Source: Practice Exam 1, Spring 2016 | Purdue University
Difficulty: Intermediate | Prerequisites: Descriptive statistics notes, probability distributions notes, comfort with z-tables.
The normal distribution is the single most important distribution in introductory statistics. It is the foundation for z-scores, confidence intervals and hypothesis tests. The Central Limit Theorem explains why sampling distributions become approximately normal regardless of the population shape, which is what makes inference possible. This section is the highest-weighted topic on the exam (30 points in the practice exam) and involves three distinct skills: computing probabilities from z-scores, working backwards from a probability to find a value (inverse normal), and applying the CLT to the sampling distribution of the sample mean.
Convert any normal-distribution problem to a z-score, then use the z-table. For "find the probability," standardise and look up. For "find the value," work backwards from the table to the z-score, then unstandardise. When the question asks about the average of n observations rather than a single observation, divide the standard deviation by the square root of n before computing the z-score. The CLT is what lets you do this even when the population is not normal, provided n is large enough (at least 30).
Normal distribution
A continuous, bell-shaped, symmetric distribution fully described by its mean (mu) and standard deviation (sigma). In simple terms, most values cluster around the centre and taper off symmetrically.
Standard normal distribution (Z)
A normal distribution with mean 0 and standard deviation 1. Any normal variable X can be converted to Z using the z-score formula.
Z-score
The number of standard deviations a value x sits from the mean: z = (x - mu) / sigma. It standardises the problem so you can use the z-table.
Z-table
A table giving P(Z < z) for the standard normal distribution. The exam requires four decimal places when reading from this table.
Standard deviation and the shape of the normal curve
A larger standard deviation makes the curve wider and flatter. A smaller standard deviation makes it taller and narrower. The mean shifts the curve left or right but does not change its shape.
Central Limit Theorem (CLT)
If you take sufficiently large random samples (n >= 30 as a rule of thumb) from any population, the distribution of the sample mean is approximately normal, regardless of the shape of the population. It fails when the sample is small and the population is not normal.
Sampling distribution of the sample mean
The distribution of X-bar across all possible samples of size n. Its mean equals the population mean (mu). Its standard deviation (the standard error) equals sigma / sqrt(n).
Standard error
The standard deviation of the sampling distribution: sigma / sqrt(n). It shrinks as n increases, which means larger samples give more precise estimates.
Bell-shaped, symmetric about the mean.
Described entirely by mu (centre) and sigma (spread).
Changing mu shifts the curve; changing sigma stretches or compresses it.
A larger sigma makes the curve wider and flatter. This is a common multiple-choice question.
Step 1: Compute the z-score: z = (x - mu) / sigma.
Step 2: The z-table gives P(Z < z). For P(Z > z), use the complement: 1 - P(Z < z).
Example: X is normal with mu = 50,000 and sigma = 9,000. Find P(X > 60,000).
z = (60,000 - 50,000) / 9,000 = 1.11.
P(Z > 1.11) = 1 - P(Z < 1.11) = 1 - 0.8665 = 0.1335.
The question gives you a probability and asks for the value of x.
Step 1: Convert the probability to a left-tail probability. If the question says "top 15%," then P(Z > b) = 0.15, so P(Z < b) = 0.85.
Step 2: Look up 0.85 in the body of the z-table to find b = 1.04.
Step 3: Unstandardise: x = mu + z × sigma = 50,000 + 1.04 × 9,000 = 59,360.
The answer means the top 15% corresponds to at least 59,360 gallons.
If the population is normal, the sampling distribution of X-bar is normal for any sample size.
If the population is not normal, the sampling distribution of X-bar is approximately normal when n >= 30.
The CLT does not apply when the sample is small (below 30) and the population is not normal. This is the scenario to watch for on the exam.
The sampling distribution of X-bar has mean mu and standard deviation sigma / sqrt(n).
Use this standard error in the z-score formula instead of the population sigma.
Example: mu = 50,000, sigma = 9,000, n = 6. Find P(X-bar > 60,000).
Standard error = 9,000 / sqrt(6) = 3,674.23.
z = (60,000 - 50,000) / 3,674.23 = 2.72.
P(Z > 2.72) = 1 - 0.9967 = 0.0033.
Notice how the probability is much smaller for the sample mean than for a single observation (0.0033 vs 0.1335). This is because the sampling distribution is tighter.
Z-score for a single observation:
z = \frac{x - \mu}{\sigma}Z-score for a sample mean:
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}Standard error of the sample mean:
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}Unstandardising (inverse normal):
x = \mu + z \cdot \sigmaCLT applicability rule of thumb: the CLT gives an approximately normal sampling distribution when n >= 30 or the population is itself normal.
Students forget to use the complement rule. The z-table gives P(Z < z). For P(Z > z), you must compute 1 - P(Z < z). Omitting the "1 minus" is one of the most common errors.
When working with the sample mean, students use sigma instead of sigma / sqrt(n). If the question asks about the average of n observations, you must divide by the square root of n.
Students sometimes think the CLT makes every sampling distribution normal. It does not apply when n is small and the population is not normal.
On inverse normal problems, students look up the wrong tail. "Top 15%" means the right tail is 0.15, so the left-tail area is 0.85. Always convert to a left-tail probability before using the z-table.
⚠️ This is the highest-weighted section on the practice exam (30 points). Expect a multi-part problem with parts asking for P(X > value), an inverse normal calculation, and a sample-mean probability.
⚠️ Show every step: the z-score calculation (2 points), the complement (2 points), the z-table lookup (2 points) and the final answer (1 point). Partial credit is generous if you show the work.
⚠️ Z-table answers must have four decimal places. All other numeric answers need two decimal places.
⚠️ The multiple-choice question about what makes the normal curve wider and flatter tests whether you know it is the standard deviation (not the mean).
The z-table gives P(Z < z). To find P(Z > 1.5), you compute ______. (1 - P(Z < 1.5).)
True or false: The mean of the sampling distribution of X-bar equals the population mean. (True.)
The standard error of X-bar when sigma = 10 and n = 25 is ______. (10 / sqrt(25) = 2.)
True or false: The CLT applies when n = 15 and the population is heavily skewed. (False, n is too small and the population is not normal.)
A larger standard deviation makes the normal curve ______ and ______. (Wider and flatter.)
Q: X is normally distributed with mean 50,000 and standard deviation 9,000. Find P(X > 60,000).
A: z = (60,000 - 50,000) / 9,000 = 1.11. P(Z > 1.11) = 1 - P(Z < 1.11) = 1 - 0.8665 = 0.1335.
Q: Using the same distribution, how much gasoline is purchased if the amount is in the top 15%?
A: P(Z > b) = 0.15, so P(Z < b) = 0.85. From the z-table, b = 1.04. x = 50,000 + 1.04 × 9,000 = 59,360.
Q: Find the probability that the average amount of gasoline purchased over 6 weeks exceeds 60,000, given mu = 50,000 and sigma = 9,000.
A: Standard error = 9,000 / sqrt(6) = 3,674.23. z = (60,000 - 50,000) / 3,674.23 = 2.72. P(Z > 2.72) = 1 - 0.9967 = 0.0033.
Q: In which situation does the CLT not apply? (a) Large sample, normal population. (b) Small sample, normal population. (c) Large sample, non-normal population. (d) Small sample, non-normal population.
A: (d). The CLT requires either a normal population or a large sample (n >= 30). A small sample from a non-normal population does not satisfy either condition.
The normal distribution is the gateway to confidence intervals and hypothesis testing, which make up the second half of STAT 350. The z-score formula reappears with slight modifications for t-tests when sigma is unknown. The CLT underpins virtually every inferential procedure you will encounter: it is the reason you can use normal-based methods on data from populations of any shape, provided the sample is large enough.
Normal distribution, Gaussian distribution, bell curve, z-score, z-table, standard normal, standardisation, complement rule, inverse normal, percentile, top percent, Central Limit Theorem, CLT, sampling distribution, sample mean, X-bar, standard error, sigma over root n, STAT 350, Purdue, introductory statistics, probability, confidence intervals preview