15 units. Find the probability that a measurement is below 85 units or above 115 units.

The two intervals, $X<85$ and $X>115$, are disjoint: a single measurement cannot be below 85 and above 115 at once. The model is specified as normal, and its mean and standard deviation are given, so standardization applies.

The standardized cutoffs are $z_{85}=(85-100)/15=-1$ and $z_{115}=(115-100)/15=1$. The probability is the sum of the left and right tails:

\begin{aligned} P(X<85\text{ or }X>115) &=\Phi(-1)+[1-\Phi(1)]\\ &=2\Phi(-1)\\ &\approx 0.3173105\\ &\approx 0.3173. \end{aligned}

The symmetry of the standard normal distribution gives $1-\Phi(1)=\Phi(-1)$. Thus, about 31.73% of measurements fall outside the interval from 85 to 115 units under this model. This probability is the combined area of both tails, not the area of either tail alone.

Adding Bounded Intervals

For a bounded interval, the cumulative probability at its lower endpoint must be subtracted from the cumulative probability at its upper endpoint. It is not enough to add the two endpoint areas: that would include values below the interval. For a union of several bounded ranges, calculate one difference for each range, then add the interval probabilities.

Worked Example: Two Separate Bounded Ranges

Let $X\sim N(50,8^2)$ represent a hypothetical measurement in units. Find the probability that $X$ is between 42 and 46 units or between 54 and 62 units.

The intervals $(42,46)$ and $(54,62)$ do not overlap, because the first ends below the second begins. The normal model and its parameters are specified. Standardizing the four cutoffs gives $42\mapsto -1$, $46\mapsto -0.5$, $54\mapsto 0.5$, and $62\mapsto 1.5$.

For the first interval, subtract the cumulative area to the left of $-1$ from the area to the left of $-0.5$. For the second, subtract the area to the left of $0.5$ from the area to the left of $1.5$. Using standard normal cumulative probabilities:

\begin{aligned} P(42<X<46\text{ or }54<X<62) &=[\Phi(-0.5)-\Phi(-1)]+[\Phi(1.5)-\Phi(0.5)]\\ &\approx 0.149882285+0.241730337\\ &\approx 0.391612622\\ &\approx 0.3916. \end{aligned}

The answer is about 39.16%. Each bracket is the area of one interval, and the two areas can be added because the intervals are separate. The displayed component values and total are rounded approximations to the corresponding normal probabilities.

Combining Tails and a Middle Interval

A union may mix one or more bounded intervals with tail intervals. The same rule still works: find the probability of each piece in the form that suits it, then add the pieces if they are disjoint. A left tail is a cumulative probability, a right tail is a complement, and a bounded range is a difference of cumulative probabilities.

Worked Example: Two Tails and a Bounded Interval

Suppose $X\sim N(100,15^2)$ measures a hypothetical quantity in units. Find the probability that $X<70$, or $95<X<110$, or $X>130$.

The three ranges do not overlap: the left tail ends below the bounded interval, and the right tail starts above it. The given normal model supplies the required assumptions and parameters. The standardized cutoffs are $70\mapsto -2$, $95\mapsto -1/3$, $110\mapsto 2/3$, and $130\mapsto 2$.

The two tails each have probability $\Phi(-2)$, by symmetry. The middle interval has probability $\Phi(2/3)-\Phi(-1/3)$. Therefore:

\begin{aligned} &P(X<70\text{ or }95<X<110\text{ or }X>130)\\ &\quad=\Phi(-2)+[\Phi(2/3)-\Phi(-1/3)]+[1-\Phi(2)]\\ &\quad=2\Phi(-2)+\Phi(2/3)-\Phi(-1/3)\\ &\quad\approx 0.02275013195+0.3780661223+0.02275013195\\ &\quad\approx 0.4235663862\\ &\quad\approx 0.4236. \end{aligned}

Thus, approximately 42.36% of values fall in at least one of the specified ranges under this model. The total is computed from the unrounded probabilities; rounding each component too early can make the last reported digits inconsistent.

Why Addition Works

The addition step follows directly from the definition of probability for disjoint events. If a value of $X$ belongs to exactly one of the listed intervals, then the event “$X$ is in at least one interval” is the union of mutually exclusive events. Probability adds across such a union. For a continuous normal variable, this also has a geometric interpretation: the probability is area under the density curve, and the areas of non-overlapping pieces add.

P(A_1\cup\cdots\cup A_k) =\sum_{i=1}^{k}P(A_i) \qquad\text{when }A_i\cap A_j=\varnothing\text{ for }i\ne j.

This argument does not require the intervals to be symmetric, equally wide, or equally likely. It requires only that their events be disjoint. If two intervals overlap, their shared region would be counted twice by simple addition. In that case, first rewrite the event as non-overlapping pieces, or use the general two-event rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

For a continuous normal distribution, the probability of any exact endpoint is zero. Consequently, changing an interval endpoint from $<$ to $\le$, or from $>$ to $\ge$, does not change its probability. This fact makes endpoint conventions immaterial for the calculations here, though the intervals should still be described precisely.

Common Mistakes and Checks

Adding endpoint cumulative areas. For $a<X<b$, the probability is the cumulative area at $b$ minus the cumulative area at $a$, not their sum.
Using addition for overlapping ranges. If ranges share values, their areas cannot simply be added; the shared probability would be counted more than once.
Forgetting a right-tail complement. The probability above a cutoff with standardized value $z$ is $1-\Phi(z)$, not $\Phi(z)$.
Rounding each component too early. Keep full table or calculator precision during the calculation and round the final total. When showing approximate components, make clear that the total uses unrounded values.
Ignoring scale or context. Standardize with $z=(c-\mu)/\sigma$, using the model’s standard deviation $\sigma$, and interpret the result as a probability for the specified quantity.

A useful reasonableness check is to compare the result with the event’s location and width. A narrow interval near the mean usually has less area than a much wider interval; a far-tail interval should not be mistaken for a central one. The total probability must be between 0 and 1. If the component areas are disjoint, their sum cannot exceed 1.

Key takeaway: For a normal variable, standardize every cutoff, calculate the area of each listed interval with cumulative probabilities, and add the areas only after confirming the intervals are disjoint. Preserve precision until the final rounding.

Check Your Understanding

Answer each question by identifying the relevant areas and explaining whether addition is justified.

For $X\sim N(20,4^2)$, write an expression using $\Phi$ for $P(X<16\text{ or }X>28)$. Are the two events disjoint?
For $X\sim N(50,10^2)$, write an expression using $\Phi$ for the probability that $X$ is between 40 and 45 or between 60 and 70.
Why must the cumulative probability at the lower endpoint be subtracted when finding the area between two cutoffs?
What problem arises if someone adds the probabilities of $X<5$ and $X<10$ to find the probability that $X<5$ or $X<10$?
For a continuous normal variable, does replacing $X< a$ with $X\le a$ change the probability? Explain briefly.