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: 55, 34, 42, 40, 19, 46, 30.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 545?

μ = 500σ = 75x = 545
P =
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 = 10, p = 0.2: P(X = 3) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 3 correct answers?

n = 10p = 0.2k = 3
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9850

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9850. Find z to two decimals.

A = 0.9850
z =
6.Normal distributionHard
μ = 32x = ?99%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 99% of them lie (the 99th percentile).

μ = 32σ = 6P = 99%
x = min
7.Linear regression & correlationMedium
xy26496108141016xy5101015

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
xy3355457569641174xy5104060ŷ = 60

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 60?

n = 5ŷ = 60
x =
9.Confidence intervalsMedium
x̄E₁ = 1195%, n = 50x̄E₂ = ?99%, n = 50

A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 11. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 50E₁ = 1199%
E₂ =
10.Confidence intervalsHard
σ = 14, n = ?x̄ − Ex̄ + Ex̄E = 490%

The population standard deviation is 14. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 4?

σ = 14E = 490%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.2622.262t = ?df = 9

10 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 6.9 and the standard deviation of the differences is 5.9. Find the paired test statistic t for H₀: μd = 0.

n = 10d̄ = 6.9sd = 5.9
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.01βσ = 7, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 7. Find the minimum sample size n.

α = 0.01power = 80%δ = 2σ = 7zα = 2.326zβ = 0.842
n =
13.Chi-square testsMedium
busbikecarTotalstudents45221380staff25132260Total703535140

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 = 1402 × 3
χ² =
14.Chi-square testsHard
yesnoTotalage 18–303268100age 31–5093140age 51+255580Total66154220dfα = 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 = 0.01 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 = 0.01
χ² =

3Solutions

1.Descriptive statistics

Answer: s = 11.65

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 = 266/7 = 38

Squared deviations

Σ(x − x̄)² = 17² + (−4)² + 4² + 2² + (−19)² + 8² + (−8)² = 814

Standard deviation

s = √(814/6) = 11.65

Answer

s = 11.65
2.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
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 = 0.2013

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) = 120·0.23·0.87 = 120·0.008·0.2097 = 0.2013

Answer

P = 0.2013
5.Normal distribution

Answer: z = −2.17

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 of z

Φ(z) = 1 − 0.9850 = 0.0150

Read the z-table backwards

Find 0.0150 inside the table and read off z.

z = −2.17

Answer

z = −2.17
6.Normal distribution

Answer: x = 45.96 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) = 0.99 → z = 2.326

Back to x

x = 32 + 2.326·6 = 45.96 min

Answer

x = 45.96 min
7.Linear regression & correlation

Answer: r = 0.9882

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 = 55/5 = 11

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 50
Syy = Σ(y − ȳ)² = 64

Correlation coefficient

r = 50/√(40·64) = 0.9882

Answer

r = 0.9882
8.Linear regression & correlation

Answer: x = 8.072

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 = 35/5 = 7
ȳ = 274/5 = 54.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 194

Regression line

b = 194/40 = 4.85
a = 54.8 − 4.85·7 = 20.85
ŷ = 20.85 + 4.85·x

Solve for x

x = (60 − 20.85)/4.85 = 8.072

Answer

x = 8.072
9.Confidence intervals

Answer: E₂ = 14.46

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)·√(50/50) = 1.314

New margin

E₂ = 11·1.314 = 14.46

Answer

E₂ = 14.46
10.Confidence intervals

Answer: n = 34

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.

Solve for n

n = (1.645·14/4)² = 33.15

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 34

Answer

n = 34
11.Hypothesis tests & confidence intervals

Answer: t = 3.698

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 of the mean difference

SE = 5.9/√10 = 1.866

Test statistic

t = 6.9/1.866 = 3.698

Answer

t = 3.698

With df = 9, the two-sided 5% critical value is 2.262; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: n = 123

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)·7/2)² = 122.9

Round up

n = 123

Answer

n = 123
13.Chi-square tests

Answer: χ² = 7.642

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

students: 80·70/140 = 40, 80·35/140 = 20, 80·35/140 = 20
staff: 60·70/140 = 30, 60·35/140 = 15, 60·35/140 = 15

Chi-square statistic

χ² = (45 − 40)²/40 + (22 − 20)²/20 + (13 − 20)²/20 + (25 − 30)²/30 + (13 − 15)²/15 + (22 − 15)²/15
χ² = 7.642

Answer

χ² = 7.642

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

14.Chi-square tests

Answer: χ² = 1.321

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

age 18–30: 30, 70
age 31–50: 12, 28
age 51+: 24, 56

Chi-square statistic

χ² = (32 − 30)²/30 + (68 − 70)²/70 + (9 − 12)²/12 + (31 − 28)²/28 + (25 − 24)²/24 + (55 − 56)²/56
χ² = 1.321

Decision

df = 2, χ²crit = 9.210
1.321 < 9.210

Answer

χ² = 1.321

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

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