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
152025303540455055

Find the sample standard deviation s of the data: 13, 24, 48, 55, 15.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 72.

μ = 60σ = 20x = 72
z =
3.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
σ =
4.ProbabilityHard
n = 6, p = 0.3: P(X = 4) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?

n = 6p = 0.3k = 4
P =
5.Normal distributionMedium
105 s108.2 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 108.2 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 108.2 s
P =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Linear regression & correlationMedium
xy357458565666780877xy468607080

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 33Σy = 403Σx² = 199Σy² = 27523Σxy = 2300
r =
8.Linear regression & correlationHard
xy122221317415510xy24101520x = 3.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.

n = 5x = 3.5
ŷ =
9.Confidence intervalsMedium
x̄E₁ = 6.990%, n = 36x̄E₂ = ?99%, n = 72

A 90% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 6.9. What is the margin of error E₂ of a 99% interval based on 72 observations?

90%n₁ = 36E₁ = 6.999%n₂ = 72
E₂ =
10.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 26. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1?

σ = 26E = 199%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 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 8.3 and the standard deviation of the differences is 11.5. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = 8.3sd = 11.5
t =
12.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 399%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 399%
n =
13.Chi-square testsMedium
busbikecarTotalstudents732324120staff27272680Total1005050200

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 = 2002 × 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 checks whether counts in 4 categories follow a Poisson distribution whose mean is estimated from the same 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 = 19.33

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 = 155/5 = 31

Squared deviations

Σ(x − x̄)² = (−18)² + (−7)² + 17² + 24² + (−16)² = 1494

Standard deviation

s = √(1494/4) = 19.33

Answer

s = 19.33
2.Descriptive statistics

Answer: z = 0.6

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 − 60)/20 = 0.6

Answer

z = 0.6
3.Probability

Answer: σ = 0.8367

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 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7
σ = √0.7 = 0.8367

Answer

σ = 0.8367
4.Probability

Answer: P = 0.05953

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 = 4) = 15·0.34·0.72 = 15·0.0081·0.49 = 0.05953

Answer

P = 0.05953
5.Normal distribution

Answer: P = 0.3085

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (108.2 − 105)/6.403 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
6.Normal distribution

Answer: P = 0.8133

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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
7.Linear regression & correlation

Answer: r = 0.9359

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 = 2300 − 33·403/6 = 83.5
Sxx = 199 − 33²/6 = 17.5
Syy = 27523 − 403²/6 = 454.8

Correlation coefficient

r = 83.5/√(17.5·454.8) = 0.9359

Answer

r = 0.9359
8.Linear regression & correlation

Answer: ŷ = 15.5

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 = 85/5 = 17

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 17 − (−3)·3 = 26
ŷ = 26 − 3·x

Prediction

ŷ = 26 + (−3)·3.5 = 15.5

Answer

ŷ = 15.5
9.Confidence intervals

Answer: E₂ = 7.64

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 = (2.576/1.645)·√(36/72) = 1.107

New margin

E₂ = 6.9·1.107 = 7.64

Answer

E₂ = 7.64
10.Confidence intervals

Answer: n = 4486

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 = (2.576·26/1)² = 4486

Round up

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

n = 4486

Answer

n = 4486
11.Hypothesis tests & confidence intervals

Answer: t = 2.5

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 = 11.5/√12 = 3.32

Test statistic

t = 8.3/3.32 = 2.5

Answer

t = 2.5

With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.

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 = (2.576·12/3)² = 106.2

Round up

n = 107

Answer

n = 107
13.Chi-square tests

Answer: χ² = 14.13

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: 120·100/200 = 60, 120·50/200 = 30, 120·50/200 = 30
staff: 80·100/200 = 40, 80·50/200 = 20, 80·50/200 = 20

Chi-square statistic

χ² = (73 − 60)²/60 + (23 − 30)²/30 + (24 − 30)²/30 + (27 − 40)²/40 + (27 − 20)²/20 + (26 − 20)²/20
χ² = 14.13

Answer

χ² = 14.13

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is 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 = 4 − 1 − 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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