AP Syllabus focus:
‘Determine the boundaries of an interval associated with a given area in a normal distribution using z-scores, calculators, standard normal tables, or computer-generated output. Understanding how to assign appropriate inequalities to the boundaries of these intervals is critical: a. The lowest p% of values lying to the left of x, b. p% of values lying between x and y, c. The highest p% of values lying to the right of x, and d. Dividing the area associated with p% into two equal parts for the most extreme values, ensuring half of the p% extreme values lie to the left and right of x respectively.’
Determining interval boundaries from specified areas in a normal distribution links probability to specific values, helping quantify how likely observations fall within, below, or above chosen cutpoints.
Determining Intervals from Given Areas in a Normal Distribution
Understanding how to translate a probability (or area under the curve) into an interval boundary is a fundamental skill when working with continuous random variables and the standard normal distribution. Because the normal curve associates every interval with a probability, locating boundaries requires interpreting areas using z-scores, calculator outputs, or standard normal tables. These tools allow us to connect proportions of a distribution to the specific values where those proportions begin or end.
The Role of Z-Scores in Interval Determination
A z-score measures how many standard deviations a value lies from the mean. It provides a standardized scale, enabling consistent interpretation across all normal distributions.
Z-score: A standardized value describing the number of standard deviations an observation lies from the mean in a normal distribution.
Using z-scores is essential because the standard normal distribution has a fixed shape, allowing any area-to-boundary question to be solved by locating the correct z-value and then translating it into the corresponding value in the original distribution.
Required Tools for Interval Construction
Students use several tools to determine interval boundaries:
Standard normal tables listing cumulative probabilities.
Graphing calculators with inverse normal functions.
Statistical software or computer-generated output for precise values.
Z-score formulas for converting between raw values and standardized values.
Regardless of the tool, the goal remains the same: identify the z-score where a chosen proportion of area lies to the left, between values, or in symmetric tails.
Understanding How Areas Translate to Interval Boundaries
Each scenario described in the syllabus corresponds to a specific relationship between area and boundary placement. Because probability area accumulates from left to right in the normal distribution, the location of an interval’s boundary depends on how much total area is specified and where that area is positioned.
Lowest p% to the Left of a Boundary
When the task is to find the value x such that the lowest p% of observations fall below it, the focus is on cumulative area.
This diagram illustrates a normal distribution where the area to the left of a cutoff value represents the lowest proportion of observations, demonstrating how cumulative area determines a boundary. The z-score shown provides the standardized location of that boundary. The specific numerical values extend beyond the syllabus but serve as an example of the concept. Source.
The boundary corresponds to a z-score where the cumulative probability equals p%.
This boundary is always located on the left portion of the distribution, and the inequality takes the form Value ≤ x.
Because cumulative area always increases from left to right, this type of boundary is determined by identifying the z-score whose cumulative table entry matches the specified probability.
Middle p% Between Two Boundaries
Another common task is to find two values x and y such that p% of all values lie between them.
The figure shows a central probability region bounded by two cutpoints, illustrating how a specified percentage of observations lies between two values. The equal unshaded tails highlight that the remaining probability lies outside the middle region. Contextual labels reflect a real-world application but extend beyond the syllabus requirements. Source.
The middle area is symmetric only when p% is centered around the mean.
When the area is not required to be symmetric, the boundaries will not be equally distant from the mean.
Determining these boundaries requires finding two z-scores: one representing the cumulative area up to x, and another representing the cumulative area up to y.
This scenario emphasizes proper assignment of inequalities, typically x ≤ Value ≤ y, so the student identifies the correct interior region of the curve.
Highest p% to the Right of a Boundary
When determining a value where the highest p% of observations lie to its right:
The area to the right equals p%, so the cumulative area to the left equals 1 − p%.
The inequality for this boundary is Value ≥ x.
Tools such as inverse normal functions require left-tail probabilities, so converting right-tail areas is a crucial step.
This scenario often leads to positive z-scores because the upper tail lies above the mean in a symmetric distribution.
Two-Tailed Extreme Values
The syllabus also emphasizes dividing p% into two equal tail areas to identify cutpoints for the most extreme values.
The diagram shows a standard normal distribution with a shaded central region bounded by two symmetric z-scores representing the middle probability area. The two unshaded tails represent extreme values equally split between the left and right sides. The specific use of z = ±1.96 reflects the classic 95% central region but slightly exceeds syllabus specificity. Source.
Each tail receives p/2%, spread symmetrically.
The resulting boundaries are located equidistant from the mean with one negative and one positive z-score.
Inequalities take the form Value ≤ x and Value ≥ y.
This structure is central to confidence intervals and hypothesis testing, where identifying extreme regions is essential.
Converting Z-Scores to Original Units
After determining z-scores, students translate them into raw values using the linear structure of normal distributions.
EQUATION
= Boundary value in the original distribution
= Mean of the distribution
= Standard deviation of the distribution
= Corresponding z-score from area
This equation ensures that boundaries determined from standard normal tools correctly map back to the original measurement scale of the population.
A clear understanding of this conversion allows students to interpret normal distribution results within real-world contexts and ensures proper assignment of interval boundaries when using given areas.
FAQ
In AP Statistics, interval boundaries derived from normal distributions are conventionally written using non-strict inequalities, such as x ≤ value, to indicate inclusion of the boundary point.
Because continuous distributions assign zero probability to any single point, strict and non-strict inequalities represent the same region in terms of probability.
However, non-strict notation is preferred for clarity and consistency in interval communication.
Standard normal tables and calculator inverse-normal functions typically work with left-tail probabilities only.
Thus, a right-tail probability must be converted by subtracting the given proportion from 1.
This ensures compatibility with statistical tools and avoids misinterpreting the area’s position on the curve.
When areas approach the extreme ends of the distribution, the z-scores become very large in magnitude.
In such cases:
• Tables may not list the exact probabilities.
• Calculators may round to infinity or fail to distinguish values close to zero or one.
Students should use technology capable of handling these edge cases, recognising that the resulting boundaries may be far from the mean.
Symmetric intervals centre the probability around the mean, giving cutpoints that are equidistant from the mean and have equal magnitude z-scores.
This means you only need one z-score to find both boundaries.
Asymmetric intervals require two cumulative probabilities, two inverse-normal lookups, and separate conversions to raw values.
Yes. The area under the standardised normal curve depends only on the z-scores, not the original mean or standard deviation.
If the same cumulative probabilities are used, the corresponding z-scores remain identical.
Only the raw interval boundaries shift, moving with the new mean while preserving the same proportion of the distribution.
Practice Questions
Question 1 (1–3 marks)
A variable follows a normal distribution with mean 50 and standard deviation 8.
Find the value x such that 10% of observations lie above x.
Give your answer to two decimal places.
Question 1
• 1 mark: Correctly identifies that the upper 10% corresponds to a left-tail probability of 0.90.
• 1 mark: Correctly obtains the z-score associated with cumulative probability 0.90 (approximately 1.28).
• 1 mark: Correctly calculates x = 50 + (1.28 × 8) = 60.24 (accept answers between 60.2 and 60.3).
Question 2 (4–6 marks)
A machine fills cereal boxes, and the amount dispensed follows a normal distribution with mean 420 g and standard deviation 12 g.
The manufacturer wants to identify the central 80% of filled amounts, creating an interval from a lower boundary L to an upper boundary U.
a) Determine the z-scores that mark the boundaries of the central 80% of a normal distribution.
b) Hence calculate the values of L and U.
c) Explain clearly what the interval [L, U] represents in the context of the problem.
Question 2
a)
• 1 mark: Recognises that the central 80% leaves 20% in the tails, with 10% in each tail.
• 1 mark: States the corresponding z-scores as approximately -1.28 and +1.28.
b)
• 1 mark: Calculates L = 420 + (-1.28 × 12) = 404.64 g (accept reasonable rounding).
• 1 mark: Calculates U = 420 + (1.28 × 12) = 435.36 g (accept reasonable rounding).
c)
• 1 mark: Provides a correct contextual interpretation, e.g., “Approximately 80% of all cereal boxes filled by the machine will contain between L and U grams of cereal.”
