Recovering an Observation from Its Z Score

Introduction

A z score expresses how far an observation is from a reference mean, measured in standard deviations. Sometimes the observation itself is missing, but its z score and the relevant mean and standard deviation are known. In that case, the calculation can be reversed to recover the observation.

The important detail is to use the same reference values that were used to calculate the z score. A sample-based z score is centered on \(\bar{x}\) and scaled by \(s\); a population-based z score is centered on \(\mu\) and scaled by \(\sigma\). The recovered value should also agree with the direction indicated by the z score.

Reversing the Z-Score Formula

Definition: For a sample-based z score, \(z=(x-\bar{x})/s\), where \(x\) is the observation, \(\bar{x}\) is the sample mean, and \(s\) is the sample standard deviation. For a population-based z score, \(z=(x-\mu)/\sigma\), where \(\mu\) is the population mean and \(\sigma\) is the population standard deviation. In either case, the standard deviation must be positive.

To recover \(x\), undo the operations used to calculate the z score. First multiply the z score by its standard deviation; then add the corresponding reference mean. Use \(\bar{x}\) and \(s\) for a sample-based z score, or \(\mu\) and \(\sigma\) for a population-based z score.

Key result: The observation corresponding to a sample-based z score is \(x=\bar{x}+zs\). The observation corresponding to a population-based z score is \(x=\mu+z\sigma\). The recovered observation has the same units as the original measurement.

Worked Example: Recovering a Sample Observation Below the Mean

Suppose a hypothetical student’s score has a sample-based z score of \(-1.25\) in a group with mean score \(\bar{x}=72\) points and sample standard deviation \(s=8\) points. Recover the score.

The z score is sample-based, so the appropriate formula uses \(\bar{x}\) and \(s\):

x=\bar{x}+zs

Substitute the given values:

$$ x=72+(-1.25)(8)=72-10=62 $$

The recovered score is 62 points. The calculation also passes a direction check: the z score is negative, so this sample-based observation must be below \(\bar{x}=72\), and \(62<72\). Its distance from the mean is \(72-62=10\) points, which is \(10/8=1.25\) sample standard deviations.

Why the Formula Works

The result follows directly from the definition of a z score. For a sample-based z score, begin with the equation that relates the observation to the sample mean and standard deviation. Multiplying both sides by \(s\) removes the division by \(s\); adding \(\bar{x}\) then isolates \(x\).

z=\frac{x-\bar{x}}{s} \quad\Longrightarrow\quad zs=x-\bar{x} \quad\Longrightarrow\quad x=\bar{x}+zs

The population-based derivation uses the same algebra with the population quantities instead:

z=\frac{x-\mu}{\sigma} \quad\Longrightarrow\quad z\sigma=x-\mu \quad\Longrightarrow\quad x=\mu+z\sigma

These rearrangements require no assumption that the data follow a normal distribution. They are algebraic identities for the z-score definitions. They do require a positive standard deviation: if the standard deviation is zero, the original z-score formula divides by zero and the z score is undefined.

Use the Matching Reference Values

The notation in the formula is not interchangeable. A population-based z score describes an observation’s position relative to \(\mu\), not necessarily relative to a separate sample mean \(\bar{x}\). Likewise, a sample-based z score is relative to \(\bar{x}\), not necessarily to a population mean. If the problem states which kind of z score is given, use its corresponding mean and standard deviation. If it does not, check the context or the definition supplied.

Worked Example: Recovering a Population Observation

In a hypothetical setting, a population measurement has mean \(\mu=500\) units and standard deviation \(\sigma=4\) units. An observation has population-based z score \(z=-0.875\). Recover the observation.

Because the z score is population-based, use \(\mu\) and \(\sigma\):

x=\mu+z\sigma

Substitution and arithmetic give:

$$ x=500+(-0.875)(4)=500-3.5=496.5 $$

The observation is 496.5 units. It is below the population mean, as expected from its negative z score: \(496.5<500\). It is \(3.5\) units below that mean, and \(3.5/4=0.875\) population standard deviations is the corresponding distance.

Suppose a separate sample from this population has sample mean \(\bar{x}=490\). The recovered observation \(496.5\) is greater than that sample mean. This does not contradict the negative z score: the score was calculated relative to the population mean of 500, not the separate sample mean of 490.

Direction and Size Checks

The sign of \(z\) tells which side of the matching reference mean the observation lies on. For a sample-based z score, a positive \(z\) means \(x>\bar{x}\), a negative \(z\) means \(x<\bar{x}\), and \(z=0\) means \(x=\bar{x}\). For a population-based z score, the comparisons are instead with \(\mu\): positive means \(x>\mu\), negative means \(x<\mu\), and zero means \(x=\mu\). These checks are useful for spotting a sign error or a mismatched reference value.

The magnitude \(|z|\) gives the distance from the reference mean in standard-deviation units. For example, if \(z=1.5\), the observation is \(1.5\) standard deviations above its matching mean; if \(z=-1.5\), it is \(1.5\) standard deviations below it. The recovered distance in the original measurement units is \(|z|\) times the relevant standard deviation.

Worked Example: Recovering a Sample Observation Above the Mean

A hypothetical sample of waiting times has mean \(\bar{x}=18.4\) minutes and standard deviation \(s=3.2\) minutes. One waiting time has a sample-based z score of \(0.75\). Find the waiting time.

Use the sample-based recovery formula and substitute the values:

x=\bar{x}+zs =18.4+(0.75)(3.2) =18.4+2.4 =20.8

The recovered waiting time is 20.8 minutes. It is above the sample mean, as a positive sample-based z score requires: \(20.8>18.4\). Its difference from the mean is \(2.4\) minutes, and \(2.4/3.2=0.75\), which checks against the given z score.

Precision, Units, and Rounding

A z score is a number of standard deviations, so it has no measurement units. The recovered observation does have units: multiplying the dimensionless \(z\) by a standard deviation in the measurement units gives a distance in those same units, and adding the mean preserves them. If a problem reports a mean in kilograms and a standard deviation in kilograms, the recovered observation is in kilograms.

The result can only be as precise as the information used to calculate it. If a z score has been rounded, the recovered value is generally an estimate based on that rounded z score, not necessarily the exact original observation. Keep the calculation unrounded until the final step when possible, and report the final value at a precision justified by the supplied data. If the measurement is known to be a whole-number count, a decimal answer may indicate that the rounded z score does not determine the exact count.

Worked Example: Recognizing the Effect of a Rounded Z Score

Suppose a hypothetical sample has \(\bar{x}=40\) seconds and \(s=6\) seconds. A reported sample-based z score is \(0.83\), rounded to two decimal places. Calculate the value implied by the reported score, and explain the limitation.

Using the reported z score:

x=\bar{x}+zs =40+(0.83)(6) =40+4.98 =44.98

The reported z score implies an observation of approximately 44.98 seconds. Since \(0.83\) is rounded, the exact z score may differ slightly from \(0.83\), so \(44.98\) need not be the exact original observation. The direction check still works: the reported z score is positive, and \(44.98>40\). Units are also consistent: seconds plus a number of standard deviations times seconds gives seconds.

Common Mistakes

Using the wrong mean. A negative population-based z score places \(x\) below \(\mu\), not automatically below \(\bar{x}\). Check the reference values used to define the score.
Subtracting when recovering \(x\) from a negative score. In \(x=\text{mean}+z(\text{standard deviation})\), a negative \(z\) already makes the adjustment negative. Subtracting a negative adjustment incorrectly moves the result above the mean.
Using the standard deviation as if it were the variance. The recovery formula multiplies \(z\) by the standard deviation \(s\) or \(\sigma\), not by a variance.
Dropping the measurement units. The z score has no units, but the recovered observation does. Carry the units of the mean and standard deviation through the calculation.
Treating a rounded z score as exact. A result calculated from a rounded score is usually only as precise as that input allows.

Key takeaway: Recover an observation by adding the z-score adjustment to the mean used to define that score: \(x=\bar{x}+zs\) for a sample-based z score and \(x=\mu+z\sigma\) for a population-based z score. Then check that the observation is on the correct side of that same reference mean and has the correct units.

Check Your Understanding

For each question, identify the appropriate reference values and check the recovered observation against the z score’s direction.

A sample-based z score is \(-0.5\), with \(\bar{x}=30\) cm and \(s=4\) cm. What is the observation?
A population-based z score is \(1.2\), with \(\mu=75\) points and \(\sigma=5\) points. Recover the observation and state the direction check.
Why should a negative population-based z score be checked against \(\mu\), rather than automatically against a sample mean \(\bar{x}\)?
A sample has mean \(12\) minutes and standard deviation \(2\) minutes. A sample-based z score is reported as \(1.35\), rounded to two decimal places. What value does the reported score imply, and why might it not be exact?
What happens to the z-score formula if the relevant standard deviation is zero?