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
10152025303540455055

Find the sample standard deviation s of the data: 58, 13, 45, 25, 44, 10, 15.

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

Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 72.5.

μ = 50σ = 15x = 72.5
z =
3.ProbabilityMedium
xp(x)10.320.2530.2540.2μ = 2.35

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4p(x): 0.3, 0.25, 0.25, 0.2
σ =
4.ProbabilityHard
€p(x)−20.7360.25380.02

A game costs €2 to play. You win €8 with probability 0.25 and €40 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€8: p = 0.25€40: p = 0.02
E(G) = €
5.Normal distributionMedium
765 kg783 kgT = X₁ + … + X₉?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 9 of them is below 783 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 9< 783 kg
P =
6.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Linear regression & correlationMedium
xy242346446550649xy2464550

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 20Σy = 233Σx² = 90Σy² = 10897Σxy = 950
r =
8.Linear regression & correlationHard
xy254136168231030xy510102030ŷ = 8

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 = 8?

n = 5ŷ = 8
x =
9.Confidence intervalsMedium
x̄E₁ = 1.495%, n = 64x̄E₂ = ?99%, n = 64

A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 1.4. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 64E₁ = 1.499%
E₂ =
10.Confidence intervalsHard
σ = 29, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 29. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?

σ = 29E = 395%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 52.88 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 52.88s = 12n = 100
t =
12.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 2.599%
n =
13.Chi-square testsMedium
yesnoTotalmen261440women344680Total6060120

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

A chi-square test of independence is carried out on a contingency table with 3 rows and 3 columns (not counting the totals). How many degrees of freedom df does the test have?

3 × 3
df =

3Solutions

1.Descriptive statistics

Answer: s = 18.9

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 = 210/7 = 30

Squared deviations

Σ(x − x̄)² = 28² + (−17)² + 15² + (−5)² + 14² + (−20)² + (−15)² = 2144

Standard deviation

s = √(2144/6) = 18.9

Answer

s = 18.9
2.Descriptive statistics

Answer: z = 1.5

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 = (72.5 − 50)/15 = 1.5

Answer

z = 1.5
3.Probability

Answer: σ = 1.108

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 = 1·0.3 + 2·0.25 + 3·0.25 + 4·0.2 = 2.35

Mean of the squares

E(X²) = 1·0.3 + 4·0.25 + 9·0.25 + 16·0.2 = 6.75

Variance

Var(X) = 6.75 − 2.35² = 1.227
σ = √1.227 = 1.108

Answer

σ = 1.108
4.Probability

Answer: E(G) = 0.8 €

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.

Net outcomes

G: 6 (p = 0.25), 38 (p = 0.02), −2 (p = 0.73)

Expected value

E(G) = 6·0.25 + 38·0.02 − 2·0.73 = 0.8 €

Answer

E(G) = 0.8 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.6915

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 total

μT = 765 kg, σT = 12·√9 = 36 kg

Standardise

z = (783 − 765)/36 = 0.5

Probability

P = Φ(0.5) = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: P = 0.7881

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 difference

μ = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Linear regression & correlation

Answer: r = 0.9091

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.

Centred sums

Sxy = 950 − 20·233/5 = 18
Sxx = 90 − 20²/5 = 10
Syy = 10897 − 233²/5 = 39.2

Correlation coefficient

r = 18/√(10·39.2) = 0.9091

Answer

r = 0.9091
8.Linear regression & correlation

Answer: x = 2.867

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 = 87/5 = 17.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 120

Regression line

b = 120/40 = 3
a = 17.4 − 3·6 = −0.6
ŷ = −0.6 + 3·x

Solve for x

x = (8 − (−0.6))/3 = 2.867

Answer

x = 2.867
9.Confidence intervals

Answer: E₂ = 1.84

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)·√(64/64) = 1.314

New margin

E₂ = 1.4·1.314 = 1.84

Answer

E₂ = 1.84
10.Confidence intervals

Answer: n = 359

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 = (1.96·29/3)² = 359

Round up

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

n = 359

Answer

n = 359
11.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 12/√100 = 12/10 = 1.2

Test statistic

t = (52.88 − 50)/1.2 = 2.4

Answer

t = 2.4
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 107

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 = (2.576·10/2.5)² = 106.2

Round up

n = 107

Answer

n = 107
13.Chi-square tests

Answer: χ² = 5.4

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

men: 40·60/120 = 20, 40·60/120 = 20
women: 80·60/120 = 40, 80·60/120 = 40

Chi-square statistic

χ² = (26 − 20)²/20 + (14 − 20)²/20 + (34 − 40)²/40 + (46 − 40)²/40
χ² = 5.4

Answer

χ² = 5.4

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

14.Chi-square tests

Answer: df = 4

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 = (3 − 1)·(3 − 1) = 4

Answer

df = 4

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