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Normal distributions · Tutorial 374 of 1000

Finding Values Between Symmetric Boundaries

Use equal tail areas and invNorm to find the symmetric cutoffs for the middle 80% and the first and third quartiles of a normal model.

Intermediate 9 min read

What You'll Learn

  • Convert the middle 80% into equal 10% tail areas.
  • Use invNorm to find both boundaries of a central normal-model interval.
  • Find the first and third quartiles using the 25th and 75th percentiles.
  • Use symmetry to check whether two cutoffs are centered at the mean.
  • Interpret the cutoffs and the interquartile range in context.

Two Cutoffs Around the Center

In Finding a Value from a Percentile with invNorm, you used a cumulative area to find one cutoff. Some questions ask for two cutoffs that enclose a central portion of a normal model: for example, the lower and upper values containing the middle 80% of measurements. Others ask for the first and third quartiles, the boundaries of the middle 50%.

For these questions, the important idea is symmetry. A normal model is centered at its mean, and equal distances on opposite sides of the mean have matching areas. So a central interval is found by dividing the area outside it equally between the two tails, then using invNorm to find the lower and upper cutoffs.

Definition: The middle \(C\) proportion of a normal model is the central area between two cutoffs, with equal tail areas on either side. The first quartile, \(Q_1\), is the 25th percentile; the third quartile, \(Q_3\), is the 75th percentile. The quartiles bound the middle 50% of the model.

Finding the Middle 80%

If the middle 80% lies between two cutoffs, the remaining 20% must lie outside them. Because the interval is symmetric, half of that 20% is in the lower tail and half is in the upper tail: 10% in each. The lower cutoff is therefore at the 10th percentile, and the upper cutoff is at the 90th percentile.

$$ \text{Middle 80\%: }\quad L=\operatorname{invNorm}(0.10,\mu,\sigma),\qquad U=\operatorname{invNorm}(0.90,\mu,\sigma) $$

Here, \(L\) is the lower cutoff and \(U\) is the upper cutoff. The two area inputs add to 1, reflecting their symmetry around the mean. The values returned by invNorm are in the original units of the random variable.

You can also think in standard-deviation units. The standard-normal cutoffs for the 10th and 90th percentiles are about \(-1.282\) and \(1.282\). Thus, the middle 80% of a normal model extends about 1.282 standard deviations below the mean and 1.282 standard deviations above it. For a particular model, invNorm applies those cutoffs to its mean and standard deviation.

Formula: For a central proportion \(C\), the two tails each have area \((1-C)/2\). The lower cutoff uses left-tail area \((1-C)/2\), and the upper cutoff uses left-tail area \(1-(1-C)/2\). For the middle 80%, these areas are 0.10 and 0.90.

An equivalent check is to confirm that the two calculated cutoffs are equally far from the mean. If the model is \(N(\mu,\sigma)\), then \(L\) should be below \(\mu\), \(U\) should be above \(\mu\), and the distances \(\mu-L\) and \(U-\mu\) should match apart from rounding.

Finding the Quartiles

The first quartile \(Q_1\) has 25% of the model’s area to its left. The third quartile \(Q_3\) has 75% to its left, which means 25% to its right. These percentiles leave equal areas in the two tails, so the interval from \(Q_1\) to \(Q_3\) is centered at the mean in a normal model. It contains the middle 50%.

$$ Q_1=\operatorname{invNorm}(0.25,\mu,\sigma),\qquad Q_3=\operatorname{invNorm}(0.75,\mu,\sigma) $$

For a standard normal distribution, the quartile z-scores are approximately \(-0.6745\) and \(0.6745\). The quartiles of any normal model therefore lie about 0.6745 standard deviations below and above its mean. The difference \(Q_3-Q_1\), called the interquartile range or IQR, measures the width of the middle 50% in the variable’s original units.

To use these results, name the random variable and its model, decide which central interval is requested, and enter the appropriate left-tail areas into invNorm. The calculator input order is area, mean, standard deviation, as in the previous tutorial. A sketch with equal tails marked can help you verify the areas before calculating.

A Procedure for Symmetric Cutoffs

1
Define the variable and model.
State what \(X\) measures, its units, and the given normal model \(N(\mu,\sigma)\).
2
Find each tail area.
For a middle 80%, use 0.10 in each tail. For quartiles, use 0.25 in each outside tail.
3
Convert to left-tail areas.
Use 0.10 and 0.90 for the middle 80%; use 0.25 and 0.75 for the quartiles.
4
Calculate and check.
Use invNorm for both cutoffs. Check that the lower value is below the mean, the upper value is above it, and their midpoint is the mean.
5
Interpret in context.
Report both cutoffs with units and explain what proportion of modeled values lies between them.

The normal model must be appropriate for the variable, as assumed or established in the problem. These calculations describe proportions under that model; they do not prove that real-world values follow a normal distribution. Keep the mean, standard deviation, and reported cutoffs in the same units.

Worked Example: The Middle 80% of Battery Life

Worked Example: The Middle 80% of Battery Life

In a fictional product-planning scenario, the operating life of a device battery is modeled as normal with mean 72 hours and standard deviation 6 hours. Find the cutoffs containing the middle 80% of modeled battery lives.

State. Let \(X\) be the operating life of a device battery, in hours. The model is \(X\sim N(72,6)\). We want two cutoffs with 80% of the model’s area between them.

Plan. The area outside the middle 80% is \(1-0.80=0.20\). By symmetry, each tail has area \(0.20/2=0.10\). The lower cutoff has 0.10 to its left; the upper cutoff has 0.90 to its left.

Do. Use invNorm with each left-tail area, followed by the mean and standard deviation:

$$ L=\operatorname{invNorm}(0.10,72,6)\approx64.3107\text{ hours} $$
$$ U=\operatorname{invNorm}(0.90,72,6)\approx79.6893\text{ hours} $$

As a symmetry check, \(72-64.3107=7.6893\) hours and \(79.6893-72=7.6893\) hours. A check using the original model gives \(\operatorname{normalcdf}(64.3107,79.6893,72,6)\approx0.8000\), rounded to four decimal places.

Conclude. The middle 80% of modeled battery lives lies between about 64.31 and 79.69 hours. Under this model, about 80% of batteries have operating lives between those values, with about 10% below the lower cutoff and 10% above the upper cutoff.

Worked Example: The Quartiles of Assessment Scores

Worked Example: The Quartiles of Assessment Scores

Scores on a fictional assessment are modeled as normal with mean 500 points and standard deviation 40 points. Find \(Q_1\) and \(Q_3\), and calculate the interquartile range.

State. Let \(X\) be an assessment score, in points, with model \(X\sim N(500,40)\). The first quartile is the 25th percentile, and the third quartile is the 75th percentile.

Plan. Use left-tail areas 0.25 and 0.75 as the invNorm inputs. The quartiles should be equally far from the mean because the normal model is symmetric. The IQR is \(Q_3-Q_1\).

Do. Calculate both quartiles with the model’s mean and standard deviation:

$$ Q_1=\operatorname{invNorm}(0.25,500,40)\approx473.0204\text{ points} $$
$$ Q_3=\operatorname{invNorm}(0.75,500,40)\approx526.9796\text{ points} $$

The IQR is \(526.9796-473.0204=53.9592\) points, or about 53.96 points. The midpoint of the quartiles is \((473.0204+526.9796)/2=500.0000\) points, matching the mean. By their percentile definitions, 25% of modeled scores lie at or below \(Q_1\), and 75% lie at or below \(Q_3\); the area between the quartiles is 0.75-0.25=0.50.

Conclude. The first and third quartiles are about 473.02 and 526.98 points. The middle 50% of modeled assessment scores lies between these values, and its interquartile range is about 53.96 points.

Worked Example: The Middle 80% of Delivery Times

Worked Example: The Middle 80% of Delivery Times

For a fictional delivery route, delivery times are modeled as normal with mean 18 minutes and standard deviation 2.5 minutes. Find and interpret the middle 80% of delivery times.

State. Let \(X\) be a delivery time, in minutes, with model \(X\sim N(18,2.5)\). We need the lower and upper boundaries of the central 80%.

Plan. The two tails together contain 20% of the model. Each has area 0.10, so the boundaries are the 10th and 90th percentiles. Because the model’s mean is 18 minutes, the interval should be centered at 18.

Do. Enter the two left-tail areas into invNorm:

$$ L=\operatorname{invNorm}(0.10,18,2.5)\approx14.7961\text{ minutes} $$
$$ U=\operatorname{invNorm}(0.90,18,2.5)\approx21.2039\text{ minutes} $$

The distances from the mean are \(18-14.7961=3.2039\) minutes and \(21.2039-18=3.2039\) minutes. Also, \(\operatorname{normalcdf}(14.7961,21.2039,18,2.5)\approx0.8000\), rounded to four decimal places.

Conclude. The middle 80% of modeled delivery times is approximately 14.80 to 21.20 minutes. The interval is centered at the mean, and the model assigns about 10% of delivery times to each tail beyond these cutoffs.

Common Mistakes and AP Exam Tips

The main challenge is translating “middle” into the correct tail areas. The middle percentage is not the area input for either individual cutoff. First find what is outside the interval, split that area equally between the tails, and then express each boundary as a left-tail area for invNorm.

  • Using 0.80 for one boundary. The middle 80% is between two values; it does not mean either boundary is the 80th percentile. Use 0.10 for the lower cutoff and 0.90 for the upper cutoff.
  • Forgetting to divide the outside area in half. For the middle 80%, 20% is outside the interval, but each tail has only 10%. For quartiles, each outside tail has 25%.
  • Reversing the lower and upper areas. A smaller left-tail area produces a lower cutoff. Thus 0.10 gives the lower middle-80% cutoff, while 0.90 gives the upper cutoff.
  • Using the wrong calculator order. The inputs are area, mean, and standard deviation. A command with the model parameters in a different order does not give the intended cutoff.
  • Reporting only z-scores. The z-scores \(\pm1.282\) and \(\pm0.6745\) describe standardized distances. If the question asks for measurement cutoffs, use the model’s mean and standard deviation to report values in the original units.
  • Giving an incomplete interpretation. State both cutoff values with units and identify the proportion of modeled values between them. For the middle 80%, also identify the 10% in each tail when useful.

A full-credit response names the random variable and normal model, shows how the requested central area determines both tail areas, gives both invNorm calculations, and interprets the results in context. As a final reasonableness check, confirm that the lower cutoff is below the mean, the upper cutoff is above it, and their midpoint is approximately the mean.

AP Exam Tip: For a central interval, calculate the two tail areas before using invNorm. The calculator needs left-tail areas, so enter the lower-tail area for the lower cutoff and its complement for the upper cutoff.

Key Takeaway

Symmetric boundaries of a normal model come from equal tail areas. For the middle 80%, find the 10th and 90th percentiles. For the quartiles, find the 25th and 75th percentiles; these bound the middle 50%, and their difference is the IQR.

Key takeaway: Divide the area outside the requested central interval equally between its two tails, convert both boundaries to left-tail areas, and use invNorm to find cutoffs in the original units. Check that the boundaries are symmetric around the mean.

Check Your Understanding

For each question, identify the left-tail areas to use and explain what the resulting cutoffs represent.

  1. A fictional plant-height model is \(X\sim N(24,3)\), with height measured in centimeters. Write the two invNorm expressions for the middle 80%.
  2. For a normal model with mean 80 and standard deviation 12, write the invNorm expressions for \(Q_1\) and \(Q_3\). What proportion of the model lies between them?
  3. A student uses areas 0.20 and 0.80 to find the middle 80% of a normal model. Explain the error and give the correct areas.
  4. For \(X\sim N(18,2.5)\), the middle-80% cutoffs are approximately 14.7961 and 21.2039. Use their midpoint to check the model’s symmetry.
  5. If a normal model has \(Q_1=42\) and \(Q_3=58\), what is its IQR, and what proportion of model values lies between the quartiles?