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
1520253035404550

Find the sample standard deviation s of the data: 43, 51, 45, 21, 13, 22.

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

A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is above 20?

μ = 50σ = 15x = 20
P =
3.ProbabilityMedium
xp(x)10.120.230.3540.35μ = 2.95

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

x: 1, 2, 3, 4p(x): 0.1, 0.2, 0.35, 0.35
σ =
4.ProbabilityHard
n = 12, p = 0.4: P(X = 3) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 3 correct answers?

n = 12p = 0.4k = 3
P =
5.Normal distributionMedium
171.35186.75P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 171.35 and 186.75 cm? Give P as a decimal.

μ = 178σ = 7x1 = 171.35x2 = 186.75
P =
6.Normal distributionHard
235240245250255260265gtop 90%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g90%
x = g
7.Linear regression & correlationMedium
xy1669187920842294249426107xy202580100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy17482046235926732989xy2025306080x = 21.5

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (21.5, 56) is made. Find its residual e = y − ŷ.

n = 5(21.5, 56)
e =
9.Confidence intervalsMedium
n = 20, s = 2.5L = ?Ux̄ = 12E99%

A sample of 20 items has mean 12 and standard deviation s = 2.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 20x̄ = 12s = 2.599%t(19) = 2.861
lower limit =
10.Confidence intervalsHard
σ = 21, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 21. 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 0.5?

σ = 21E = 0.599%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 47.12 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 47.12s = 8n = 25
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 45 of 300 patients on treatment A recover, and 4 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 45n₁ = 300x₂ = 4n₂ = 80
z =
13.Chi-square testsMedium
no defectminormajorTotalday shift12222054night shift37322998Total495449152

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “night shift, major”.

row total = 98column total = 49N = 152
E =
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 7 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 = 15.69

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 = 195/6 = 32.5

Squared deviations

Σ(x − x̄)² = 10.5² + 18.5² + 12.5² + (−11.5)² + (−19.5)² + (−10.5)² = 1232

Standard deviation

s = √(1232/5) = 15.69

Answer

s = 15.69
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) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
3.Probability

Answer: σ = 0.9734

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 = 1·0.1 + 2·0.2 + 3·0.35 + 4·0.35 = 2.95

Mean of the squares

E(X²) = 1·0.1 + 4·0.2 + 9·0.35 + 16·0.35 = 9.65

Variance

Var(X) = 9.65 − 2.95² = 0.9475
σ = √0.9475 = 0.9734

Answer

σ = 0.9734
4.Probability

Answer: P = 0.1419

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) = 220·0.43·0.69 = 220·0.064·0.01008 = 0.1419

Answer

P = 0.1419
5.Normal distribution

Answer: P = 0.7233

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 = (171.35 − 178)/7 = −0.95
z₂ = (186.75 − 178)/7 = 1.25

Area between

P = Φ(1.25) − Φ(−0.95) = 0.8944 − 0.1711 = 0.7233

Answer

P = 0.7233
6.Normal distribution

Answer: x = 243.6 g

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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 250 − 1.28·5 = 243.6 g

Answer

x = 243.6 g
7.Linear regression & correlation

Answer: r = 0.9811

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 = 126/6 = 21
ȳ = 527/6 = 87.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 245
Syy = Σ(y − ȳ)² = 890.8

Correlation coefficient

r = 245/√(70·890.8) = 0.9811

Answer

r = 0.9811
8.Linear regression & correlation

Answer: e = −1.55

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 = 115/5 = 23
ȳ = 315/5 = 63

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 327

Regression line

b = 327/90 = 3.633
a = 63 − 3.633·23 = −20.57
ŷ = −20.57 + 3.633·x

Predicted value

ŷ = −20.57 + 3.633·21.5 = 57.55

Residual

e = 56 − 57.55 = −1.55

Answer

e = −1.55

The point lies below the line.

9.Confidence intervals

Answer: lower limit = 10.4

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.

Critical value

df = n − 1 = 19, t = 2.861

Margin of error

E = 2.861·2.5/√20 = 1.599

Interval

12 ± 1.599 → [10.4, 13.6]

Answer

lower limit = 10.4
10.Confidence intervals

Answer: n = 11706

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 = (2.576·21/0.5)² = 11706

Round up

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

n = 11706

Answer

n = 11706
11.Hypothesis tests & confidence intervals

Answer: t = −1.8

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 = 8/√25 = 8/5 = 1.6

Test statistic

t = (47.12 − 50)/1.6 = −1.8

Answer

t = −1.8
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = 2.371

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.

Sample proportions

p̂₁ = 45/300 = 0.15
p̂₂ = 4/80 = 0.05
p̂ = 49/380 = 0.1289

Standard error under H₀

SE = √(0.1289·0.8711·(1/300 + 1/80)) = 0.04217

Test statistic

z = (0.15 − 0.05)/0.04217 = 2.371

Answer

z = 2.371

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

13.Chi-square tests

Answer: E = 31.59

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 count

E = 98·49/152 = 31.59

Answer

E = 31.59
14.Chi-square tests

Answer: χ²crit = 11.07

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 = 7 − 1 − 1 = 5

Table

χ²crit(df = 5, α = 0.05) = 11.070

Answer

χ²crit = 11.070

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

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