To solve z score practice problems, subtract the mean from the observed value and divide by the standard deviation: z = (x – mean) / standard deviation. Then interpret the sign and size. A result of -1.5 means the observation is 1.5 standard deviations below its reference mean.
The ten original problems below move from straightforward calculations to comparisons, reverse calculations, and situations where a numerical answer would be misleading. All campus examples are fictional. Try each question before reading its answer, and write an interpretation even when the arithmetic feels easy.
Know what each number represents
For population information, the usual notation is z = (x – μ) / σ. For a descriptive standardization within a sample, use z = (x – x̄) / s. The mean and standard deviation must describe the same reference group. Penn State’s lesson on describing data explains these forms and standardized comparisons.
A z-score has no physical unit because the units in the numerator and denominator cancel. Positive means above the reference mean, negative means below, and zero means equal to it. The sign alone doesn’t say whether the outcome is desirable.
Don’t automatically divide by a standard error. These questions concern individual observations. A test statistic about a sample mean is a different task with different assumptions. Write “individual value” at the top of your page if your course has recently introduced both topics.
A quick setup check before calculating
- Identify the observation, its reference mean, and that group’s standard deviation.
- Predict the sign by comparing the observation with the mean.
- Subtract first, using parentheses in your calculator.
- Divide by the standard deviation, then round the final result.
- Explain the answer in the original context.
Keep this answer frame nearby: “The [observation] is [distance] standard deviations [above or below] the mean for [reference group].” Naming the group catches mistakes that a calculator can’t detect, such as using last semester’s mean with this semester’s standard deviation.
If you need a refresher on averages before starting, work through the mean, median, and mode practice problems. Here, every required mean and standard deviation is supplied so you can focus on standardizing and interpreting.
Ten z score practice problems
1. An above-average quiz score
A class quiz has mean 68 points and standard deviation 8 points. Elena scores 80. Calculate her z-score and interpret it relative to this class.
2. A below-average reading time
A group’s completion times for a reading exercise have mean 32 minutes and standard deviation 6 minutes. One student finishes in 23 minutes. Calculate the z-score. Does a negative sign mean the student took longer?
3. Exactly at the mean
A fictional campus survey records a mean weekly library time of 5 hours, with standard deviation 2 hours. A respondent reports 5 hours. Find the z-score. Is this the same as reporting no library time?
4. Recover the original value
A project score has z = -0.75 in a group with mean 84 points and standard deviation 4 points. What was the original score? Check whether it falls on the correct side of 84.
5. Compare two quiz results
Sam earns 78 on Quiz A, where the class mean is 66 and standard deviation is 6. Sam earns 90 on Quiz B, where the mean is 81 and standard deviation is 9. Which result is farther above its class mean in standard-deviation units?
6. Compare outcomes where lower is preferable
Two practice tasks measure typing errors. On Task A, a student makes 2 errors; the reference mean is 6 with standard deviation 2. On Task B, the student makes 1 error; the reference mean is 3 with standard deviation 2. On which task is the student’s error count farther below its reference mean?
7. Repair a sign mistake
A student computes a z-score for x = 42, mean = 50, and standard deviation = 5. Their working is (50 – 42) / 5 = 1.6. Identify the error, calculate the correct answer, and explain why you could spot the problem without finishing the arithmetic.
8. Handle a zero standard deviation
Every student in a five-person group reports exactly 3 completed exercises. The group mean is 3 and standard deviation is 0. What happens if you try to calculate a z-score for a group member?
9. Separate a z-score from a percentile
A value has z = 1.00. Nothing is stated about the shape of the reference distribution. Can you conclude that approximately 84.13% of observations are below it? What additional model would make that normal-table calculation appropriate?
10. Use a normal model explicitly
For this exercise only, assume a population of scores follows a normal distribution with mean 70 and standard deviation 10. What z-score corresponds to 80? Approximately what proportion of the population scores above 80? Use the supplied standard-normal left-tail area P(Z ≤ 1) = 0.8413.
Worked answers with interpretation checks
1. Elena: z = 1.50
z = (80 – 68) / 8 = 12 / 8 = 1.50. Elena’s score is 1.5 standard deviations above this class’s mean. The positive sign matches the fact that 80 is greater than 68.
Don’t describe this as 1.5 points above average. The raw difference is 12 points. Dividing by the 8-point standard deviation changes the scale of that difference.
2. Reading time: z = -1.50
z = (23 – 32) / 6 = -9 / 6 = -1.50. This student’s completion time is 1.5 standard deviations below the group mean. Here, below means fewer minutes, so the student finished faster.
That doesn’t establish better comprehension. The variable measures time, and the prompt gives no accuracy or understanding score. Keep your interpretation limited to the quantity measured.
3. Library time: z = 0
z = (5 – 5) / 2 = 0. The respondent’s reported time equals the reference mean. A zero z-score describes the difference from the mean, not the original amount of time.
For contrast, someone reporting zero hours would have z = (0 – 5) / 2 = -2.50 using this reference. Keeping the raw value and standardized value in separate columns prevents that mix-up.
4. Original project score: 81
Rearrange the formula to x = mean + z(standard deviation). Here, x = 84 + (-0.75)(4) = 84 – 3 = 81 points. Substituting back gives (81 – 84) / 4 = -0.75.
The score is below 84, as the negative z-score requires. OpenStax’s standard normal lesson shows the relationship between original values, standardized values, and distance from the mean.
5. Quiz A has the larger standardized distance
Quiz A: (78 – 66) / 6 = 2. Quiz B: (90 – 81) / 9 = 1. Sam is two standard deviations above the mean on A and one above the mean on B. A is farther above its reference mean despite the lower raw score.
This comparison doesn’t prove that Quiz A measured more knowledge or was harder. It answers the stated relative-position question. Different content, grading rules, or reference groups can matter for other comparisons.
6. Task A has the farther-below result
Task A: (2 – 6) / 2 = -2. Task B: (1 – 3) / 2 = -1. Task A is two standard deviations below its reference mean, compared with one on B.
Because the variable counts errors, the more negative standardized result represents fewer errors relative to that task’s reference distribution. Simply choosing the larger z-score would reverse the intended comparison. State the direction of a desirable outcome before ranking anything.
7. Correct answer: z = -1.60
The numerator should be observation minus mean: (42 – 50) / 5 = -8 / 5 = -1.60. The original working reversed the subtraction. Since 42 is below 50, its z-score must be negative when the standard deviation is positive.
On a calculator, enter the subtraction inside parentheses. Typing 42 – 50 / 5 would divide before subtracting and produce a different calculation. Predicting the sign first is a quick independent check.
8. The z-score is undefined
The formula produces (3 – 3) / 0 = 0 / 0, which is undefined. Don’t label the result zero. There is no positive spread to serve as the unit of standardized distance.
You can still state the ordinary descriptive fact: all five reported values equal 3. If software returns a missing value or error, inspect the denominator instead of replacing the result with a convenient number.
9. A normal percentile is not justified yet
No. A z-score of 1 says the value is one standard deviation above the mean. It doesn’t specify how observations are distributed around that mean. Standardizing a skewed distribution doesn’t make its shape normal.
The approximately 84.13% statement uses a normal model. Without a suitable distributional assumption or actual data for a rank calculation, report the standardized distance and leave the percentile undetermined.
10. Above 80: approximately 15.87%
z = (80 – 70) / 10 = 1. The supplied normal area is to the left. The area above 80 is therefore 1 – 0.8413 = 0.1587, or approximately 15.87%.
Draw a small bell curve, mark 80 to the right of 70, and shade the right-hand tail. The answer should be less than one half. This is a population proportion under the exercise’s normal model, not a claim about an actual campus dataset.
Make your practice notes useful to someone else
Create four columns: original information, substitution, result, and interpretation. For missed questions, add the error type, such as reversed subtraction, wrong reference group, or unjustified percentile. Keep any normal-distribution assumption beside the problem instead of hiding it in a general heading.
Use statistics study notes to compare notation, or browse study documents for complementary practice. Check that a shared solution concerns individual observations before borrowing its denominator.
Upload your own notes to help other students. Share original examples and corrections you have permission to publish. Explain what each answer means, and omit private student scores or restricted course assessments.
Frequently asked questions
Do negative z-scores mean negative raw values?
No. The reading time of 23 minutes had a negative z-score because it was below the 32-minute mean. Its original value remained positive.
Does a z-score need a normal distribution?
You can calculate a descriptive z-score with a suitable mean and nonzero standard deviation without assuming normality. Using a standard normal table to turn it into a probability requires the relevant normal model.
How many decimal places should I use?
Follow your course instructions. These exercises display most z-scores to two decimal places. Keep unrounded intermediate values when a later calculation depends on them, and label any rounded probability as approximate.