From a Value to Its Position in a Distribution
In Notation \(N(\mu,\sigma)\) and Parameters, you learned that \(\mu\) gives the center of a normal model and \(\sigma\) gives its standard deviation. A z-score uses those parameters to describe where a particular value \(x\) sits relative to the model’s center. It expresses the value’s distance from the mean in standard-deviation units.
The calculation is useful because a raw difference depends on the units and scale of the variable. Being 8 points above one exam’s mean may be a large distance if scores usually vary by 4 points, but a smaller distance if they usually vary by 16 points. The z-score accounts for that difference in spread.
Here, \(x-\mu\) is the signed distance between the value and the mean, measured in the original units. Dividing by \(\sigma\), which has those same units, converts the distance to a number of standard deviations. Therefore, a z-score has no units. Keep the subtraction in the order shown: value minus mean.
- If \(z>0\), the value is above the mean.
- If \(z<0\), the value is below the mean.
- If \(z=0\), the value equals the mean.
The absolute value \(|z|\) describes how many standard deviations separate the value from the mean, without regard to direction. For example, a score of \(-1.5\) is 1.5 standard deviations below the mean; a score of \(1.5\) is the same distance above it. A larger absolute z-score means a greater distance from the mean in standard-deviation units.
A Reliable Calculation Routine
Before using the formula, read the model carefully. In this course, \(N(\mu,\sigma)\) lists the mean first and the standard deviation second. Do not treat the second parameter as a variance or use a sample statistic when the problem gives a model’s parameters. Write the three inputs with their meanings so that the calculation is easy to check.
Record the particular value \(x\), the model mean \(\mu\), and the model standard deviation \(\sigma\). Confirm that \(\sigma>0\).
Calculate \(x-\mu\), keeping the order as value minus mean. A positive difference means \(x\) is above the mean; a negative difference means it is below.
Compute \(z=(x-\mu)/\sigma\). Keep enough calculator precision during the calculation, then round the final score as appropriate for the question.
Confirm that the sign matches the value’s position relative to the mean. State the distance in standard-deviation units, not in the variable’s original units.
A z-score is a standardized distance, not a probability. By itself, it does not tell you the probability of being below, above, or between values. Later work with normal distributions will use z-scores as a step toward finding areas and probabilities. For now, focus on calculating and describing the location.
Worked Example: Several Scores in One Normal Model
Worked Example: Several Scores in One Normal Model
A fictional school models scores on a placement assessment as normal with mean 72 points and standard deviation 8 points. Let \(X\) be the score of a randomly selected student, so \(X\sim N(72,8)\). Calculate and compare z-scores for scores of 88, 60, 72, and 76 points.
State. The model gives \(\mu=72\) points and \(\sigma=8\) points. Each score \(x\) will be standardized using the same mean and standard deviation. The standard deviation is positive, so division by \(\sigma\) is valid.
Plan. For each score, subtract 72 from the score and divide by 8. Then interpret the sign and absolute value. Since the model is explicitly normal, these scores describe positions in that stated normal model; the arithmetic itself uses the given parameters.
Do. For a score of 88 points:
A score of 88 is 2 standard deviations above the model mean. For a score of 60 points:
A score of 60 is 1.5 standard deviations below the mean. For a score equal to the mean, 72 points:
A score of 72 is exactly at the mean. Finally, for a score of 76 points:
A score of 76 is half a standard deviation above the mean. Conclude. In this model, 88 is the farthest of the four scores from the mean because its absolute z-score, 2, is largest. The scores 60 and 76 are on opposite sides of the mean, but 60 is farther away: its absolute z-score is 1.5 compared with 0.5 for 76. These comparisons describe standardized locations, not probabilities or guarantees about how often the scores occur.
Worked Example: Standardize a Measurement in Context
Worked Example: Standardize a Measurement in Context
A fictional greenhouse models the time for a certain seed variety to sprout as normal, with mean 14 days and standard deviation 2.5 days. Let \(T\) be the sprouting time, in days, for a randomly selected seed under this model. Calculate the z-score for a seed that sprouts in 19 days and for one that sprouts in 11 days.
Identify the inputs. For both calculations, \(\mu=14\) days and \(\sigma=2.5\) days. The standard deviation is positive. The two values are \(x=19\) days and \(x=11\) days.
Calculate for 19 days. Subtract the model mean from the observed value, then divide by the standard deviation:
The positive z-score shows that 19 days is above the mean. Its size says the sprouting time is 2 standard deviations above the model mean. Notice that the raw difference is 5 days, but the z-score reports that difference relative to a typical spread of 2.5 days.
Calculate for 11 days.
The negative sign shows that 11 days is below the mean. It is 1.2 standard deviations below the mean. Compare. The 19-day time is farther from the mean than the 11-day time because \(2>|-1.2|\). Both answers are dimensionless, even though the original measurements and parameters are in days.
Worked Example: Compare Values on Different Scales
Worked Example: Compare Values on Different Scales
Two fictional machines fill containers with different products. For Machine A, the amount \(A\), in milliliters, is modeled by \(A\sim N(500,6)\). For Machine B, the amount \(B\), in milliliters, is modeled by \(B\sim N(120,4)\). Compare a 512-milliliter fill from Machine A with a 128-milliliter fill from Machine B. Which fill is farther above its own model mean?
Calculate the score for Machine A. Here, the value is 512 milliliters, the mean is 500 milliliters, and the standard deviation is 6 milliliters.
The 512-milliliter fill is 2 standard deviations above Machine A’s mean.
Calculate the score for Machine B. Here, the value is 128 milliliters, the mean is 120 milliliters, and the standard deviation is 4 milliliters.
The 128-milliliter fill is also 2 standard deviations above its model mean. Conclude. Both values have the same relative position above their own means, even though their raw distances from the means are different: 12 milliliters for A and 8 milliliters for B. The z-scores allow a comparison that accounts for each model’s different center and spread. Neither raw value is farther from its own mean in standardized units; their z-scores are equal.
Common Mistakes and AP Exam Tips
- Reversing the subtraction. Use \(x-\mu\), not \(\mu-x\). Reversing it changes the sign and therefore changes whether the value is described as above or below the mean.
- Dividing by the wrong number. Divide by the standard deviation \(\sigma\), not by the variance \(\sigma^2\), and not by the mean.
- Dropping the negative sign. If \(x<\mu\), then \(x-\mu<0\), so the z-score should be negative. Reporting only its absolute value loses the direction.
- Interpreting the score in original units. A z-score of 2 does not mean 2 points, centimeters, or days. It means 2 standard deviations above the mean.
- Comparing raw differences when spreads differ. A larger raw difference does not automatically mean a value is farther from its mean in standardized terms. Calculate and compare absolute z-scores when the models have different standard deviations.
- Calling a z-score a probability. The score locates a value relative to a mean and standard deviation. It is not the percentage of observations below that value; finding an area or probability requires an additional normal-distribution calculation.
- Rounding too early. If a division does not come out exactly, keep the calculator’s precision until the final answer and then round consistently. Include enough digits to support the comparison being made.
A clear AP response shows the formula, substitutes the value and both parameters, and reports the result with its direction and standardized distance. For instance, “\(z=-1.2\), so this sprouting time is 1.2 standard deviations below the model mean” communicates the calculation and its meaning. A bare number may show arithmetic, but it does not explain the sign or the scale.
Key Takeaway
Calculating a z-score converts a value’s signed distance from its mean into standard-deviation units. The sign tells which side of the mean the value is on, and the absolute value tells how far away it is. This standardized location makes values easier to compare when the models have different centers, spreads, or original units.
Check Your Understanding
For each question, show the subtraction and division, then state what the sign and size mean.
- A model has mean 40 and standard deviation 5. Calculate the z-score for \(x=50\), and describe its position relative to the mean.
- A normal model for a device’s operating time has mean 18 hours and standard deviation 3 hours. Find the z-score for a device that operates for 13.5 hours.
- In a model with mean 250 grams and standard deviation 10 grams, what is the z-score for a measurement of 250 grams? Explain why.
- One value has z-score \(-2.1\), and another has z-score \(1.4\). Which is farther from its mean in standard-deviation units, and which is below its mean?
- Two models have different standard deviations. Why can comparing the values’ raw distances from their means give a different conclusion from comparing their absolute z-scores?