Tutorials › AP Statistics › Working Backward to Find Mean or Standard Deviation

Home  /  Tutorials  /  Normal distributions

Normal distributions · Tutorial 375 of 1000

Working Backward to Find Mean or Standard Deviation

Use a known percentile and one known normal-model parameter to solve for the unknown mean or standard deviation.

Intermediate 9 min read

What You'll Learn

  • Translate a percentile into a left-tail area and find its standard-normal z-score.
  • Rearrange the normal quantile equation to solve for an unknown mean.
  • Rearrange the same equation to solve for an unknown standard deviation.
  • Check whether the percentile’s position agrees with the sign of the z-score.
  • Recognize why the 50th percentile cannot determine a standard deviation.

Working Backward from a Percentile

In Finding a Value from a Percentile with invNorm, you used a normal model’s mean and standard deviation to find a cutoff. Sometimes the cutoff and percentile are known, but one of the model’s parameters is missing. You can work backward to find the unknown parameter.

The key is to express the percentile cutoff in standard-deviation units. The z-score for a percentile tells how far its cutoff is from the mean, measured in standard deviations. This turns the relationship among the cutoff, mean, and standard deviation into an equation you can rearrange.

Definition: The \(p\)th percentile of a normal model is a value with proportion \(p\) of the model’s area to its left, where \(p\) is written as a decimal. Its standard-normal z-score, \(z_p\), satisfies \(P(Z\leq z_p)=p\). The corresponding cutoff \(x_p\) in a model \(N(\mu,\sigma)\) satisfies \(x_p=\mu+z_p\sigma\).

For example, the 90th percentile uses a left-tail area of 0.90. Its standard-normal z-score is about 1.2816, so its cutoff is about 1.2816 standard deviations above the mean. The 20th percentile uses a left-tail area of 0.20 and has a z-score of about \(-0.8416\), so its cutoff is about 0.8416 standard deviations below the mean.

$$ x_p=\mu+z_p\sigma $$

If the percentile and standard deviation are known, rearrange the equation to find the mean. If the percentile and mean are known, rearrange it to find the standard deviation. Find \(z_p\) from the percentile using \(\operatorname{invNorm}(p,0,1)\), or use a supplied z-score if the question provides one.

$$ \mu=x_p-z_p\sigma \qquad\qquad \sigma=\frac{x_p-\mu}{z_p}\quad (z_p\ne 0) $$

The sign of \(z_p\) matters. A percentile below 50% has a negative z-score, and a percentile above 50% has a positive z-score. The standard deviation must be positive. Therefore, a proposed percentile cutoff must be below the mean when its z-score is negative and above the mean when its z-score is positive.

A Backward-Solving Strategy

Before calculating, identify the percentile as a left-tail area, then find its standard-normal z-score. Keep the cutoff, mean, and standard deviation in the variable’s original units; the z-score has no units. Use the equation to solve for the one unknown parameter.

1
Define the variable and model information.
State what \(X\) measures and its units. Identify the known percentile cutoff and the known parameter.
2
Find the percentile z-score.
Enter the percentile as a left-tail area in \(\operatorname{invNorm}(p,0,1)\). Note whether the result is positive, negative, or zero.
3
Rearrange and calculate.
Use \(x_p=\mu+z_p\sigma\). Solve for \(\mu\) or \(\sigma\), showing the substitution and result.
4
Check and interpret.
Confirm that the percentile cutoff is on the correct side of the mean and that the standard deviation is positive. Interpret the parameter with units in context.

These steps assume the variable is appropriately modeled by the stated normal distribution and the percentile information refers to that model. They do not establish that real-world data are normal. If the percentile is exactly the 50th percentile, its z-score is zero: the equation identifies the mean, but it contains no information about the standard deviation.

Worked Example: Finding a Mean from the 90th Percentile

Worked Example: Finding a Mean from the 90th Percentile

In a fictional packaging scenario, the amount of liquid in a container is modeled as normal. The 90th percentile is 530 milliliters, and the standard deviation is 15 milliliters. Find the mean amount.

State. Let \(X\) be the amount of liquid in a container, in milliliters. The model is \(X\sim N(\mu,15)\), and its 90th-percentile cutoff is \(x_{0.90}=530\) milliliters.

Plan. The 90th percentile has left-tail area 0.90. Find its standard-normal z-score, then use \(\mu=x_p-z_p\sigma\). Since the 90th percentile is above the 50th percentile, its z-score should be positive and the cutoff should be above the mean.

Do. First find the z-score, then substitute the known cutoff and standard deviation:

$$ z_{0.90}=\operatorname{invNorm}(0.90,0,1)\approx1.2815516 $$
$$ \mu=530-(1.2815516)(15) =530-19.223274 \approx510.7767\text{ milliliters} $$

The result is below 530 milliliters, as expected for a 90th-percentile cutoff. A check gives \((530-510.7767)/15\approx1.2816\), matching the 90th-percentile z-score.

Conclude. The model’s mean liquid amount is about 510.78 milliliters. Under the model, 530 milliliters is approximately 1.2816 standard deviations above that mean, with 90% of modeled amounts at or below it.

Worked Example: Finding a Standard Deviation Above the Mean

Worked Example: Finding a Standard Deviation Above the Mean

For a fictional sensor reading, the readings are modeled as normal with mean 52 units. The 90th percentile is 64.1 units. Find the standard deviation.

State. Let \(X\) be a sensor reading, in units. The model is \(X\sim N(52,\sigma)\), and \(x_{0.90}=64.1\).

Plan. The 90th-percentile z-score is positive. Solve \(\sigma=(x_p-\mu)/z_p\). Because the cutoff is above the mean, the numerator and z-score should both be positive, giving a positive standard deviation.

Do. Find the z-score and substitute:

$$ z_{0.90}=\operatorname{invNorm}(0.90,0,1)\approx1.2815516 $$
$$ \sigma=\frac{64.1-52}{1.2815516} =\frac{12.1}{1.2815516} \approx9.4417\text{ units} $$

As a check, \(52+(1.2815516)(9.4417)\approx64.1\) units. Also, using this mean and standard deviation, \(\operatorname{normalcdf}(-1E99,64.1,52,9.4417)\approx0.9000\), rounded to four decimal places.

Conclude. The standard deviation of the modeled sensor readings is about 9.44 units. With that spread, 64.1 units is the 90th-percentile cutoff.

Worked Example: Finding a Standard Deviation Below the Mean

Worked Example: Finding a Standard Deviation Below the Mean

In a fictional environmental model, a measurement is normally distributed with mean 42 units. Its 20th percentile is 36 units. Find the standard deviation.

State. Let \(X\) be the environmental measurement, in units. The model is \(X\sim N(42,\sigma)\), and \(x_{0.20}=36\).

Plan. The 20th percentile has a negative z-score because it is below the median. Use \(\sigma=(x_p-\mu)/z_p\). Both the numerator and z-score should be negative, so their ratio should be positive.

Do. Find the 20th-percentile z-score and substitute:

$$ z_{0.20}=\operatorname{invNorm}(0.20,0,1)\approx-0.8416212 $$
$$ \sigma=\frac{36-42}{-0.8416212} =\frac{-6}{-0.8416212} \approx7.1291\text{ units} $$

The standard deviation is positive, as it must be. Checking the standardized distance, \((36-42)/7.1291\approx-0.8416\), which agrees with the z-score for the 20th percentile. A calculator check gives \(\operatorname{normalcdf}(-1E99,36,42,7.1291)\approx0.2000\), rounded to four decimal places.

Conclude. The model’s standard deviation is about 7.13 units. The 20th-percentile measurement is about 0.8416 standard deviations below the mean.

When the Percentile Is the Median

The 50th percentile is a special case. Its standard-normal z-score is 0, so the quantile equation becomes \(x_{0.50}=\mu+0\sigma=\mu\). The median of a normal model equals its mean, so a known 50th-percentile cutoff gives the mean directly.

Important: If \(p=0.50\), then \(z_p=0\). You can conclude that \(\mu=x_{0.50}\), but you cannot solve for \(\sigma\) from that percentile alone. Dividing by zero in the standard-deviation formula is not valid.

For any other percentile, the standard-deviation equation can be used if the mean and cutoff are known. Still check the signs. For a percentile below 50%, both \(x_p-\mu\) and \(z_p\) should be negative; for a percentile above 50%, both should be positive. If the signs do not match, recheck the percentile, cutoff, and subtraction order.

Common Mistakes and AP Exam Tips

  • Entering a percentile as a whole-number percent. Use 0.90 for the 90th percentile, not 90. The area input is a proportion to the left.
  • Using the wrong tail. A stated percentile is a left-tail area. For example, the 20th percentile uses 0.20, not 0.80.
  • Dropping the z-score’s sign. The 20th-percentile z-score is negative. Keeping that sign is essential when solving for \(\mu\) or \(\sigma\).
  • Subtracting in the wrong order. In \(\sigma=(x_p-\mu)/z_p\), use cutoff minus mean. Reversing the numerator without also changing the denominator can give an incorrect sign.
  • Reporting a negative standard deviation. A standard deviation cannot be negative. A negative answer usually signals a sign error or inconsistent percentile information.
  • Rounding too early. Keep the calculator’s full z-score precision through the calculation, then round the final parameter. Early rounding can slightly change the answer.
  • Giving a number without context. State what the parameter describes and include the variable’s units. A z-score has no units, but \(\mu\), \(\sigma\), and the cutoff do.

A full-credit response identifies the variable and its units, converts the stated percentile to a left-tail area, shows the z-score and rearranged equation, and reports the parameter in context. When finding a standard deviation, explicitly note that the result is positive and verify that the cutoff lies on the side of the mean indicated by the percentile.

AP Exam Tip: Write \(x_p=\mu+z_p\sigma\) before rearranging. This makes the sign and role of each quantity visible and helps you catch an incorrect subtraction or tail choice.

Key Takeaway

A normal-model percentile links its cutoff to the mean and standard deviation through \(x_p=\mu+z_p\sigma\). Find the percentile’s standard-normal z-score, rearrange for the unknown parameter, and check the direction and units of the result.

Key takeaway: Use \(\mu=x_p-z_p\sigma\) to find an unknown mean and \(\sigma=(x_p-\mu)/z_p\) to find an unknown standard deviation when \(z_p\ne0\). The 50th percentile identifies the mean but cannot identify the standard deviation.

Check Your Understanding

For each question, show how the percentile determines the z-score and explain what your result means.

  1. A fictional model has a 90th-percentile cutoff of 118 seconds and a standard deviation of 8 seconds. Write the equation you would use to find its mean.
  2. A normal model has mean 30 centimeters and a 20th-percentile cutoff of 25 centimeters. Should the standard deviation calculation use a positive or negative z-score? Explain.
  3. For a normal model with mean 64 points, the 75th-percentile cutoff is 70 points. Write the expression for the standard deviation using \(z_{0.75}=\operatorname{invNorm}(0.75,0,1)\).
  4. A student uses \(\operatorname{invNorm}(0.80,0,1)\) for the 20th percentile. Identify the error and state the correct area input.
  5. A normal model’s 50th-percentile cutoff is 16 minutes. What can you conclude about its mean, and can you determine its standard deviation from this fact alone?

Previous

Next

Back to Tutorials