Statistics Reference

Normal Distributions and Confidence Intervals

A normal model turns a value into a standardized z-score and an estimated percentile. Confidence intervals use sample data to estimate a population value with an explicit margin of error.

Solved Examples

Jump directly to a worked example by normal-model or inference task.

Fact Table

IdeaMeaningUse It When
Normal distributionA symmetric bell-shaped model determined by mean $\mu$ and standard deviation $\sigma$.The data are approximately unimodal and symmetric without strong outliers.
Z-scoreThe number of standard deviations a value is above or below the mean.You need to compare a value to a normal model.
PercentileThe percentage of values at or below a score.You need to interpret area under a normal curve.
Margin of errorThe amount added to and subtracted from an estimate.You need to state uncertainty in a sample estimate.
Confidence intervalA plausible range for a population parameter from a sampling method.You are estimating a mean or proportion from random data.

Content Formulas

Standardize
$$z=\frac{x-\mu}{\sigma}$$
Empirical Rule
$$68\%\text{ within }1\sigma,\quad95\%\text{ within }2\sigma$$
Confidence Interval
$$\text{estimate}\pm\text{margin of error}$$
A normal curve is a model, not a promise. Check shape and outliers before using normal-distribution percentages.

Normal Distribution Lab

Change the model and score. The graph stays on a z-score scale so the same curve works for any mean and standard deviation.

Classic Examples

Find a Z-score

A test score of $80$ comes from a normal model with mean $70$ and standard deviation $5$. Find the z-score.

Solution Steps
  1. Subtract the mean from the score.
  2. Divide by the standard deviation.
  3. Interpret the positive result.
$$80-70=10$$$$z=\frac{10}{5}=2$$$$\boxed{\text{The score is 2 standard deviations above the mean.}}$$

Use the Empirical Rule

A normal distribution has mean $70$ and standard deviation $5$. Estimate the percentage of scores between $60$ and $80$.

Solution Steps
  1. Measure each endpoint in standard deviations from the mean.
  2. Recognize the interval from $\mu-2\sigma$ to $\mu+2\sigma$.
  3. Use the empirical-rule percentage.
$$60=70-2(5),\qquad80=70+2(5)$$$$60\text{ to }80\text{ is within }2\sigma\text{ of the mean}$$$$\boxed{\text{about }95\%}$$

Interpret a Percentile

A normal-table or calculator result gives $P(Z\le1.28)\approx0.900$. Interpret this result.

Solution Steps
  1. Read the probability as the area to the left of the z-score.
  2. Convert the decimal probability to a percent.
  3. State the percentile meaning.
$$P(Z\le1.28)\approx0.900$$$$0.900=90.0\%$$$$\boxed{\text{A z-score of }1.28\text{ is about the 90th percentile.}}$$

Build a Confidence Interval

A random sample estimates that $60\%$ of students prefer a new schedule, with margin of error $7$ percentage points. State the confidence interval.

Solution Steps
  1. Write the estimate and margin of error in the same units.
  2. Subtract the margin for the lower endpoint.
  3. Add the margin for the upper endpoint.
$$60\%\pm7\%$$$$60\%-7\%=53\%$$$$\boxed{53\%\text{ to }67\%}$$

Interpret an Interval Correctly

How should a $95\%$ confidence interval of $53\%$ to $67\%$ be described?

Solution Steps
  1. Name the population quantity being estimated.
  2. State the interval as an estimate from this sample.
  3. Connect $95\%$ to the long-run success of the sampling method.
$$\text{population preference proportion}$$$$\text{the sample estimates }53\%\text{ to }67\%$$$$\boxed{\text{In repeated samples, this method captures the true proportion about 95\% of the time.}}$$

Standards Connection

This page supports the normal-distribution and model-fitting work in CCSS S-ID and sample-based inference and margin-of-error work in CCSS S-IC.