Descriptive statistics
- Mean
- x̄ = Σx/n
- Sample standard deviation
- s = √(Σ(x − x̄)²/(n − 1))
- Median
- middle value of the sorted data
- z-score
- z = (x − μ)/σ
- Normal distribution
- 68% within ±1σ, 95% within ±2σ
AP Statistics practice problems: cram pack
Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
1Formulas and methods
Probability
- Complement
- P(not A) = 1 − P(A)
- Addition rule
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- Independent events
- P(A ∩ B) = P(A)·P(B)
- Conditional probability
- P(A|B) = P(A ∩ B)/P(B)
- Binomial
- P(X = k) = C(n,k)·pk(1 − p)n−k
Normal distribution
- z-score
- z = (x − μ)/σ
- Below x
- P(X < x) = Φ(z)
- Above x
- P(X > x) = 1 − Φ(z)
- Between
- Φ(z2) − Φ(z1)
- Percentile
- x = μ + z·σ
Method
- Convert the value to a z-score: z = (x − μ)/σ.
- Look up Φ(z), the area to the left of z.
- For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
- For a percentile, work backwards: find z for the area, then x = μ + z·σ.
Linear regression & correlation
- Sums of squares
- Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
- Slope
- b = Sxy/Sxx
- Intercept
- a = ȳ − b·x̄
- Correlation
- r = Sxy/√(SxxSyy)
- Residual
- e = y − ŷ
Confidence intervals
- σ known
- x̄ ± z·σ/√n
- σ unknown
- x̄ ± t·s/√n, df = n − 1
- Critical z
- 90%: 1.645, 95%: 1.96, 99%: 2.576
- Sample size
- n = (z·σ/E)², rounded up
Method
- Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
- Find the critical value for the confidence level.
- Margin of error: E = critical value × standard deviation/√n.
- Interval: x̄ ± E.
Hypothesis tests & confidence intervals
- z-test for a mean
- z = (x̄ − μ0)/(σ/√n)
- t-test for a mean
- t = (x̄ − μ0)/(s/√n), df = n − 1
- Two-sided at 5%
- reject H0 if |z| > 1.96
- Confidence interval
- x̄ ± z·σ/√n
Method
- State H₀ and H₁ and decide whether the test is one- or two-sided.
- Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
- Compare with the critical value, or find the p-value.
- Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.
Chi-square tests
- Test statistic
- χ² = Σ(O − E)²/E
- Goodness of fit
- E = n·p, df = k − 1 − (estimated parameters)
- Independence
- E = row total × column total / grand total, df = (r − 1)(c − 1)
- Critical values (α = 0.05)
- df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
- Decision
- reject H0 if χ² > χ²crit
2Practice problems
1.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 30, 28, 17, 17, 16.
n = 6
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 103.
μ = 100σ = 5x = 103
z =
3.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 13? The order does not matter.
n = 13k = 2
N =
4.ProbabilityHard
12 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 12p = 0.2
P =
5.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.85 and 188.5 cm? Give P as a decimal.
μ = 178σ = 7x1 = 167.85x2 = 188.5
P =
6.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).
μ = 32σ = 6P = 5%
x = min
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 16?
n = 5ŷ = 16
x =
9.Confidence intervalsMedium
The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7.7 and standard deviation 1.3. Find the lower limit of the 90% confidence interval for the mean difference.
n = 20d̄ = 7.7sd = 1.390%t(19) = 1.729
lower limit = mmHg
10.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 11 plots, mean yield 27.5 kg, standard deviation 8.5 kg. Plot group 2: 10 plots, mean 36 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
n₁ = 11x̄₁ = 27.5s₁ = 8.5n₂ = 10x̄₂ = 36s₂ = 1095%t(19) = 2.093
lower limit = kg
11.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 118.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 118.1s = 5n = 49
t =
12.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 199%
n =
13.Chi-square testsMedium
The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.
N = 1602 × 3
χ² =
14.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 10 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 10
df =
3Solutions
1.Descriptive statistics
Answer: s = 8.183
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Mean
x̄ = 143/6 = 23.83
Squared deviations
Σ(x − x̄)² = 11.17² + 6.167² + 4.167² + (−6.833)² + (−6.833)² + (−7.833)² = 334.8
Standard deviation
s = √(334.8/5) = 8.183
Answer
s = 8.183
2.Descriptive statistics
Answer: z = 0.6
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = (103 − 100)/5 = 0.6
Answer
z = 0.6
3.Probability
Answer: N = 78
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Combinations
C(13, 2) = (13·12)/(2·1) = 78
Answer
N = 78
4.Probability
Answer: P = 0.2749
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Zero and one failure
P(X = 0) = 0.812 = 0.06872
P(X = 1) = 12·0.2·0.811 = 0.2062
Add them
P = 0.06872 + 0.2062 = 0.2749
Answer
P = 0.2749
5.Normal distribution
Answer: P = 0.8597
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (167.85 − 178)/7 = −1.45
z₂ = (188.5 − 178)/7 = 1.5
Area between
P = Φ(1.5) − Φ(−1.45) = 0.9332 − 0.07353 = 0.8597
Answer
P = 0.8597
6.Normal distribution
Answer: x = 22.13 min
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z for the percentile
Φ(z) = 0.05 → z = −1.645
Back to x
x = 32 + −1.645·6 = 22.13 min
Answer
x = 22.13 min
7.Linear regression & correlation
Answer: r² = 0.9378
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 15/5 = 3
ȳ = 339/5 = 67.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 72
Syy = Σ(y − ȳ)² = 552.8
Coefficient of determination
r² = 72²/(10·552.8) = 0.9378
Answer
r² = 0.9378
About 93.78% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 6.846
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 30/5 = 6
ȳ = 69/5 = 13.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 104
Regression line
b = 104/40 = 2.6
a = 13.8 − 2.6·6 = −1.8
ŷ = −1.8 + 2.6·x
Solve for x
x = (16 − (−1.8))/2.6 = 6.846
Answer
x = 6.846
9.Confidence intervals
Answer: lower limit = 7.197 mmHg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Critical value
df = 19, t = 1.729
Margin of error
E = 1.729·1.3/√20 = 0.5026
Interval
7.7 ± 0.5026 → [7.197, 8.203]
Answer
lower limit = 7.197 mmHg
10.Confidence intervals
Answer: lower limit = −16.95 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (10·8.5² + 9·10²)/19 = 85.39
sp = 9.241
Margin of error
E = 2.093·9.241·√(1/11 + 1/10) = 8.451
Interval
−8.5 ± 8.451 → [−16.95, −0.04919]
Answer
lower limit = −16.95 kg
11.Hypothesis tests & confidence intervals
Answer: t = −2.66
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Standard error
SE = 5/√49 = 5/7 = 0.7143
Test statistic
t = (118.1 − 120)/0.7143 = −2.66
Answer
t = −2.66
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
12.Hypothesis tests & confidence intervals
Answer: n = 2655
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Solve for n
n = (2.576·20/1)² = 2654
Round up
n = 2655
Answer
n = 2655
13.Chi-square tests
Answer: χ² = 4.347
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.
Expected counts
day shift: 100·32/160 = 20, 100·48/160 = 30, 100·80/160 = 50
night shift: 60·32/160 = 12, 60·48/160 = 18, 60·80/160 = 30
Chi-square statistic
χ² = (15 − 20)²/20 + (33 − 30)²/30 + (52 − 50)²/50 + (17 − 12)²/12 + (15 − 18)²/18 + (28 − 30)²/30
χ² = 4.347
Answer
χ² = 4.347
df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
14.Chi-square tests
Answer: df = 9
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.
Degrees of freedom
df = 10 − 1 = 9
Answer
df = 9
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