Normal Probability

Find probabilities and percentiles for normal data in real units: standardize, read the z-table, and work backwards to cut-off values.

Advanced30 min18 problems
Practice Problems

Definition

Normal probability answers questions like "What fraction of bottles hold less than 494 mL?" when a quantity XX follows a normal distribution with mean μ\mu and standard deviation σ\sigma, measured in real units (cm, points, hours, mL).
The probability is the area under the bell curve over the values you ask about. Every normal curve becomes the standard normal curve after one step, so you only ever need one z-table.

The four-step method

  1. 1.Sketch and shade the region: left of aa, right of aa, or between aa and bb.
  2. 2.Standardize: turn each value into a z-score
z=x−μσz = \frac{x - \mu}{\sigma}
  1. 3.Read the z-table: it gives the area to the left of zz, that is P(Z<z)P(Z < z) (see the lesson on the standard normal distribution).
  2. 4.Adjust for the region:
  • Left: P(X<a)=P(Z<za)P(X < a) = P(Z < z_a)
  • Right: P(X>a)=1−P(Z<za)P(X > a) = 1 - P(Z < z_a)
  • Between: P(a<X<b)=P(Z<zb)−P(Z<za)P(a < X < b) = P(Z < z_b) - P(Z < z_a)

Working backwards: percentiles and cut-offs

Sometimes the area is given and the value is missing ("Which score marks the top 10%?"). Find the zz whose left area matches in the table, then go back to real units:
x=μ+zσx = \mu + z\sigma
A probability of 0.93320.9332 also means 93.32% of the population, so every answer can be read as a proportion of people, bottles or batteries.

Try it now

Test scores are normally distributed with μ=60\mu = 60 and σ=10\sigma = 10. Which expression gives the z-score of a score of 75?

Worked Examples

Women's heights in a country are normally distributed with μ=165\mu = 165 cm and σ=6\sigma = 6 cm. What proportion of women are shorter than 174 cm?
1

Sketch and shade

Shade the area to the left of 174 cm, which lies to the right of the mean 165 cm → The answer should be more than 0.5
2

Standardize

z=174−1656=96z = \frac{174 - 165}{6} = \frac{9}{6} → z=1.5z = 1.5
3

Read the z-table

Row 1.5, column 0.00 → P(Z<1.5)=0.9332P(Z < 1.5) = 0.9332
4

Interpret

A left area needs no adjustment → P(X<174)=0.9332P(X < 174) = 0.9332

Common Mistakes

Using the table value directly for a "greater than" question, for example P(X>14)=0.8944P(X > 14) = 0.8944

Why it's wrong: The z-table always gives the area to the LEFT of zz. A right tail is everything else.

Correct: Subtract from 1: P(X>14)=1−0.8944=0.1056P(X > 14) = 1 - 0.8944 = 0.1056. A quick sketch shows the right tail is small, so 0.8944 cannot be right.

Subtracting the z-scores first: P(−1<Z<2)=P(Z<2−(−1))=P(Z<3)P(-1 < Z < 2) = P(Z < 2 - (-1)) = P(Z < 3)

Why it's wrong: Areas under a curve do not work like that. You need the area up to each bound, then the difference between those areas.

Correct: Look up both left areas and subtract them: 0.9772−0.1587=0.81850.9772 - 0.1587 = 0.8185.

Standardizing in the wrong order or dividing by the wrong number: z=μ−xσz = \frac{\mu - x}{\sigma} or z=x−μσ2z = \frac{x - \mu}{\sigma^2}

Why it's wrong: Reversing the subtraction flips the sign, so you land on the wrong side of the mean. Dividing by the variance gives the wrong distance.

Correct: Always z=x−μσz = \frac{x - \mu}{\sigma}: values above the mean get a positive zz, values below get a negative zz.

In inverse problems, using the percentage as the z-score (z = 0.90) or using the wrong tail (z = -1.28 for the top 10%)

Why it's wrong: The percentage is an AREA, which you look for inside the table. The z-score is read from the row and column edges.

Correct: Top 10% means 0.90 to the left, which gives z≈1.28z \approx 1.28 (positive, above the mean). Bottom 10% gives z≈−1.28z \approx -1.28.

Interactive Visual

Women's heights: μ = 165 cm, σ = 6 cm. Drag the bound or switch the shading

174 · z = 1.5xz147-3153-2159-11650171117721833Height (cm)

μ = 165 cm, σ = 6 cm · Drag a handle or type x or z.

z = (x − μ)/σ = (174 − 165)/6 = 1.5

P(X ≤ 174) = P(Z ≤ 1.5) ≈ 0.9332 = 93.32%

About 93.32% of all values fall in the shaded region.

Exam scores: μ = 70, σ = 8. Area between 62 and 86

62 · z = -186 · z = 2xz46-354-262-1700781862943Exam score (points)

μ = 70 points, σ = 8 points · Drag a handle or type x or z.

z = (x − μ)/σ = (62 − 70)/8 = -1

z = (x − μ)/σ = (86 − 70)/8 = 2

P(62 ≤ X ≤ 86) = P(-1 ≤ Z ≤ 2) ≈ 0.8186 = 81.86%

About 81.86% of all values fall in the shaded region.

Cut-off: which score marks the top 10%?

70 · z = 0xz46-354-262-1700781862943Exam score (points)

μ = 70 points, σ = 8 points · Drag a handle or type x or z.

Move the handle until the shaded area is 0.1000 (10%).

P(Z ≥ 0) ≈ 0.5000

x = μ + zσ = 70 + 0 · 8 ≈ 70 points

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Practice Problems

18 problems
Problem 1 of 18
Easy
Test scores are normally distributed with μ=60\mu = 60 and σ=10\sigma = 10. Which expression gives the z-score of a score of 75?

Why It Matters

Normal probabilities turn a mean and a standard deviation into concrete decisions:
  • Manufacturing: what share of bottles is underfilled, and where to set the machine so almost none are
  • Health: doctors compare a child's height or weight to growth charts, which are percentiles of a normal curve
  • Education: exam boards set the score for the top 10% or decide how many students pass
  • Products: a company can promise "90% of batteries last at least 18 hours" only after this calculation
  • Sports and science: deciding whether a result is unusual means asking how much area lies beyond it
The same four steps answer all of these, whatever the units.

Real World Applications

Quality Control: Underfilled Cans

Drink factories check what share of cans fall below the amount printed on the label, and adjust the filling machines.

Example:

A line fills cans with μ=330\mu = 330 mL and σ=2\sigma = 2 mL. Any can below 327 mL is flagged as underfilled.

1Try It Yourself

Cans are filled with μ=330\mu = 330 mL and σ=2\sigma = 2 mL, normally distributed. The table gives P(Z<−1.5)=0.0668P(Z < -1.5) = 0.0668.

What proportion of cans contain less than 327 mL? Give a decimal.

Step 1: Write the mathematical expression

Standardize 327 mL, then read the left area:

Growth Charts at the Doctor

Pediatricians compare babies with growth charts. Each curve on the chart is a percentile of a normal distribution.

Example:

Weights of 6-month-old babies in a study are about normal with μ=8\mu = 8 kg and σ=0.5\sigma = 0.5 kg. A baby below the 10th percentile gets an extra check-up.

2Try It Yourself

Baby weights at 6 months: μ=8\mu = 8 kg, σ=0.5\sigma = 0.5 kg. The table shows that z≈−1.28z \approx -1.28 has 10% of the area to its left.

Below what weight (in kg) is a baby in the lowest 10%?

Step 1: Write the mathematical expression

Use x=μ+zσx = \mu + z\sigma:

Planning a City Marathon

Race organizers estimate how many runners will be on the course at certain times, to plan water stations and staff.

Example:

Finishing times are about normal with μ=240\mu = 240 minutes and σ=30\sigma = 30 minutes. Organizers want to know the share of runners finishing between 225 and 285 minutes.

3Try It Yourself

Finishing times: μ=240\mu = 240 min, σ=30\sigma = 30 min. Table values: P(Z<1.5)=0.9332P(Z < 1.5) = 0.9332 and P(Z<−0.5)=0.3085P(Z < -0.5) = 0.3085.

What proportion of runners finish between 225 and 285 minutes?

Step 1: Write the mathematical expression

Subtract the two left areas:

Key Takeaways

  • 1To find a normal probability, sketch the region, standardize with z=x−μσz = \frac{x - \mu}{\sigma}, read the z-table, then adjust
  • 2The z-table gives the LEFT area: P(X<a)=P(Z<za)P(X < a) = P(Z < z_a)
  • 3Right tail: P(X>a)=1−P(Z<za)P(X > a) = 1 - P(Z < z_a)
  • 4Between two values: P(a<X<b)=P(Z<zb)−P(Z<za)P(a < X < b) = P(Z < z_b) - P(Z < z_a)
  • 5Inverse problems: find zz for the given area inside the table, then x=μ+zσx = \mu + z\sigma (top 10% means z≈1.28z \approx 1.28)
  • 6A probability is also a proportion of the population: 0.06680.0668 of 2000 bottles is about 134 bottles

Frequently Asked Questions

No. For a continuous variable a single value has probability 0, so << and ≤\le give the same area. This is different from the binomial distribution, where it matters.
No. For a continuous variable a single value has probability 0, so << and ≤\le give the same area. This is different from the binomial distribution, where it matters.
Round to two decimals for the table, or take a value between the two neighbouring entries. For z=1.645z = 1.645 the areas 0.9495 and 0.9505 are equally close, so 0.95 is a good answer.
Tables usually stop at about 3.49. Beyond that, the left area is practically 1 (for large positive z) or practically 0 (for large negative z). Such values are extremely rare.
The 90th percentile is the value with 90% of the population below it. "Below" is the left side of the curve, which is exactly what the z-table measures.

Glossary

Standardizing
Turning a value xx into a z-score with z=x−μσz = \frac{x - \mu}{\sigma}, so that the standard normal table can be used
Z-table
Table of the standard normal distribution: for each zz it gives the area to the left, P(Z<z)P(Z < z)
Left tail / right tail
The area below a value (left) or above a value (right) under the normal curve
Complement rule
P(X>a)=1−P(X<a)P(X > a) = 1 - P(X < a), because the whole area under the curve is 1
Percentile
The value with a given percentage of the population below it, for example the 90th percentile
Cut-off value
The value x=μ+zσx = \mu + z\sigma that separates a given area, for example the top 10%

Formula Card

Standardize

z=x−μσz = \frac{x - \mu}{\sigma}

From real units to a z-score

Left area

P(X<a)=P(Z<za)P(X < a) = P(Z < z_a)

Read straight from the z-table

Right area

P(X>a)=1−P(Z<za)P(X > a) = 1 - P(Z < z_a)

Complement of the table value

Area between

P(a<X<b)=P(Z<zb)−P(Z<za)P(a < X < b) = P(Z < z_b) - P(Z < z_a)

Difference of two left areas

Cut-off value

x=μ+zσx = \mu + z\sigma

From a z-score back to real units (percentiles)

Today's challengeOne problem, three tries

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