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: 39, 51, 33, 54, 14, 36, 45.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 52.5?

μ = 60σ = 15x = 52.5
P =
3.ProbabilityMedium
n = 7k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.

n = 7k = 3
N =
4.ProbabilityHard
n = 9, p = 0.1: P(X = 2) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 9p = 0.1k = 2
P =
5.Normal distributionMedium
60 kg59.1 kgT = X₁ + … + X₅?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 5 of them is below 59.1 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 5< 59.1 kg
P =
6.Normal distributionHard
157164171178185192199cmtop 60%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm60%
x = cm
7.Linear regression & correlationMedium
xy294156228321036xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
8.Linear regression & correlationHard
xy239347447558664766xy246405060x = 5.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.

n = 6x = 5.5
ŷ =
9.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 100 phones, mean 95 h, known standard deviation 4 h. Model 2: 60 phones, mean 87 h, known standard deviation 18 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 95 hσ₁ = 4 hn₂ = 60x̄₂ = 87 hσ₂ = 18 h99%
lower limit = h
10.Confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 8. 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 0.5?

σ = 8E = 0.590%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.88p = ?

The mean of a population is claimed to be 80; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 72.48. Test H₀: μ = 80 against H₁: μ < 80. Find the p-value p.

μ₀ = 80σ = 40n = 100x̄ = 72.48H₁: μ < μ₀
p =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 194.2. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 194.2σ = 10n = 2590%
U =
13.Chi-square testsMedium
coughno coughTotalsmokers182240non-smokers471360Total6535100

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 × 2
χ² =
14.Chi-square testsHard
passfailTotalline 1322860line 2501060line 3301040Total11248160dfα = 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 = 1603 × 2α = 0.01
χ² =

3Solutions

1.Descriptive statistics

Answer: s = 13.38

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 = 272/7 = 38.86

Squared deviations

Σ(x − x̄)² = 0.1429² + 12.14² + (−5.857)² + 15.14² + (−24.86)² + (−2.857)² + 6.143² = 1075

Standard deviation

s = √(1075/6) = 13.38

Answer

s = 13.38
2.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
3.Probability

Answer: N = 35

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(7, 3) = (7·6·5)/(3·2·1) = 35

Answer

N = 35
4.Probability

Answer: P = 0.1722

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 = 2) = 36·0.12·0.97 = 36·0.01·0.4783 = 0.1722

Answer

P = 0.1722
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 total

μT = 60 kg, σT = 0.8·√5 = 1.789 kg

Standardise

z = (59.1 − 60)/1.789 = −0.5

Probability

P = Φ(−0.5) = 0.3085

Answer

P = 0.3085
6.Normal distribution

Answer: x = 176.3 cm

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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 178 − 0.25·7 = 176.3 cm

Answer

x = 176.3 cm
7.Linear regression & correlation

Answer: r² = 0.9869

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 = 114/5 = 22.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 142
Syy = Σ(y − ȳ)² = 510.8

Coefficient of determination

r² = 142²/(40·510.8) = 0.9869

Answer

r² = 0.9869

About 98.69% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: ŷ = 59.13

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 = 27/6 = 4.5
ȳ = 321/6 = 53.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 98.5

Regression line

b = 98.5/17.5 = 5.629
a = 53.5 − 5.629·4.5 = 28.17
ŷ = 28.17 + 5.629·x

Prediction

ŷ = 28.17 + 5.629·5.5 = 59.13

Answer

ŷ = 59.13
9.Confidence intervals

Answer: lower limit = 1.926 h

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.

Point estimate

x̄₁ − x̄₂ = 95 − 87 = 8

Standard error

SE = √(4²/100 + 18²/60) = 2.358

Margin of error

E = 2.576·2.358 = 6.074

Interval

8 ± 6.074 → [1.926, 14.07]

Answer

lower limit = 1.926 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 693

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 = (1.645·8/0.5)² = 692.7

Round up

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

n = 693

Answer

n = 693
11.Hypothesis tests & confidence intervals

Answer: p = 0.03005

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.

Test statistic

SE = σ/√n = 40/√100 = 4
z = (72.48 − 80)/4 = −1.88

Tail area

Φ(1.88) = 0.9699
1 − Φ(1.88) = 0.0301

p-value

p = 0.03005

Answer

p = 0.03005

p ≤ 0.05, so H₀ is rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: U = 197.5

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).

Margin of error

E = 1.645·10/√25 = 3.29

Interval limit

U = 194.2 + 3.29 = 197.5

Answer

U = 197.5
13.Chi-square tests

Answer: χ² = 11.72

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

smokers: 40·65/100 = 26, 40·35/100 = 14
non-smokers: 60·65/100 = 39, 60·35/100 = 21

Chi-square statistic

χ² = (18 − 26)²/26 + (22 − 14)²/14 + (47 − 39)²/39 + (13 − 21)²/21
χ² = 11.72

Answer

χ² = 11.72

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

14.Chi-square tests

Answer: χ² = 13.49

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

line 1: 42, 18
line 2: 42, 18
line 3: 28, 12

Chi-square statistic

χ² = (32 − 42)²/42 + (28 − 18)²/18 + (50 − 42)²/42 + (10 − 18)²/18 + (30 − 28)²/28 + (10 − 12)²/12
χ² = 13.49

Decision

df = 2, χ²crit = 9.210
13.49 > 9.210

Answer

χ² = 13.49

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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