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: 52, 32, 58, 58, 20, 42.

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

A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is above 20?

μ = 50σ = 20x = 20
P =
3.ProbabilityMedium
red: 8blue: 3

A bag holds 8 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 3
P =
4.ProbabilityHard
n = 7, p = 0.3: P(X = 3) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 3 correct answers?

n = 7p = 0.3k = 3
P =
5.Normal distributionMedium
32 min34 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 34 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34 min
P =
6.Normal distributionHard
380420460500540580620hbottom 20%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 20% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h20%
x = h
7.Linear regression & correlationMedium
xy122221320418520xy24182022

The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy240345456560659762xy246405060ŷ = 63

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 63?

n = 6ŷ = 63
x =
9.Confidence intervalsMedium
x̄E₁ = 10.799%, n = 64x̄E₂ = ?95%, n = 64

A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 10.7. What is the margin of error E₂ of a 95% interval from the same data?

99%n₁ = 64E₁ = 10.795%
E₂ =
10.Confidence intervalsHard
p ≈ 0.4, n = ?p̂ − Ep̂ + Ep̂E = 0.0295%

A pilot study suggests that about 0.4 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.02?

p ≈ 0.4E = 0.0295%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 132.5 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 30 parts has mean 135.7 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 132.5 mmσ₁ = 18 mmn₂ = 30x̄₂ = 135.7 mmσ₂ = 5 mm
z =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 18, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 18. Find the minimum sample size n.

α = 0.01power = 80%δ = 4σ = 18zα = 2.326zβ = 0.842
n =
13.Chi-square testsMedium
coughno coughTotalsmokers271340non-smokers572380Total8436120

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 = 1202 × 2
χ² =
14.Chi-square testsHard
passfailTotalplant A402060plant B481260plant C6634100Total15466220dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.05 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 2203 × 2α = 0.05
χ² =

3Solutions

1.Descriptive statistics

Answer: s = 15.36

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 = 262/6 = 43.67

Squared deviations

Σ(x − x̄)² = 8.333² + (−11.67)² + 14.33² + 14.33² + (−23.67)² + (−1.667)² = 1179

Standard deviation

s = √(1179/5) = 15.36

Answer

s = 15.36
2.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
3.Probability

Answer: P = 0.5091

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.

First and second draw

P = 8/11 · 7/10 = 56/110 = 0.5091

Answer

P = 0.5091
4.Probability

Answer: P = 0.2269

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.

Binomial formula

P(X = 3) = 35·0.33·0.74 = 35·0.027·0.2401 = 0.2269

Answer

P = 0.2269
5.Normal distribution

Answer: P = 0.2119

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 = (34 − 32)/2.5 = 0.8

Upper tail

P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119

Answer

P = 0.2119
6.Normal distribution

Answer: x = 466 h

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left

Φ(z) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 500 − 0.84·40 = 466 h

Answer

x = 466 h
7.Linear regression & correlation

Answer: r = −0.7462

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 = 101/5 = 20.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −7
Syy = Σ(y − ȳ)² = 8.8

Correlation coefficient

r = −7/√(10·8.8) = −0.7462

Answer

r = −0.7462
8.Linear regression & correlation

Answer: x = 6.594

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 = 27/6 = 4.5
ȳ = 322/6 = 53.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 78

Regression line

b = 78/17.5 = 4.457
a = 53.67 − 4.457·4.5 = 33.61
ŷ = 33.61 + 4.457·x

Solve for x

x = (63 − 33.61)/4.457 = 6.594

Answer

x = 6.594
9.Confidence intervals

Answer: E₂ = 8.141

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 = (1.96/2.576)·√(64/64) = 0.7609

New margin

E₂ = 10.7·0.7609 = 8.141

Answer

E₂ = 8.141
10.Confidence intervals

Answer: n = 2305

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 = 1.96²·0.4·0.6/0.02² = 2305

Round up

n = 2305

Answer

n = 2305
11.Hypothesis tests & confidence intervals

Answer: z = −1.282

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.

Standard error of the difference

SE = √(18²/60 + 5²/30) = √6.233 = 2.497

Test statistic

z = (132.5 − 135.7)/2.497 = −1.282

Answer

z = −1.282

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 204

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 0.842)·18/4)² = 203.2

Round up

n = 204

Answer

n = 204
13.Chi-square tests

Answer: χ² = 0.1786

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

smokers: 40·84/120 = 28, 40·36/120 = 12
non-smokers: 80·84/120 = 56, 80·36/120 = 24

Chi-square statistic

χ² = (27 − 28)²/28 + (13 − 12)²/12 + (57 − 56)²/56 + (23 − 24)²/24
χ² = 0.1786

Answer

χ² = 0.1786

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

14.Chi-square tests

Answer: χ² = 3.937

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

plant A: 42, 18
plant B: 42, 18
plant C: 70, 30

Chi-square statistic

χ² = (40 − 42)²/42 + (20 − 18)²/18 + (48 − 42)²/42 + (12 − 18)²/18 + (66 − 70)²/70 + (34 − 30)²/30
χ² = 3.937

Decision

df = 2, χ²crit = 5.991
3.937 < 5.991

Answer

χ² = 3.937

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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