Unit 2 · Topic 2.11 Beta

The Normal Distribution

A density curve models a distribution with areas as probabilities.

Practice 3: Analyze DataPractice 4: Interpret Results

Part 1 · Hook

Why this matters

A farm sorts its eggs by weight: small, large, jumbo. It cannot weigh every possible egg in advance, but egg weights follow a smooth, bell-shaped pattern. With a curve, a mean of 58 grams and a standard deviation of 4 grams, the farm can predict what share of tomorrow's eggs will be jumbo.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A value is 6 above the mean and the standard deviation is 4. What is its z-score?

  1. 1.5
  2. 6
  3. 0.67
  4. 24
Show the answer

z=64=1.5z = \frac{6}{4} = 1.5 standard deviations above the mean.

  • Correct: 1.5:
  • 6:
  • 0.67:
  • 24:

2. Which describes a distribution with a single peak and matching tails on both sides?

  1. Roughly symmetric
  2. Skewed right
  3. Skewed left
  4. Uniform
Show the answer

Matching tails on both sides mean symmetric.

  • Correct: Roughly symmetric:
  • Skewed right:
  • Skewed left:
  • Uniform:

Part 4 · See it

See it first

Normal density curve for egg weights with mean 58 grams and standard deviation 4 grams, axis marked from 46 to 70 grams. Brackets show that about 68% of weights fall within 1 standard deviation (54 to 62 g), about 95% within 2 (50 to 66 g) and about 99.7% within 3 (46 to 70 g).
Egg weights modeled by a normal curve with mean 58 g and SD 4 g: about 68%, 95% and 99.7% within 1, 2 and 3 SDs. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A distribution is roughly symmetric and bell-shapeda normal curve N(μ,σ)N(\mu, \sigma) can model it, and areas under the curve are probabilities
  2. Every normal curve has the same shape in standard-deviation unitsabout 68%, 95% and 99.7% of values lie within 1, 2 and 3 SDs (the empirical rule)
  3. Standardize a value with z=x−μσz = \frac{x - \mu}{\sigma}any normal question becomes one about the standard normal N(0,1)N(0, 1), answered by Table A or technology
  4. Work backward from an areainverse normal gives the z, then x=μ+zσx = \mu + z\sigma gives the boundary

Part 6 · Key ideas

Key ideas

  • A density curve is on or above the axis with total area 1; the area over an interval is the probability of that interval.
  • A normal distribution N(μ,σ)N(\mu, \sigma) is symmetric and bell-shaped, centered at μ\mu with spread σ\sigma. The standard normal is N(0,1)N(0, 1).
  • The empirical rule: about 68%, 95% and 99.7% within 1, 2 and 3 standard deviations of the mean.
  • Show work: name the distribution and parameters, the boundary and the direction, for example "N(58,4)N(58, 4), P(W>66)P(W > 66)", with a sketch.
  • Forward: z=x−μσz = \frac{x - \mu}{\sigma}, then the area. Backward: the area, then z, then x=μ+zσx = \mu + z\sigma.

Part 7 · Misconception

A common mistake

The wrong idea: A normal model works for any data set once you know its mean and standard deviation.

What actually happens: A normal model fits only roughly symmetric, mound-shaped data. For strongly skewed data, normal areas give the wrong percents.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Battery life of a phone model

Under a standard test, the battery life of a new phone model is approximately normal with mean 18 hours and standard deviation 1.5 hours (invented data). Let B = the battery life of a randomly chosen phone.
1416182022hours16

1. What is the probability that a randomly chosen phone's battery lasts less than 16 hours?

  1. 0.9088
  2. 0.16
  3. 0.0912
  4. 0.025
Show the answer

B is N(18,1.5)N(18, 1.5). zz=16−181.5= \frac{16 - 18}{1.5}≈−1.33\approx -1.33, and P(B<16)≈0.0912P(B < 16) \approx 0.0912 by technology (Table A with z=−1.33z = -1.33 gives 0.0918).

  • 0.9088: That is the area to the right of 16 hours, the complement.
  • 0.16: That uses the empirical rule for 1 standard deviation, but 16 hours is 1.33 standard deviations below the mean.
  • Correct: 0.0912: Right: the area to the left of z = −1.33.
  • 0.025: That is the empirical rule for 2 standard deviations; 16 hours is less than 2 SDs away.

2. About what percent of phones have a battery life of more than 20.5 hours?

  1. About 95.2%
  2. About 16%
  3. About 2.5%
  4. About 4.8%
Show the answer

zz=20.5−181.5= \frac{20.5 - 18}{1.5}≈1.67\approx 1.67, and P(B>20.5)P(B > 20.5)=1−P(Z<1.67)= 1 - P(Z < 1.67)≈0.0478\approx 0.0478, about 4.8%.

  • About 95.2%: That is the area below 20.5 hours.
  • About 16%: That is the empirical rule for 1 SD; 20.5 is 1.67 SDs above the mean.
  • About 2.5%: That would be more than 2 SDs above the mean, 21 hours.
  • Correct: About 4.8%: Right: the area to the right of z = 1.67.

3. What battery life is the 90th percentile?

  1. 19.92 hours
  2. 16.08 hours
  3. 19.35 hours
  4. 21.00 hours
Show the answer

The 90th percentile has 0.90 of the area below it: z≈1.28z \approx 1.28, so 18+1.28(1.5)≈19.9218 + 1.28(1.5) \approx 19.92 hours.

  • Correct: 19.92 hours: Right: 1.28 standard deviations above the mean.
  • 16.08 hours: That is the 10th percentile, 1.28 SDs below the mean.
  • 19.35 hours: That uses z = 0.9 in place of the z-score with 0.90 of the area below it.
  • 21.00 hours: That is 2 standard deviations above the mean, about the 97.7th percentile.

Graph

Egg weights at a farm

The weights of eggs from a farm's hens are approximately normal with mean 58 grams and standard deviation 4 grams (invented data).
455055606570grams5054586266

4. By the empirical rule, about what percent of eggs weigh between 54 and 62 grams?

  1. About 95%
  2. About 68%
  3. About 34%
  4. About 50%
Show the answer

54 and 62 are 1 standard deviation below and above the mean, 58±458 \pm 4, so about 68% of eggs fall between them.

  • About 95%: That is within 2 standard deviations, 50 to 66 grams.
  • Correct: About 68%: Right: within 1 standard deviation.
  • About 34%: That is one side only, 58 to 62 grams.
  • About 50%: Half of the eggs are below the mean, 58 grams.

5. By the empirical rule, about what percent of eggs weigh more than 66 grams?

  1. About 5%
  2. About 16%
  3. About 2.5%
  4. About 0.15%
Show the answer

66 is 2 standard deviations above the mean. About 95% lie within 2 SDs, leaving 5% in the two tails, so about 2.5% above 66 grams.

  • About 5%: That is both tails together; above 66 is one tail.
  • About 16%: That is above 1 SD, 62 grams.
  • Correct: About 2.5%: Right: half of the 5% outside 2 SDs.
  • About 0.15%: That is above 3 SDs, 70 grams.

6. What is the z-score of a 63-gram egg?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

zz=x−μσ= \frac{x - \mu}{\sigma}=63−584= \frac{63 - 58}{4}=1.25= 1.25: the egg is 1.25 standard deviations above the mean.

  • Answer: 1.25

Part 9 · Summary

Summary

A density curve models a distribution with areas as probabilities. Normal distributions N(μ,σ)N(\mu, \sigma) are symmetric and bell-shaped, with about 68%, 95% and 99.7% of values within 1, 2 and 3 SDs. Standardize with z=x−μσz = \frac{x - \mu}{\sigma} to find areas, or work back from an area to a boundary with x=μ+zσx = \mu + z\sigma. Always name the distribution, parameters, boundary and direction.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

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