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.

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1Formulas and methods

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

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

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. 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

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. 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

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. 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
2025303540455055

Find the sample standard deviation s of the data: 38, 18, 52, 58, 49, 21.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 120.

μ = 100σ = 10x = 120
z =
3.ProbabilityMedium
xp(x)00.3510.3520.3μ = 0.95

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3
Var(X) =
4.ProbabilityHard
n = 12, p = 0.3: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 12p = 0.3
P =
5.Normal distributionMedium
32 min30.8 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30.8 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30.8 min
P =
6.Normal distributionHard
μ = 32x = ?97.5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 32σ = 6P = 97.5%
x = min
7.Linear regression & correlationMedium
xy2248688121016xy51051015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy146251352459561xy2445505560x = 2.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 57) is made. Find its residual e = y − ŷ.

n = 5(2.5, 57)
e =
9.Confidence intervalsMedium
x̄E₁ = 1095%, n = 64x̄E₂ = ?99%, n = 576

A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 10. What is the margin of error E₂ of a 99% interval based on 576 observations?

95%n₁ = 64E₁ = 1099%n₂ = 576
E₂ =
10.Confidence intervalsHard
p ≈ 0.4, n = ?p̂ − Ep̂ + Ep̂E = 0.0599%

A pilot study suggests that about 0.4 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.4E = 0.0599%
n =
11.Hypothesis tests & confidence intervalsMedium
1.22p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 15. A random sample of 25 observations has a mean of 53.66. Test H₀: μ = 50 against H₁: μ > 50. Find the p-value p.

μ₀ = 50σ = 15n = 25x̄ = 53.66H₁: μ > μ₀
p =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 25 measurements has mean 99.3. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 99.3σ = 5n = 2599%
L =
13.Chi-square testsMedium
no defectminormajorTotalday shift323831101night shift22453299Total548363200

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “night shift, major”.

row total = 99column total = 63N = 200
E =
14.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.Descriptive statistics

Answer: s = 16.71

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 = 236/6 = 39.33

Squared deviations

Σ(x − x̄)² = (−1.333)² + (−21.33)² + 12.67² + 18.67² + 9.667² + (−18.33)² = 1395

Standard deviation

s = √(1395/5) = 16.71

Answer

s = 16.71
2.Descriptive statistics

Answer: z = 2

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 = (120 − 100)/10 = 2

Answer

z = 2
3.Probability

Answer: Var(X) = 0.6475

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.35 + 1·0.35 + 2·0.3 = 0.95

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55

Variance

Var(X) = 1.55 − 0.95² = 0.6475

Answer

Var(X) = 0.6475
4.Probability

Answer: P = 0.08503

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) = 0.712 = 0.01384
P(X = 1) = 12·0.3·0.711 = 0.07118

Add them

P = 0.01384 + 0.07118 = 0.08503

Answer

P = 0.08503
5.Normal distribution

Answer: P = 0.6844

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (30.8 − 32)/2.5 = −0.48

Upper tail

P = 1 − Φ(−0.48) = 1 − 0.3156 = 0.6844

Answer

P = 0.6844
6.Normal distribution

Answer: x = 43.76 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) = 0.975 → z = 1.96

Back to x

x = 32 + 1.96·6 = 43.76 min

Answer

x = 43.76 min
7.Linear regression & correlation

Answer: r = 0.9701

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 = 46/5 = 9.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 64
Syy = Σ(y − ȳ)² = 108.8

Correlation coefficient

r = 64/√(40·108.8) = 0.9701

Answer

r = 0.9701
8.Linear regression & correlation

Answer: e = 5.1

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 = 269/5 = 53.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 38

Regression line

b = 38/10 = 3.8
a = 53.8 − 3.8·3 = 42.4
ŷ = 42.4 + 3.8·x

Predicted value

ŷ = 42.4 + 3.8·2.5 = 51.9

Residual

e = 57 − 51.9 = 5.1

Answer

e = 5.1

The point lies above the line.

9.Confidence intervals

Answer: E₂ = 4.381

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Ratio of margins

E₂/E₁ = (2.576/1.96)·√(64/576) = 0.4381

New margin

E₂ = 10·0.4381 = 4.381

Answer

E₂ = 4.381
10.Confidence intervals

Answer: n = 638

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.4·0.6/0.05² = 637

Round up

n = 638

Answer

n = 638
11.Hypothesis tests & confidence intervals

Answer: p = 0.1112

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Test statistic

SE = σ/√n = 15/√25 = 3
z = (53.66 − 50)/3 = 1.22

Tail area

Φ(1.22) = 0.8888
1 − Φ(1.22) = 0.1112

p-value

p = 0.1112

Answer

p = 0.1112

p > 0.05, so H₀ is not rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: L = 96.72

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).

Margin of error

E = 2.576·5/√25 = 2.576

Interval limit

L = 99.3 − 2.576 = 96.72

Answer

L = 96.72
13.Chi-square tests

Answer: E = 31.18

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 count

E = 99·63/200 = 31.18

Answer

E = 31.18
14.Chi-square tests

Answer: df = 5

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 = 7 − 1 − 1 = 5

Answer

df = 5

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