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Quantitative Aptitude · Theoretical Distributions

Standard Normal Variate and Using Z-Tables

Updated 1 October 2026 · Fact-checked

The standard normal variate Z = (X − μ) ÷ σ converts any normal value into a distance from the mean measured in standard deviations. To solve a question, compute Z, sketch the curve, read the table area for |Z|, then add or subtract areas using symmetry and the total area of 1.

Understand Standard Normal Variate and Using Z-Tables

A normal distribution can have any mean and any standard deviation. Marks may have mean 60 and SD 8. Heights may have mean 165 and SD 6. Printing a table for every pair is impossible.

The fix is to standardise. You subtract the mean and divide by the standard deviation. The result is Z, the standard normal variate. It tells you how many standard deviations a value lies from the mean. Z = 2 means two SDs above the mean. Z = −1 means one SD below.

After standardising, every normal distribution becomes one distribution: the standard normal distribution, with mean 0 and SD 1. One table serves all questions.

The Z-table gives the area under the curve. Area equals probability, or the proportion of items. The total area is 1. The curve is symmetric about 0, so each half has area 0.5. Many tables give the area between 0 and Z. Check which kind your table is before you read it. This guide mainly uses the 0-to-Z type.

Some tables are cumulative and give P(Z < z). With these, read P(Z < z) directly. For a right tail use 1 − value, and for an interval subtract the two cumulative values.

Because of symmetry, the table lists positive Z only. For a negative Z, use its positive value to read the area. Then draw the curve and decide whether to add, subtract or use 0.5.

Key formulas to remember

Standard normal variate
Z = (X − μ) ÷ σ
μ is the mean and σ is the standard deviation of X. Z has mean 0 and SD 1.
Symmetry
P(Z < 0) = P(Z > 0) = 0.5 and P(Z > a) = P(Z < −a)
Total area under the curve is 1. Use this to handle negative Z.
Area from the mean (0-to-Z table)
P(0 < Z < a) = P(−a < Z < 0) = table value at a
The table value for a negative Z is read at |Z|.
Right tail
P(Z > a) = 0.5 − P(0 < Z < a), for a ≥ 0
For a < 0, P(Z > a) = 0.5 + P(0 < Z < |a|).
Interval on opposite sides of the mean
P(−a < Z < b) = P(0 < Z < a) + P(0 < Z < b), for a, b > 0
Add the two areas.
Interval on the same side of the mean
P(a < Z < b) = P(0 < Z < b) − P(0 < Z < a), for 0 ≤ a < b
Subtract the smaller area from the larger.
Number of items
Expected number = N × probability
N is the total number of items in the group.
Standard reference areas
P(−1 < Z < 1) ≈ 0.6826, P(−2 < Z < 2) ≈ 0.9545, P(−3 < Z < 3) ≈ 0.9973
Also P(−1.96 < Z < 1.96) ≈ 0.95. Use these to check or skip table lookups.

How to solve Standard Normal Variate and Using Z-Tables questions

This method works for any Z-table question, whichever way the area is asked.

  1. 1Write down μ, σ and the required event, such as X > 76 or 35 < X < 50.
  2. 2Convert each X limit to Z using Z = (X − μ) ÷ σ. Keep the sign.
  3. 3Sketch a quick bell curve. Mark 0 and your Z values. Shade the required region.
  4. 4Read the table area for each |Z|. Confirm whether your table gives area from 0 to Z.
  5. 5Decide the operation from the sketch. Add if the region crosses 0. Subtract if both limits are on the same side. Use 0.5 minus the area for a tail.
  6. 6If the question asks for a number of items, multiply the probability by N.
  7. 7Match the result to the options. Check that it is between 0 and 1 and that its size looks reasonable against your sketch.

Quickest way: Sketch, symmetry and benchmark Z values

When to use it: Use it in the objective paper, where each question has about a minute and the options are far apart.

  1. Compute Z first. Many questions give a clean Z such as 1, 2 or 1.5.
  2. Compare with the benchmarks: Z = 1 gives 0.3413, Z = 2 gives 0.4772, Z = 3 gives 0.4987 from the mean. Use them to eliminate options.
  3. Sketch the shaded region in two seconds. A tail beyond Z = 2 must be tiny, under 0.03. An answer like 0.9772 is then clearly the opposite region.
  4. For a 0-to-Z table, use 'tail = 0.5 − area' and 'both sides = add'. If the table is cumulative (P(Z < z)), read P(Z < z) directly. For a right tail use 1 − value, and for an interval subtract the two cumulative values.
  5. Check the options. Probabilities above 1 or negative are impossible. Options sometimes include the complement of the right answer, so check which region your sketch shades.
  6. If your Z value is not in the table you remember and the options are close, skip it. A wrong answer costs 0.25 marks.

Common mistakes in Standard Normal Variate and Using Z-Tables

  • Forgetting to divide by σ, or using σ² (variance) instead of σ.

    Students rush and remember only 'subtract the mean'. Problems also sometimes give variance.

    Fix: Always write σ separately. If variance is given, take its square root first.

  • Reading the table value as the full probability for a tail, such as P(Z > 2) = 0.4772.

    The 0-to-Z table gives area from the centre, not from the end.

    Fix: For a right tail from a positive Z, compute 0.5 − table value. Sketch first.

  • Dropping the negative sign and treating Z = −1 like Z = +1 in the final answer.

    The table lists positive values only, so students ignore the sign altogether.

    Fix: Keep the sign while drawing the curve. Use |Z| only for the table lookup. Then place the region on the correct side of 0.

  • Subtracting areas when the limits are on opposite sides of the mean.

    Students memorise 'subtract' from same-side questions.

    Fix: If one Z is negative and the other positive, the region crosses 0 and you add the two areas.

  • Giving the probability when the question asks for a number of items.

    The calculation ends at the table and students stop.

    Fix: Re-read the last line of the question. If it asks 'how many', multiply by N.

  • Using a cumulative table as if it were a 0-to-Z table.

    Tables differ. A cumulative table gives P(Z < z), so the value at Z = 0 is 0.5.

    Fix: Check the table at Z = 0. If it shows 0.5, it is cumulative. Read P(Z < z) directly, use 1 − value for a right tail, and subtract the two cumulative values for an interval. If it shows 0, it is a 0-to-Z table.

Worked examples

Example 1

The marks of a group of students are normally distributed with mean 60 and standard deviation 8. Using the table area P(0 < Z < 2) = 0.4772, what is the probability that a student scores more than 76? Options: (A) 0.4772 (B) 0.9772 (C) 0.0228 (D) 0.0456

Show the solution
  1. μ = 60, σ = 8, X = 76.
  2. Z = (76 − 60) ÷ 8 = 16 ÷ 8 = 2.
  3. We need the right tail P(Z > 2).
  4. P(Z > 2) = 0.5 − P(0 < Z < 2) = 0.5 − 0.4772 = 0.0228.
  5. Option B is the left-side area P(Z < 2). Option A is only the area from the mean. Option D is the two-tailed area, which is not asked.

Answer: (C) 0.0228

Example 2

A variable X is normal with mean 40 and standard deviation 5. Given P(0 < Z < 1) = 0.3413 and P(0 < Z < 2) = 0.4772, find P(35 < X < 50). Options: (A) 0.1359 (B) 0.6826 (C) 0.8185 (D) 0.9772

Show the solution
  1. μ = 40, σ = 5.
  2. For X = 35: Z = (35 − 40) ÷ 5 = −1.
  3. For X = 50: Z = (50 − 40) ÷ 5 = 2.
  4. The limits are on opposite sides of 0, so add the areas.
  5. P(−1 < Z < 2) = P(0 < Z < 1) + P(0 < Z < 2) = 0.3413 + 0.4772 = 0.8185.
  6. Option A (0.1359) is the same-side difference 0.4772 − 0.3413. Option B is P(−1 < Z < 1).

Answer: (C) 0.8185

Example 3

The scores of 1,000 candidates are normally distributed with mean 70 and standard deviation 10. Given P(0 < Z < 1.5) = 0.4332, about how many candidates score between 70 and 85? Options: (A) 341 (B) 433 (C) 477 (D) 933

Show the solution
  1. μ = 70, σ = 10, N = 1,000.
  2. For X = 70: Z = 0. For X = 85: Z = (85 − 70) ÷ 10 = 1.5.
  3. P(70 < X < 85) = P(0 < Z < 1.5) = 0.4332.
  4. Number of candidates = 1,000 × 0.4332 = 433.2, which is about 433.
  5. Option D (933) comes from wrongly adding 0.5 to the area.

Answer: (B) 433

Exam tips

  • Learn the benchmark areas 0.3413 (Z = 1), 0.4332 (Z = 1.5), 0.4772 (Z = 2) and 0.4987 (Z = 3). Many questions use only these.
  • Check what your exam table shows. Look at the value at Z = 0 to see whether it is a 0-to-Z table or a cumulative table.
  • Always sketch. A tail, a middle band and a two-sided region each have a different operation. The sketch also catches sign errors.
  • Watch the last line: probability, percentage or number of items. Options often include all three forms of the same working.
  • Use symmetry to avoid extra lookups. P(Z < −a) equals P(Z > a), so one table value is enough.

Practice questions from Theoretical Distributions

Standard Normal Variate and Using Z-Tables: frequently asked questions

What is a Z-score in simple words?

A Z-score tells you how many standard deviations a value is above or below the mean. You find it with Z = (X − μ) ÷ σ. A positive Z means above the mean and a negative Z means below.

How do I find probability using a Z-table?

Convert X to Z, sketch the curve and shade the required region. Read the table area for |Z|. Then add, subtract or use 0.5 depending on the sketch. For a right tail from a positive Z with a 0-to-Z table, use 0.5 minus the table value.

What do I do with a negative Z value?

Read the table at the positive value of Z, because the curve is symmetric. Then use the sketch to place the area on the left of 0. For example, P(−1 < Z < 0) equals P(0 < Z < 1).

Why are the mean and SD of Z equal to 0 and 1?

Subtracting the mean shifts the distribution so its centre is at 0. Dividing by σ rescales the spread so the standard deviation becomes 1. The shape stays normal.