Normal Probability Calculations

Normal probability calculations answer two questions about a normal distribution: what is the probability that a value falls in a range, and what value cuts off a given percent? For a normal distribution probability, change each value to a z-score and use a z-table or a calculator. For a cutoff, work backward.

Before this lesson: Normal Distribution, Standard Normal Distribution and Percentiles and Z-Scores

The idea: turn any normal curve into the standard one

A probability for a normal variable is an area under its curve, just like on any density curve. The z-table only covers N(0, 1). So you change each value x into a z-score, which counts standard deviations from the mean. The area to the left of x is the same as the area to the left of its z-score.

z = \dfrac{x - \mu}{\sigma} \qquad x = \mu + z\sigma

Use the first formula to find a probability. Use the second one, solved for x, to find a cutoff value.

How to find a normal distribution probability

1. Draw the curve, mark the mean, and shade the area you want.
2. Change each value to a z-score.
3. Find the area with the z-table or a calculator. Subtract from 1 for a right area.
4. Answer the question in context.

How to find a cutoff value

1. Shade the area you know. Find the area to the left of the cutoff.
2. Find the z-score for that left area in the z-table, or use invNorm.
3. Change z back to x with x = \mu + z\sigma.
4. Answer the question in context.

z-scores used in this lesson

Calculator: on a TI-84, normalcdf(lower, upper, μ, σ) finds an area without a z-score, and invNorm(left area, μ, σ) finds a cutoff. Both are under 2nd, VARS (DISTR). In Desmos, normaldist(μ, σ).cdf(a, b) gives the area between a and b. Many teachers still want the z-score work, so show it first.

Normal probability examples

Example 1: Probability less than a value

Women’s heights are approximately normal with mean 64 inches and standard deviation 2.5 inches. What proportion of women are shorter than 68 inches?

z = \dfrac{68 - 64}{2.5}

= \dfrac{4}{2.5}

= 1.6

P(X \lt 68) = P(Z \lt 1.6)

= 0.9452

Answer: about 0.9452, so about 94.5% of women are shorter than 68 inches. Calculator check: normalcdf(−1E99, 68, 64, 2.5) ≈ 0.9452. In Desmos, normaldist(64, 2.5).cdf(68) gives the same value.

Example 2: Probability greater than a value

Scores on a college entrance test are approximately normal with mean 500 and standard deviation 100. What proportion of scores are above 650?

z = \dfrac{650 - 500}{100}

= \dfrac{150}{100}

= 1.5

P(Z \lt 1.5) = 0.9332

P(X \gt 650) = 1 - 0.9332

= 0.0668

Answer: about 0.0668, or 6.68% of scores. With a calculator, normalcdf(650, 1E99, 500, 100) ≈ 0.0668.

Example 3: Probability between two values

Car battery lifetimes are approximately normal with mean 4.5 years and standard deviation 0.8 years. What is the probability that a battery lasts between 3.5 and 5.5 years?

z_1 = \dfrac{3.5 - 4.5}{0.8}

= \dfrac{-1}{0.8}

= -1.25

z_2 = \dfrac{5.5 - 4.5}{0.8}

= \dfrac{1}{0.8}

= 1.25

P(-1.25 \lt Z \lt 1.25) = 0.8944 - 0.1056

= 0.7888

Answer: about 0.7888. A calculator gives normalcdf(3.5, 5.5, 4.5, 0.8) ≈ 0.7887. The last digit differs because the table rounds each area.

Example 4: Find a cutoff for the top 10%

Use the test scores from Example 2. A scholarship goes to the top 10% of scores. What is the lowest score that qualifies?

1 - 0.10 = 0.90

z \approx 1.28

x = 500 + 1.28(100)

= 500 + 128

= 628

Answer: a score of about 628 or higher qualifies. On a TI-84, invNorm(0.90, 500, 100) ≈ 628.16.

Example 5: Filling cereal boxes (word problem)

A machine fills cereal boxes labeled 500 grams. The fill weights are approximately normal with mean 510 grams and standard deviation 4 grams. (a) What proportion of boxes are underweight? (b) The company wants only 1% of boxes underweight. If σ stays at 4 grams, what mean fill should it set?

z = \dfrac{500 - 510}{4}

= \dfrac{-10}{4}

= -2.5

P(X \lt 500) = 0.0062

z \approx -2.33

500 = \mu + (-2.33)(4)

500 = \mu - 9.32

509.32 = \mu

Answer: (a) about 0.0062, or 0.62% of boxes; (b) a mean fill of about 509.3 grams. With invNorm, the mean comes out to about 509.31 grams.

Common mistakes

Related methods

Check with the empirical rule

Before you trust an answer, compare it with the empirical rule. In Example 2, 650 is 1.5σ above the mean. That is between 1σ (16% above) and 2σ (2.5% above), so 6.68% is reasonable.

Is the distribution close to normal?

These steps only work for a distribution that is close to normal. For data, check the shape first. The next lesson, Normal Probability Plots, shows how.

Practice problems

Use the table above or a calculator. Women’s heights are N(64, 2.5) inches, test scores are N(500, 100), and battery lifetimes are N(4.5, 0.8) years.

1. What proportion of women are shorter than 60 inches?
2. What proportion of women are taller than 66 inches?
3. What proportion of women are between 61 and 67 inches tall?
4. What proportion of test scores are below 380?
5. What test score is at the 75th percentile?
6. What proportion of batteries last more than 6.1 years?
7. A store wants only 5% of batteries to fail before the warranty ends. How long should the warranty be?
8. What height separates the shortest 20% of women from the rest?
9. Cereal fills are N(510, 4) grams. What proportion of boxes hold more than 515 grams?
10. Bolt lengths are N(30, 0.2) mm. Bolts from 29.6 to 30.4 mm pass inspection. What proportion fail?

z = \dfrac{60 - 64}{2.5}

= \dfrac{-4}{2.5}

= -1.6

P(Z \lt -1.6) = 0.0548

z = \dfrac{66 - 64}{2.5}

= \dfrac{2}{2.5}

= 0.8

1 - 0.7881 = 0.2119

z_1 = \dfrac{61 - 64}{2.5} = \dfrac{-3}{2.5}

= -1.2

z_2 = \dfrac{67 - 64}{2.5} = \dfrac{3}{2.5}

= 1.2

0.8849 - 0.1151 = 0.7698

z = \dfrac{380 - 500}{100}

= \dfrac{-120}{100}

= -1.2

P(Z \lt -1.2) = 0.1151

z \approx 0.67

x = 500 + 0.67(100)

= 500 + 67

= 567

z = \dfrac{6.1 - 4.5}{0.8}

= \dfrac{1.6}{0.8}

= 2

1 - 0.9772 = 0.0228

z \approx -1.645

x = 4.5 + (-1.645)(0.8)

= 4.5 - 1.316

= 3.184

z \approx -0.84

x = 64 + (-0.84)(2.5)

= 64 - 2.1

= 61.9

z = \dfrac{515 - 510}{4}

= \dfrac{5}{4}

= 1.25

1 - 0.8944 = 0.1056

z = \dfrac{29.6 - 30}{0.2} = \dfrac{-0.4}{0.2}

= -2

z = \dfrac{30.4 - 30}{0.2} = \dfrac{0.4}{0.2}

= 2

0.9772 - 0.0228 = 0.9544

1 - 0.9544 = 0.0456

Normal distribution probability FAQ

How do you find a probability in a normal distribution?

Change the value to a z-score with z = (x − μ)/σ. Then find the area to the left of z in a z-table, and subtract from 1 if you need the area to the right.

How do you find a value from a percentile in a normal distribution?

Find the z-score with that percent of the area to its left, then use x = μ + zσ. On a TI-84, invNorm(percent, μ, σ) does both steps.

What does normalcdf do on a TI-84?

It finds the area under a normal curve between a lower and an upper bound. Enter the bounds, then the mean and standard deviation.

Why do the table and the calculator give slightly different answers?

The table rounds z to two decimal places and each area to four. The calculator does not round, so the last digit can differ.

Summary

• To find a probability, change x to z = (x − μ)/σ, then use the z-table.
• Subtract from 1 for a right area. Subtract two left areas for an area between.
• To find a cutoff, get z from the left area, then use x = μ + zσ.
• On a TI-84, normalcdf finds areas and invNorm finds cutoffs. invNorm always uses the left area.
• Check that your answer makes sense with the empirical rule.

Previous lesson: Standard Normal Distribution

Next lesson: Normal Probability Plots

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