Probabilities Between Two Normal Cutoffs

Why Two Cutoffs Require Two Cumulative Areas

Many questions about a normally distributed quantity ask whether it falls within a range: whether a delivery takes between two times, a measurement lies between two limits, or a score falls between two values. The interval has a lower cutoff and an upper cutoff. To find the probability inside it, we need the cumulative area up to the upper cutoff and must remove the area that lies below the lower cutoff.

The calculation works by expressing both cutoffs on the standard normal scale. The resulting probability is the difference between two left-tail cumulative probabilities. The order matters: subtract the cumulative area at the lower cutoff from the cumulative area at the upper cutoff.

Definition: Suppose $X\sim N(\mu,\sigma^2)$, where $\sigma>0$, and let $a<b$ be two cutoffs. The probability that $X$ falls between them is $$ P(a<X<b) = \Phi\!\left(\frac{b-\mu}{\sigma}\right) - \Phi\!\left(\frac{a-\mu}{\sigma}\right), $$ where $\Phi(z)=P(Z\le z)$ is the cumulative distribution function of a standard normal variable $Z\sim N(0,1)$.

This formula assumes that the normal model is an appropriate description of the variable of interest and that the specified $\mu$ and $\sigma$ are the model's mean and standard deviation in the original units. Standardizing a value $x$ gives $z=(x-\mu)/\sigma$: the number of standard deviations $x$ is above or below the mean. Since $\sigma$ is positive, standardizing preserves the order of the cutoffs.

From an Interval to a Difference

The standard normal cumulative function gives the area to the left of a specified $z$-score. Write $z_a=(a-\mu)/\sigma$ and $z_b=(b-\mu)/\sigma$. Then $z_a<z_b$, and the event that $X$ lies between $a$ and $b$ is the same event as $Z$ lying between $z_a$ and $z_b$. The area between those scores is the area to the left of the upper score minus the area to the left of the lower score:

\begin{aligned} P(a<X<b) &=P(z_a<Z<z_b)\\ &=P(Z<z_b)-P(Z\le z_a)\\ &=\Phi(z_b)-\Phi(z_a). \end{aligned}

The subtraction follows because the area to the left of $z_b$ consists of two non-overlapping pieces: the area to the left of $z_a$, plus the area between $z_a$ and $z_b$. Removing the first piece leaves the desired interval area.

A normal distribution is continuous, so the probability of any single exact value is zero. Thus $P(a<X<b)$, $P(a\le X\le b)$, and versions that include just one endpoint all have the same value. In practice, the cutoffs still need to be interpreted carefully: for example, “between 10 and 20” refers to the interval bounded by those two values, not to a tail beyond either one.

Worked Example: An Interval That Straddles the Mean

Suppose a hypothetical measurement $X$ follows $N(100,15^2)$, so its mean is 100 units and its standard deviation is 15 units. Find the probability that a measurement is between 85 and 130 units.

Check the model and cutoffs. The problem specifies a normal model with $\mu=100$ units and $\sigma=15$ units. The lower cutoff $a=85$ is less than the upper cutoff $b=130$, so the interval is in the required order.

Standardize both cutoffs.

z_a=\frac{85-100}{15}=-1, \qquad z_b=\frac{130-100}{15}=2.

Subtract the cumulative areas. Using unrounded standard normal cumulative values, $\Phi(2)=0.9772498681$ and $\Phi(-1)=0.1586552539$.

\begin{aligned} P(85<X<130) &=\Phi(2)-\Phi(-1)\\ &=0.9772498681-0.1586552539\\ &=0.8185946142\\ &\approx 0.8186. \end{aligned}

Under this hypothetical model, about 81.86% of measurements fall between 85 and 130 units. The probability is the area inside the specified interval; it is not the probability of observing either exact endpoint.

Intervals on One Side of the Mean

The same difference formula applies whether the interval straddles the mean, lies above it, or lies below it. If both standardized cutoffs are positive, both cumulative areas may be greater than 0.5, but their difference still gives the narrow area between them. If both are negative, each cumulative area is less than 0.5; the upper cutoff still has the larger cumulative area because it is farther to the right.

A standard normal table commonly reports $\Phi(z)$, the left-tail area, though some tables use a different layout. Use the table convention stated with the table. For a left-tail table, look up the cumulative area for each standardized cutoff, then subtract the lower-cutoff area from the upper-cutoff area.

Worked Example: An Interval Above the Mean

A hypothetical package-filling process is modeled by $X\sim N(50,8^2)$, with fill amounts measured in milliliters. Find the probability that a package contains between 54 and 64 milliliters.

The model specifies $\mu=50$ milliliters and $\sigma=8$ milliliters, and the cutoffs are ordered $54<64$. Standardize both using the same $\mu$ and $\sigma$:

z_a=\frac{54-50}{8}=0.50, \qquad z_b=\frac{64-50}{8}=1.75.

The cumulative standard normal values are $\Phi(0.50)=0.6914624613$ and $\Phi(1.75)=0.9599408431$. Subtract the lower cumulative area from the upper one:

\begin{aligned} P(54<X<64) &=\Phi(1.75)-\Phi(0.50)\\ &=0.9599408431-0.6914624613\\ &=0.2684783818\\ &\approx 0.2685. \end{aligned}

The modeled probability of a fill amount between 54 and 64 milliliters is about 0.2685, or 26.85%. Both cutoffs are above the mean, but their cumulative areas include all values below them; subtracting removes the area below 54 milliliters.

Worked Example: An Interval Below the Mean

Suppose a hypothetical task completion time $X$ follows $N(40,6^2)$, with times measured in minutes. Calculate the probability that completion takes between 31 and 37 minutes.

Here $\mu=40$ minutes and $\sigma=6$ minutes. The two cutoffs satisfy $31<37$. Their standardized values are:

z_a=\frac{31-40}{6}=-1.50, \qquad z_b=\frac{37-40}{6}=-0.50.

Using cumulative standard normal values $\Phi(-0.50)=0.3085375387$ and $\Phi(-1.50)=0.0668072013$, calculate:

\begin{aligned} P(31<X<37) &=\Phi(-0.50)-\Phi(-1.50)\\ &=0.3085375387-0.0668072013\\ &=0.2417303374\\ &\approx 0.2417. \end{aligned}

The model assigns about a 24.17% probability to a completion time between 31 and 37 minutes. Both $z$-scores are negative, but the upper cutoff, $-0.50$, has the greater cumulative area.

Why the Calculation Is Useful

A probability between two cutoffs answers a range question directly. It can describe the chance that a value lies within acceptable limits, belongs to a specified score band, or falls inside a measurement range. Standardizing makes the calculation independent of the original measurement units: the same standard normal areas can be used for different normally modeled variables, provided each cutoff is standardized with its own model's mean and standard deviation.

The result remains tied to the assumptions of the model. A calculated normal probability is not, by itself, evidence that the real-world variable follows a normal distribution. It is the probability implied by the stated normal model. Also, the result refers to the probability for an individual value $X$; it is not automatically a probability about an average of several observations.

Common Mistakes

Subtracting in the wrong order. The wrong reasoning is to subtract the upper-cutoff cumulative area from the lower-cutoff area, producing a negative value. Because $a<b$ implies $z_a<z_b$ and $\Phi(z_a)<\Phi(z_b)$, use $\Phi(z_b)-\Phi(z_a)$.
Using original cutoffs in a standard normal table. A table for $Z\sim N(0,1)$ takes standardized scores, not values in minutes, grams, or other original units. Calculate $z_a$ and $z_b$ first.
Taking a tail instead of the interval. The area to the left of the upper cutoff includes the area below the lower cutoff. Subtract that lower area to leave only the interval.
Rounding too early. Rounded table entries may make the final result differ in the last reported digit. Keep adequate precision in standardized values and cumulative areas, subtract before rounding, and report a suitably rounded probability.
Interpreting the probability as an observed percentage guaranteed in every sample. The model probability describes long-run chances under the stated normal model; it does not ensure that a particular finite group will have exactly that percentage in the interval.

Key takeaway: For a normal variable and ordered cutoffs $a<b$, standardize both endpoints and subtract the lower endpoint's cumulative area from the upper endpoint's cumulative area: $P(a<X<b)=\Phi((b-\mu)/\sigma)-\Phi((a-\mu)/\sigma)$.

Check Your Understanding

For each question, identify the two standardized cutoffs before finding or describing the interval probability.

A variable follows $N(20,4^2)$. Write the expression for the probability that it lies between 18 and 26.
Why must the cumulative area for the lower cutoff be subtracted from the area for the upper cutoff?
A variable has standard normal cutoffs $-1.2$ and $0.4$. Write the probability between them using $\Phi$, without evaluating it.
For a continuous normal variable, does including an endpoint change the probability of an interval? Explain briefly.
A student subtracts $\Phi(z_a)$ from $\Phi(z_b)$ but obtains a negative result, even though $a<b$. What should the student check first?