Normal Distribution: z-Scores, Tables and Worked Examples

By Engr. Henry Akpan

The Bell Curve

Measure the heights of a thousand students, the marks in a large examination, or the lifetimes of a batch of light bulbs, and plot the results. Again and again the same shape appears: most values cluster near the average, with fewer and fewer further away, in a symmetrical bell-shaped curve. This is the normal distribution, the most important probability distribution in statistics. This tutorial, for STA 111/112, MTH 102 and Further Mathematics, covers its properties, the 68–95–99.7 rule, standardising with z-scores, using the standard normal table, finding probabilities and proportions, working backwards from a probability to a value, finding an unknown mean or standard deviation, and approximating the binomial distribution, with worked examples throughout.

Properties of the Normal Distribution

A normally distributed variable X is written X ~ N(μ, σ²), where μ is the mean and σ is the standard deviation (so σ² is the variance). Its curve:

  • is symmetrical about the mean, so the mean, median and mode are all equal;
  • is bell-shaped, highest at the mean, and falls away on both sides, approaching but never touching the horizontal axis;
  • has a total area of 1 underneath it, and the area under the curve between two values gives the probability that X lies between them;
  • is wider and flatter for a larger σ, and narrower and taller for a smaller σ, while μ sets its position.

Because X is continuous, the probability of any single exact value is zero, so P(X < a) and P(X ≤ a) are the same.

The 68–95–99.7 Rule

For every normal distribution:

  • about 68% of values lie within 1 standard deviation of the mean;
  • about 95% lie within 2 standard deviations;
  • about 99.7% lie within 3 standard deviations.

Worked example 1

Examination marks are normally distributed with mean 55 and standard deviation 10. Roughly what percentage of candidates scored (a) between 45 and 65 (b) between 35 and 75 (c) above 75?

(a) 45 to 65 is within 1 standard deviation of the mean: about 68%.

(b) 35 to 75 is within 2 standard deviations: about 95%.

(c) The remaining 5% lies outside 35 to 75, split equally between the two tails by symmetry, so about 2.5% scored above 75.

Standardising: z-Scores

There are infinitely many normal distributions, but they can all be converted into one, the standard normal distribution Z ~ N(0, 1), with mean 0 and standard deviation 1. The z-score of a value x is:

z = (x − μ) ÷ σ

It tells you how many standard deviations x is above (positive z) or below (negative z) the mean. A mark of 70 in the examination above has z = (70 − 55) ÷ 10 = 1.5, so it is 1.5 standard deviations above the mean.

The standard normal table

Tables (and calculators) give Φ(z) = P(Z < z), the area to the left of z. Some useful values:

  • Φ(0) = 0.5000
  • Φ(0.5) = 0.6915
  • Φ(1) = 0.8413
  • Φ(1.5) = 0.9332
  • Φ(1.645) = 0.9500
  • Φ(1.96) = 0.9750
  • Φ(2) = 0.9772
  • Φ(2.5) = 0.9938

Tables give only positive z. Use symmetry for the rest:

  • P(Z > z) = 1 − Φ(z)
  • Φ(−z) = 1 − Φ(z)
  • P(a < Z < b) = Φ(b) − Φ(a)

Always sketch the curve and shade the area you want. It prevents most mistakes.

Finding Probabilities

Worked example 2

For the examination marks X ~ N(55, 10²), find the probability that a randomly chosen candidate scored more than 70.

z = (70 − 55) ÷ 10 = 1.5. P(X > 70) = P(Z > 1.5) = 1 − 0.9332 = 0.0668, about 6.7% of candidates.

Worked example 3

Find P(45 < X < 65) exactly.

z-values: −1 and 1. P = Φ(1) − Φ(−1) = 0.8413 − (1 − 0.8413) = 0.6826, confirming the 68% rule.

Worked example 4

The heights of adult women in a town are normally distributed with mean 165 cm and standard deviation 8 cm. Find the probability that a woman is shorter than 150 cm.

z = (150 − 165) ÷ 8 = −1.875. P(Z < −1.875) = 1 − Φ(1.875) ≈ 1 − 0.9696 = 0.0304, about 3%.

Worked example 5: expected numbers

The lifetimes of a type of bulb are normally distributed with mean 1200 hours and standard deviation 100 hours. Out of 1000 bulbs, how many would you expect to last more than 1350 hours?

z = (1350 − 1200) ÷ 100 = 1.5. P = 1 − 0.9332 = 0.0668. Expected number = 1000 × 0.0668 ≈ 67 bulbs.

Working Backwards: from a Probability to a Value

Sometimes the probability is known, and the value is wanted. Find the z-score with the right area from the table (an inverse look-up), then convert back with x = μ + zσ.

Worked example 6

A scholarship is offered to the top 10% of candidates in the examination with X ~ N(55, 10²). What is the minimum mark needed?

We need the z with 90% of the area to its left. From the table, Φ(1.2816) = 0.90, so z ≈ 1.28.

x = 55 + 1.28 × 10 ≈ 67.8, so a mark of 68 or more is needed.

Worked example 7: finding an unknown σ

A machine fills bags of rice with a mean mass of 50 kg. 20% of bags contain more than 60 kg. Find the standard deviation, assuming the masses are normally distributed.

P(X > 60) = 0.20, so P(Z < z) = 0.80, which gives z ≈ 0.8416.

Then (60 − 50) ÷ σ = 0.8416, so σ = 10 ÷ 0.8416 ≈ 11.9 kg.

The same approach finds an unknown mean, or (with two conditions and two equations) both μ and σ.

Approximating the Binomial Distribution

When n is large and p is not too close to 0 or 1 (a common guide is that np and n(1 − p) are both greater than 5), the binomial distribution B(n, p) is very close to a normal distribution with:

μ = np and σ² = np(1 − p)

Because the binomial distribution is discrete and the normal distribution is continuous, apply a continuity correction: treat each whole number k as the interval from k − 0.5 to k + 0.5.

Worked example 8

A fair coin is tossed 100 times. Estimate the probability of getting 60 or more heads.

μ = 100 × 0.5 = 50, σ² = 100 × 0.5 × 0.5 = 25, so σ = 5.

"60 or more" becomes X ≥ 59.5 after the continuity correction. z = (59.5 − 50) ÷ 5 = 1.9.

P(Z > 1.9) = 1 − 0.9713 = 0.0287. The exact binomial probability is 0.0284, so the approximation is excellent, and far quicker than adding 41 binomial terms.

Why the Normal Distribution Is Everywhere

The central limit theorem explains why so many quantities are approximately normal. When a result is the sum of many small, independent random effects, as height is the combined result of many genes and environmental factors, its distribution tends towards the normal shape, whatever the shapes of the individual effects. The same theorem says that the means of large samples are approximately normally distributed, which is the basis of confidence intervals and hypothesis testing in statistics.

Common Mistakes with the Normal Distribution

  • Using the variance instead of the standard deviation in z = (x − μ)/σ. N(55, 100) means σ = 10.
  • Forgetting to subtract from 1 for "greater than" probabilities, since the table gives the area to the left.
  • Mishandling negative z-values. Use Φ(−z) = 1 − Φ(z), and sketch the curve.
  • Forgetting the continuity correction when approximating the binomial distribution.
  • Expecting a probability greater than 1, or a negative one. Areas lie between 0 and 1.

Normal Distribution: Practice Questions

  1. X ~ N(40, 5²). Find the z-score of x = 47.5.
  2. For the variable in question 1, find P(X < 47.5) using Φ(1.5) = 0.9332.
  3. Scores are normally distributed with mean 60 and standard deviation 8. What percentage of scores lie between 52 and 68?
  4. For Z ~ N(0, 1), find P(−0.5 < Z < 1), using Φ(0.5) = 0.6915 and Φ(1) = 0.8413.
  5. The masses of eggs are normally distributed with mean 60 g and standard deviation 4 g. Above what mass are the heaviest 2.5% of eggs? (Use Φ(1.96) = 0.975.)
  6. A die is rolled 180 times. Using a normal approximation, find the mean and standard deviation of the number of sixes.

Answers: 1. z = (47.5 − 40) ÷ 5 = 1.5. 2. 0.9332. 3. Within 1 standard deviation of the mean, so about 68%. 4. 0.8413 − (1 − 0.6915) = 0.5328. 5. 60 + 1.96 × 4 ≈ 67.8 g. 6. μ = 180 × 1/6 = 30, and σ = √(180 × 1/6 × 5/6) = √25 = 5.

Conclusion

The normal distribution N(μ, σ²) is a symmetrical bell curve, and areas under it give probabilities: about 68%, 95% and 99.7% of values lie within 1, 2 and 3 standard deviations of the mean. Standardise with z = (x − μ)/σ, read Φ(z) from tables, and use 1 − Φ(z) and symmetry for other areas. Work backwards with x = μ + zσ to find cut-off values or an unknown parameter. For large n, the binomial distribution is approximated by N(np, np(1 − p)) with a continuity correction, and the central limit theorem explains why normal distributions appear throughout science and statistics.

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