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Statistics practice problems with answers

This page has 15 statistics practice problems with full worked solutions, ordered from easy to hard. They cover descriptive statistics and box plots, probability rules and Bayes' theorem, discrete random variables, the binomial and normal distributions, sampling distributions, confidence intervals, t-tests and z-tests, p-values and error types, and linear regression. Each solution shows the formula, the numbers substituted into it and how to state the result in context.

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Problem 1: Mean, median, mode and the effect of an outlier

Difficulty: Easy

Seven students report the hours of paid work they did last week: 3, 5, 5, 6, 7, 8, 22. Find the mean, median, mode and range. Then remove the value 22 and recompute the mean and median. Which measure of centre is more affected, and what does the comparison suggest about the shape of the distribution?
Show worked solution
Mean: xˉ=3+5+5+6+7+8+227=567=8\bar{x} = \frac{3 + 5 + 5 + 6 + 7 + 8 + 22}{7} = \frac{56}{7} = 8 hours. Median: the 4th of the 7 ordered values, 6 hours. Mode: 5 hours (the only repeated value). Range: 22−3=1922 - 3 = 19 hours. Without 22: the mean is 346≈5.67\frac{34}{6} \approx 5.67 hours, and the median is the average of the 3rd and 4th of the six values, 5+62=5.5\frac{5 + 6}{2} = 5.5 hours. Removing one value moved the mean by 2.33 hours but the median by only 0.5: the median is resistant to outliers and the mean is not. With the outlier included, the mean (8) is well above the median (6), which is typical of a right-skewed distribution. For skewed data, report the median and interquartile range rather than the mean and standard deviation.

Problem 2: Sample variance and standard deviation

Difficulty: Easy

For the sample {4,8,6,10,12}\{4, 8, 6, 10, 12\}, calculate the sample mean, the sample variance and the sample standard deviation. Why do we divide by n−1n - 1 rather than nn?
Show worked solution
Mean: xˉ=4+8+6+10+125=405=8\bar{x} = \frac{4 + 8 + 6 + 10 + 12}{5} = \frac{40}{5} = 8. Squared deviations from the mean: (4−8)2+(8−8)2+(6−8)2+(10−8)2+(12−8)2=16+0+4+4+16=40(4-8)^2 + (8-8)^2 + (6-8)^2 + (10-8)^2 + (12-8)^2 = 16 + 0 + 4 + 4 + 16 = 40. Sample variance: s2=40n−1=404=10s^2 = \frac{40}{n - 1} = \frac{40}{4} = 10. Sample standard deviation: s=10≈3.16s = \sqrt{10} \approx 3.16. The deviations are measured from the sample mean, which is calculated from the same data, so they are on average slightly smaller than deviations from the true population mean. Dividing by n−1n - 1 (the degrees of freedom) corrects for this and makes s2s^2 an unbiased estimator of the population variance. Dividing by nn would give 8≈2.83\sqrt{8} \approx 2.83, an underestimate.

Problem 3: Five-number summary, IQR and outliers

Difficulty: Easy

The times, in minutes, that 11 students took to finish a quiz are: 12, 15, 17, 18, 20, 21, 22, 24, 25, 27, 45. Find the five-number summary and the interquartile range, use the 1.5×IQR1.5 \times IQR rule to check for outliers, and describe what a box plot of these data would look like.
Show worked solution
The data are already in order and n=11n = 11. Median: the 6th value, 21. Lower quartile Q1Q_1: the median of the five values below the median (12, 15, 17, 18, 20), which is 17. Upper quartile Q3Q_3: the median of the five values above it (22, 24, 25, 27, 45), which is 25. Five-number summary: minimum 12, Q1=17Q_1 = 17, median 21, Q3=25Q_3 = 25, maximum 45. IQR=25−17=8IQR = 25 - 17 = 8. Outlier fences: Q1−1.5×IQR=17−12=5Q_1 - 1.5 \times IQR = 17 - 12 = 5 and Q3+1.5×IQR=25+12=37Q_3 + 1.5 \times IQR = 25 + 12 = 37. No value is below 5, and 45 is above 37, so 45 is an outlier. Box plot: a box from 17 to 25 with a line at 21, a left whisker down to 12, a right whisker up to 27 (the largest value that is not an outlier), and 45 plotted as a separate point. The long right tail means the distribution is right-skewed. (Some software uses a slightly different quartile method, which can shift Q1Q_1 and Q3Q_3 a little.)

Problem 4: Addition rule, conditional probability and independence

Difficulty: Easy

In a group of 200 students, 120 take economics, 80 take statistics and 50 take both. A student is chosen at random. Find (a) the probability that they take economics or statistics, (b) the probability that they take neither, and (c) the probability that they take statistics given that they take economics. (d) Are taking economics and taking statistics independent events? Are they mutually exclusive?
Show worked solution
Let EE = takes economics and SS = takes statistics, so P(E)=120200=0.6P(E) = \frac{120}{200} = 0.6, P(S)=80200=0.4P(S) = \frac{80}{200} = 0.4 and P(E∩S)=50200=0.25P(E \cap S) = \frac{50}{200} = 0.25. (a) Addition rule: P(E∪S)=P(E)+P(S)−P(E∩S)=0.6+0.4−0.25=0.75P(E \cup S) = P(E) + P(S) - P(E \cap S) = 0.6 + 0.4 - 0.25 = 0.75. The overlap is subtracted because students taking both would otherwise be counted twice. (b) Complement: P(neither)=1−0.75=0.25P(\text{neither}) = 1 - 0.75 = 0.25, which is 50 students. (c) P(S∣E)=P(E∩S)P(E)=0.250.6≈0.417P(S \mid E) = \frac{P(E \cap S)}{P(E)} = \frac{0.25}{0.6} \approx 0.417, or directly 50120\frac{50}{120}. (d) Independence requires P(S∣E)=P(S)P(S \mid E) = P(S), or equivalently P(E∩S)=P(E)P(S)P(E \cap S) = P(E)P(S). Here 0.417≠0.40.417 \neq 0.4 and 0.25≠0.6×0.4=0.240.25 \neq 0.6 \times 0.4 = 0.24, so the events are not independent, although the dependence is weak. They are not mutually exclusive either, because 50 students take both; mutually exclusive events have P(E∩S)=0P(E \cap S) = 0.

Problem 5: Bayes' theorem: a positive screening test

Difficulty: Medium

A condition affects 2 percent of a population. A screening test detects it in 95 percent of people who have it and gives a false positive for 10 percent of people who do not have it. A randomly chosen person tests positive. What is the probability that they have the condition?
Show worked solution
Let DD = has the condition and ++ = tests positive. We know P(D)=0.02P(D) = 0.02, P(+∣D)=0.95P(+ \mid D) = 0.95 and P(+∣Dc)=0.10P(+ \mid D^c) = 0.10. Total probability of a positive test: P(+)=0.95×0.02+0.10×0.98=0.019+0.098=0.117P(+) = 0.95 \times 0.02 + 0.10 \times 0.98 = 0.019 + 0.098 = 0.117. Bayes' theorem: P(D∣+)=P(+∣D)P(D)P(+)=0.0190.117≈0.162P(D \mid +) = \frac{P(+ \mid D)P(D)}{P(+)} = \frac{0.019}{0.117} \approx 0.162. Only about 16 percent of people who test positive have the condition. In counts: out of 10,000 people, 200 have it and 190 of them test positive; 9,800 do not, and 980 of them also test positive. So only 190 of the 1,170 positive results are true positives. When a condition is rare, false positives from the large unaffected group outnumber true positives, even with a test that sounds accurate.

Problem 6: Expected value and standard deviation of a discrete random variable

Difficulty: Medium

Let XX be the number of practice exams a student completes in a week, with P(X=0)=0.1P(X=0) = 0.1, P(X=1)=0.3P(X=1) = 0.3, P(X=2)=0.4P(X=2) = 0.4 and P(X=3)=0.2P(X=3) = 0.2. Find the mean and standard deviation of XX. If each practice exam takes 1.5 hours, what are the mean and standard deviation of the weekly time spent, T=1.5XT = 1.5X?
Show worked solution
Check that the probabilities sum to 1: 0.1+0.3+0.4+0.2=10.1 + 0.3 + 0.4 + 0.2 = 1. Mean: E(X)=∑x P(x)=0(0.1)+1(0.3)+2(0.4)+3(0.2)=0.3+0.8+0.6=1.7E(X) = \sum x \, P(x) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0.3 + 0.8 + 0.6 = 1.7. Variance: E(X2)=0(0.1)+1(0.3)+4(0.4)+9(0.2)=0.3+1.6+1.8=3.7E(X^2) = 0(0.1) + 1(0.3) + 4(0.4) + 9(0.2) = 0.3 + 1.6 + 1.8 = 3.7, so Var(X)=E(X2)−[E(X)]2=3.7−1.72=3.7−2.89=0.81\text{Var}(X) = E(X^2) - [E(X)]^2 = 3.7 - 1.7^2 = 3.7 - 2.89 = 0.81 and SD(X)=0.81=0.9SD(X) = \sqrt{0.81} = 0.9. For T=1.5XT = 1.5X: multiplying a random variable by a constant multiplies both its mean and its standard deviation by that constant, so E(T)=1.5×1.7=2.55E(T) = 1.5 \times 1.7 = 2.55 hours and SD(T)=1.5×0.9=1.35SD(T) = 1.5 \times 0.9 = 1.35 hours. The variance is multiplied by 1.52=2.251.5^2 = 2.25.

Problem 7: Binomial probabilities

Difficulty: Medium

A production process makes defective items with probability 0.2, independently of each other. In a random sample of 10 items, find (a) the probability of exactly 2 defects, (b) the probability of at least 2 defects, and (c) the mean and standard deviation of the number of defects.
Show worked solution
Let XX = number of defects. The binomial conditions hold: a fixed number of trials (n=10n = 10), two outcomes per trial, independent trials and a constant p=0.2p = 0.2. So P(X=k)=(10k)(0.2)k(0.8)10−kP(X = k) = \binom{10}{k}(0.2)^k(0.8)^{10-k}. (a) P(X=2)=(102)(0.2)2(0.8)8=45×0.04×0.1678≈0.302P(X = 2) = \binom{10}{2}(0.2)^2(0.8)^8 = 45 \times 0.04 \times 0.1678 \approx 0.302. (b) Use the complement: P(X≥2)=1−P(X=0)−P(X=1)P(X \ge 2) = 1 - P(X = 0) - P(X = 1), where P(X=0)=0.810≈0.1074P(X = 0) = 0.8^{10} \approx 0.1074 and P(X=1)=10×0.2×0.89≈0.2684P(X = 1) = 10 \times 0.2 \times 0.8^9 \approx 0.2684. So P(X≥2)≈1−0.1074−0.2684=0.6242P(X \ge 2) \approx 1 - 0.1074 - 0.2684 = 0.6242. (c) Mean: μ=np=10×0.2=2\mu = np = 10 \times 0.2 = 2 defects. Standard deviation: σ=np(1−p)=10×0.2×0.8=1.6≈1.26\sigma = \sqrt{np(1-p)} = \sqrt{10 \times 0.2 \times 0.8} = \sqrt{1.6} \approx 1.26 defects.

Problem 8: Normal distribution: probabilities and percentiles

Difficulty: Medium

Exam scores are normally distributed with mean 70 and standard deviation 8. Find (a) the proportion of students scoring above 86, (b) the proportion scoring between 62 and 86, and (c) the score needed to be in the top 10 percent.
Show worked solution
Standardise with z=x−μσz = \frac{x - \mu}{\sigma}. (a) z=86−708=2z = \frac{86 - 70}{8} = 2, and P(Z>2)=1−0.9772=0.0228P(Z > 2) = 1 - 0.9772 = 0.0228. About 2.3 percent of students score above 86. (b) For 62: z=62−708=−1z = \frac{62 - 70}{8} = -1. P(−1<Z<2)=Φ(2)−Φ(−1)=0.9772−0.1587=0.8185P(-1 < Z < 2) = \Phi(2) - \Phi(-1) = 0.9772 - 0.1587 = 0.8185. About 82 percent score between 62 and 86. (c) Work backwards. The top 10 percent starts at the 90th percentile, where Φ(z)=0.90\Phi(z) = 0.90, so z≈1.2816z \approx 1.2816. Then x=μ+zσ=70+1.2816×8≈80.3x = \mu + z\sigma = 70 + 1.2816 \times 8 \approx 80.3. A student needs a score of about 80.3 or more.

Problem 9: Sampling distribution of the mean

Difficulty: Medium

The time students spend on a weekly problem set has mean 50 minutes and standard deviation 12 minutes. (a) For a random sample of 36 students, describe the sampling distribution of the sample mean xˉ\bar{x} and find P(xˉ>53)P(\bar{x} > 53). (b) Why can you not find the probability that a single student takes more than 53 minutes without more information? (c) How large a sample would halve the standard error?
Show worked solution
(a) The sampling distribution of xˉ\bar{x} has mean μxˉ=μ=50\mu_{\bar{x}} = \mu = 50 and standard deviation (the standard error) σxˉ=σn=1236=2\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = 2. With n=36n = 36, the central limit theorem says xˉ\bar{x} is approximately normal whatever the shape of the population. So z=53−502=1.5z = \frac{53 - 50}{2} = 1.5 and P(xˉ>53)=P(Z>1.5)≈0.0668P(\bar{x} > 53) = P(Z > 1.5) \approx 0.0668. (b) The central limit theorem applies to sample means, not to individual observations. Without knowing that the population itself is normal, the distribution of one student's time is unknown. (If it were normal, P(X>53)=P(Z>0.25)≈0.40P(X > 53) = P(Z > 0.25) \approx 0.40, a far larger probability than for the mean, because averages vary much less than individuals.) (c) The standard error is σn\frac{\sigma}{\sqrt{n}}, so halving it requires doubling n\sqrt{n}, which means multiplying nn by 4: a sample of 144 students gives 12144=1\frac{12}{\sqrt{144}} = 1.

Problem 10: Confidence interval for a mean (t-interval)

Difficulty: Medium

A random sample of n=36n = 36 students has a mean weekly study time of xˉ=24\bar{x} = 24 hours with sample standard deviation s=6s = 6 hours. Construct a 95 percent confidence interval for the population mean and interpret it.
Show worked solution
The population standard deviation is unknown and is estimated by ss, so use the tt-distribution with df=n−1=35df = n - 1 = 35. The critical value is t∗≈2.030t^* \approx 2.030. Standard error: SE=sn=636=1SE = \frac{s}{\sqrt{n}} = \frac{6}{\sqrt{36}} = 1. Interval: xˉ±t∗×SE=24±2.030×1\bar{x} \pm t^* \times SE = 24 \pm 2.030 \times 1, giving (21.97, 26.03)(21.97,\ 26.03) hours. Interpretation: we are 95 percent confident that the mean weekly study time of all students in the population is between about 22.0 and 26.0 hours. The 95 percent describes the method: intervals built this way capture the true mean in 95 percent of repeated samples. It does not mean there is a 95 percent probability that this particular interval contains the mean, and it says nothing about where 95 percent of individual students fall. Using z∗=1.96z^* = 1.96 instead would give (22.04, 25.96)(22.04,\ 25.96). With n=36n = 36 the difference is small, but the tt-interval is the correct one when σ\sigma is estimated.

Problem 11: Confidence interval for a proportion and sample size

Difficulty: Medium

In a random sample of 400 students, 248 say they revise mainly in the week before an exam. (a) Construct a 95 percent confidence interval for the population proportion. (b) How many students would you need to survey to estimate the proportion to within 3 percentage points with 95 percent confidence, if you had no prior estimate?
Show worked solution
(a) Sample proportion: p^=248400=0.62\hat{p} = \frac{248}{400} = 0.62. Large-counts condition: np^=248n\hat{p} = 248 and n(1−p^)=152n(1 - \hat{p}) = 152 are both at least 10. Standard error: SE=p^(1−p^)n=0.62×0.38400=0.000589≈0.0243SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.62 \times 0.38}{400}} = \sqrt{0.000589} \approx 0.0243. Margin of error: 1.96×0.0243≈0.0481.96 \times 0.0243 \approx 0.048. Interval: 0.62±0.0480.62 \pm 0.048, or about (0.572, 0.668)(0.572,\ 0.668). We are 95 percent confident that between about 57 and 67 percent of all students in the population revise mainly in the week before an exam. (b) Solve 1.96p(1−p)n≤0.031.96\sqrt{\frac{p(1-p)}{n}} \le 0.03 for nn. With no prior estimate, use p=0.5p = 0.5, which gives the largest possible standard error: n≥(1.960.03)2×0.25≈1,067.1n \ge \left(\frac{1.96}{0.03}\right)^2 \times 0.25 \approx 1{,}067.1. Always round up, so survey at least 1,068 students.

Problem 12: One-sample t-test for a mean

Difficulty: Medium

A course coordinator claims that the mean score on a placement test is 50. A random sample of 36 students has xˉ=53\bar{x} = 53 and s=9s = 9. Test H0:μ=50H_0: \mu = 50 against H1:μ≠50H_1: \mu \neq 50 at the 5 percent significance level. Would the conclusion change at the 10 percent level?
Show worked solution
Test statistic: t=xˉ−μ0s/n=53−509/36=31.5=2.0t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{53 - 50}{9/\sqrt{36}} = \frac{3}{1.5} = 2.0, with df=35df = 35. At the 5 percent level, the two-sided critical value is t∗≈2.030t^* \approx 2.030. Since ∣t∣=2.0<2.030|t| = 2.0 < 2.030, we fail to reject H0H_0. Equivalently, the two-sided p-value is about 0.053, just above 0.05. Conclusion in context: at the 5 percent level, the sample does not give convincing evidence that the mean placement score differs from 50. This is not evidence that the mean equals 50; the data are simply not strong enough to rule it out. At the 10 percent level, the critical value is t∗≈1.690t^* \approx 1.690, and 2.0>1.6902.0 > 1.690 (equivalently 0.053<0.100.053 < 0.10), so we would reject H0H_0. The same data can lead to different decisions at different significance levels, which is why the level should be chosen before looking at the data and why reporting the p-value itself is more informative.

Problem 13: Interpreting p-values and Type I and Type II errors

Difficulty: Medium

(a) A study reports a p-value of 0.03. Which statement is correct? (i) There is a 3 percent probability that the null hypothesis is true. (ii) If the null hypothesis were true, a result at least this extreme would occur in about 3 percent of samples. (iii) There is a 97 percent probability that the effect is real. (b) A factory tests each batch of components with H0H_0: the batch meets specification. Describe a Type I error and a Type II error in this context, and explain how lowering the significance level affects each.
Show worked solution
(a) Statement (ii). A p-value is calculated assuming H0H_0 is true: it is the probability of data at least as extreme as those observed if the null hypothesis holds. It is not the probability that H0H_0 is true, and one minus the p-value is not the probability that the effect is real; statements of that kind need prior probabilities and a Bayesian analysis. (b) Type I error: rejecting H0H_0 when it is true, which here means rejecting a batch that actually meets specification (a false alarm that wastes good components). Type II error: failing to reject H0H_0 when it is false, which here means accepting a defective batch (a miss). The significance level α\alpha is the probability of a Type I error that the test allows. Lowering α\alpha, say from 0.05 to 0.01, makes rejection harder: fewer good batches are rejected, but more defective batches get through, so the Type II error probability β\beta rises and the power, 1−β1 - \beta, falls. The way to reduce both error probabilities at once is a larger sample.

Problem 14: One-proportion z-test

Difficulty: Hard

A lecturer believes that more than half of the students at her university prefer weekly online quizzes to a single midterm exam. In a random sample of 400 of these students, 224 prefer the quizzes. Carry out a test at the 5 percent significance level, checking the conditions and stating the conclusion in context.
Show worked solution
Hypotheses: let pp be the proportion of all students at the university who prefer weekly quizzes. H0:p=0.5H_0: p = 0.5 and H1:p>0.5H_1: p > 0.5 (one-sided, because the claim is "more than half"). Conditions: the sample is random; 400 is less than 10 percent of the student population (assumed large), so observations are close to independent; and np0=n(1−p0)=200≥10np_0 = n(1 - p_0) = 200 \ge 10, so the normal approximation is reasonable. Sample proportion: p^=224400=0.56\hat{p} = \frac{224}{400} = 0.56. The standard error under H0H_0 uses p0p_0, not p^\hat{p}: SE=0.5×0.5400=0.025SE = \sqrt{\frac{0.5 \times 0.5}{400}} = 0.025. Test statistic: z=0.56−0.50.025=2.4z = \frac{0.56 - 0.5}{0.025} = 2.4. One-sided p-value: P(Z>2.4)≈0.0082P(Z > 2.4) \approx 0.0082. Conclusion: since 0.0082<0.050.0082 < 0.05, reject H0H_0. There is convincing evidence that more than half of all students at this university prefer weekly online quizzes. The p-value measures the strength of the evidence against H0H_0, not the size of the preference; the point estimate is 56 percent.

Problem 15: Linear regression: slope, prediction, residual and R-squared

Difficulty: Hard

For a sample of students, hours studied per week (xx) have mean 10 and standard deviation 4, exam scores out of 100 (yy) have mean 77 and standard deviation 20, and the correlation is r=0.7r = 0.7. (a) Find the least-squares regression line. (b) Interpret the slope. (c) A student who studied 8 hours scored 75; find their residual. (d) Interpret R2R^2. (e) Why would a prediction for 40 hours a week be unreliable, and can we conclude that studying more causes higher scores?
Show worked solution
(a) Slope: b=rsysx=0.7×204=3.5b = r\frac{s_y}{s_x} = 0.7 \times \frac{20}{4} = 3.5. The line passes through (xˉ,yˉ)(\bar{x}, \bar{y}), so the intercept is a=yˉ−bxˉ=77−3.5×10=42a = \bar{y} - b\bar{x} = 77 - 3.5 \times 10 = 42. The regression line is y^=42+3.5x\hat{y} = 42 + 3.5x. (b) Each additional hour of study per week is associated with a predicted exam score 3.5 points higher, on average. (c) Predicted score: y^=42+3.5×8=70\hat{y} = 42 + 3.5 \times 8 = 70. Residual =y−y^=75−70=5= y - \hat{y} = 75 - 70 = 5: this student scored 5 points above what the line predicts. (d) R2=r2=0.72=0.49R^2 = r^2 = 0.7^2 = 0.49: about 49 percent of the variation in exam scores is explained by the linear relationship with hours studied; the remaining 51 percent is due to other factors. (e) The line predicts 42+3.5×40=18242 + 3.5 \times 40 = 182, which is impossible on a test out of 100. Forty hours is 40−104=7.5\frac{40 - 10}{4} = 7.5 standard deviations above the mean, far outside the data, so the prediction assumes the straight-line pattern continues where it has never been observed (extrapolation). And the data are observational: students who study more may also have more prior knowledge or motivation, so the association alone does not show that extra study causes the higher scores.

Frequently asked questions

When should I use a t-test instead of a z-test?

Use t procedures for a population mean when the population standard deviation is unknown and you estimate it with the sample standard deviation, which is almost always the case in practice. The t-distribution has n - 1 degrees of freedom and heavier tails than the normal distribution, so its critical values are larger, especially for small samples. Tests and intervals for a proportion use z, because the standard error is calculated from the proportion itself.

What is the difference between standard deviation and standard error?

The standard deviation measures how spread out individual observations are. The standard error measures how much a statistic, such as the sample mean, varies from one sample to another. For the mean, the standard error equals the standard deviation divided by the square root of the sample size, so it shrinks as the sample grows while the standard deviation does not.

What does a p-value mean?

A p-value is the probability of getting a result at least as extreme as the one observed, assuming the null hypothesis is true. A small p-value means the data would be surprising if the null hypothesis were true. It is not the probability that the null hypothesis is true, and it does not measure how large or important an effect is.

How do I interpret a 95 percent confidence interval?

Say that you are 95 percent confident the population parameter lies within the interval, and state it in the context of the problem. The 95 percent describes the method: if you repeated the sampling many times, about 95 percent of the intervals built this way would contain the true value. Avoid saying there is a 95 percent probability that the parameter is in your particular interval.

Are these problems suitable for AP Statistics or an introductory university course?

Yes. They cover the core of AP Statistics and of a typical first statistics course at university: exploring data, probability, random variables, the binomial and normal distributions, sampling distributions, confidence intervals, significance tests and linear regression. Chi-square tests and inference for a regression slope are not included here. For a full timed paper, try the free AP Statistics practice exam.

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