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
25303540455055

Find the sample standard deviation s of the data: 35, 32, 53, 22, 47, 58, 43.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 100?

μ = 70σ = 15x = 100
P =
3.ProbabilityMedium
n = 13k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 13? The order does not matter.

n = 13k = 3
N =
4.ProbabilityHard
€p(x)−30.75120.21470.05

A game costs €3 to play. You win €15 with probability 0.2 and €150 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.2€150: p = 0.05
E(G) = €
5.Normal distributionMedium
27.233.2P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 27.2 and 33.2 min? Give P as a decimal.

μ = 32σ = 6x1 = 27.2x2 = 33.2
P =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Linear regression & correlationMedium
xy17621966216923742580xy20257080

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy364457565673768875xy4686070ŷ = 64

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 = 64?

n = 6ŷ = 64
x =
9.Confidence intervalsMedium
n = 9, sd​ = 2.20L = ?Ud̄ = 10E95%

The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 10 and standard deviation 2.2. Find the lower limit of the 95% confidence interval for the mean difference.

n = 9d̄ = 10sd = 2.295%t(8) = 2.306
lower limit = mmHg
10.Confidence intervalsHard
μ₁ − μ₂, t(12) = 3.0550LU = ?x̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 8 plots, mean yield 89.5 kg, standard deviation 1.5 kg. Plot group 2: 6 plots, mean 101 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 8x̄₁ = 89.5s₁ = 1.5n₂ = 6x̄₂ = 101s₂ = 599%t(12) = 3.055
upper limit = kg
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 104 mm, and the line's known standard deviation is 8 mm. Line 2: a sample of 40 parts has mean 105.5 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 104 mmσ₁ = 8 mmn₂ = 40x̄₂ = 105.5 mmσ₂ = 10 mm
z =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 43.5. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 43.5σ = 10n = 1690%
U =
13.Chi-square testsMedium
busbikecarTotalstudents11101940staff9203160Total203050100

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 = 1002 × 3
χ² =
14.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test compares observed counts in 3 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Descriptive statistics

Answer: s = 12.58

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 = 290/7 = 41.43

Squared deviations

Σ(x − x̄)² = (−6.429)² + (−9.429)² + 11.57² + (−19.43)² + 5.571² + 16.57² + 1.571² = 949.7

Standard deviation

s = √(949.7/6) = 12.58

Answer

s = 12.58
2.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
3.Probability

Answer: N = 286

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, 3) = (13·12·11)/(3·2·1) = 286

Answer

N = 286
4.Probability

Answer: E(G) = 7.5 €

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: 12 (p = 0.2), 147 (p = 0.05), −3 (p = 0.75)

Expected value

E(G) = 12·0.2 + 147·0.05 − 3·0.75 = 7.5 €

Answer

E(G) = 7.5 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.3674

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 = (27.2 − 32)/6 = −0.8
z₂ = (33.2 − 32)/6 = 0.2

Area between

P = Φ(0.2) − Φ(−0.8) = 0.5793 − 0.2119 = 0.3674

Answer

P = 0.3674
6.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
7.Linear regression & correlation

Answer: r = 0.9918

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 = 105/5 = 21
ȳ = 351/5 = 70.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 88
Syy = Σ(y − ȳ)² = 196.8

Correlation coefficient

r = 88/√(40·196.8) = 0.9918

Answer

r = 0.9918
8.Linear regression & correlation

Answer: x = 4.406

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 = 33/6 = 5.5
ȳ = 402/6 = 67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 48

Regression line

b = 48/17.5 = 2.743
a = 67 − 2.743·5.5 = 51.91
ŷ = 51.91 + 2.743·x

Solve for x

x = (64 − 51.91)/2.743 = 4.406

Answer

x = 4.406
9.Confidence intervals

Answer: lower limit = 8.309 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 = 8, t = 2.306

Margin of error

E = 2.306·2.2/√9 = 1.691

Interval

10 ± 1.691 → [8.309, 11.69]

Answer

lower limit = 8.309 mmHg
10.Confidence intervals

Answer: upper limit = −5.849 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 = (7·1.5² + 5·5²)/12 = 11.73
sp = 3.425

Margin of error

E = 3.055·3.425·√(1/8 + 1/6) = 5.651

Interval

−11.5 ± 5.651 → [−17.15, −5.849]

Answer

upper limit = −5.849 kg
11.Hypothesis tests & confidence intervals

Answer: z = −0.6969

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 = √(8²/30 + 10²/40) = √4.633 = 2.153

Test statistic

z = (104 − 105.5)/2.153 = −0.6969

Answer

z = −0.6969

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

12.Hypothesis tests & confidence intervals

Answer: U = 47.61

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 = 1.645·10/√16 = 4.112

Interval limit

U = 43.5 + 4.112 = 47.61

Answer

U = 47.61
13.Chi-square tests

Answer: χ² = 2.514

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: 40·20/100 = 8, 40·30/100 = 12, 40·50/100 = 20
staff: 60·20/100 = 12, 60·30/100 = 18, 60·50/100 = 30

Chi-square statistic

χ² = (11 − 8)²/8 + (10 − 12)²/12 + (19 − 20)²/20 + (9 − 12)²/12 + (20 − 18)²/18 + (31 − 30)²/30
χ² = 2.514

Answer

χ² = 2.514

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

14.Chi-square tests

Answer: χ²crit = 5.991

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 = 2

Table

χ²crit(df = 2, α = 0.05) = 5.991

Answer

χ²crit = 5.991

Reject H₀ if the computed χ² is larger than this value.

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