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
Introductory Statistics: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
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
- Convert the value to a z-score: z = (x − μ)/σ.
- Look up Φ(z), the area to the left of z.
- For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
- For a percentile, work backwards: find z for the area, then x = μ + z·σ.
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
- Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
- Find the critical value for the confidence level.
- Margin of error: E = critical value × standard deviation/√n.
- 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
- State H₀ and H₁ and decide whether the test is one- or two-sided.
- Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
- Compare with the critical value, or find the p-value.
- Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.
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
2Practice problems
1.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 31, 48, 33, 35, 55, 42.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 70?
μ = 100σ = 20x = 70
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.15, 0.3, 0.35, 0.2
σ =
4.ProbabilityHard
A game costs €3 to play. You win €12 with probability 0.3 and €30 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€12: p = 0.3€30: p = 0.02
E(G) = €
5.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 247.75 and 256 g? Give P as a decimal.
μ = 250σ = 5x1 = 247.75x2 = 256
P =
6.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points10%
x = points
7.Confidence intervalsMedium
A sample of 30 items has mean 52.5 and standard deviation s = 15. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 30x̄ = 52.5s = 1595%t(29) = 2.045
lower limit =
8.Confidence intervalsHard
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 =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 73.88 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 73.88s = 4n = 25
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 299%
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 21?
n = 6ŷ = 21
x =
3Solutions
1.Descriptive statistics
Answer: s = 9.438
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 = 244/6 = 40.67
Squared deviations
Σ(x − x̄)² = (−9.667)² + 7.333² + (−7.667)² + (−5.667)² + 14.33² + 1.333² = 445.3
Standard deviation
s = √(445.3/5) = 9.438
Answer
s = 9.438
2.Descriptive statistics
Answer: P = 0.06681
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 0.06681
Answer
P = 0.06681
3.Probability
Answer: σ = 0.9695
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 = 1·0.15 + 2·0.3 + 3·0.35 + 4·0.2 = 2.6
Mean of the squares
E(X²) = 1·0.15 + 4·0.3 + 9·0.35 + 16·0.2 = 7.7
Variance
Var(X) = 7.7 − 2.6² = 0.94
σ = √0.94 = 0.9695
Answer
σ = 0.9695
4.Probability
Answer: E(G) = 1.2 €
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: 9 (p = 0.3), 27 (p = 0.02), −3 (p = 0.68)
Expected value
E(G) = 9·0.3 + 27·0.02 − 3·0.68 = 1.2 €
Answer
E(G) = 1.2 €
A positive value means you win money on average.
5.Normal distribution
Answer: P = 0.5586
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 = (247.75 − 250)/5 = −0.45
z₂ = (256 − 250)/5 = 1.2
Area between
P = Φ(1.2) − Φ(−0.45) = 0.8849 − 0.3264 = 0.5586
Answer
P = 0.5586
6.Normal distribution
Answer: x = 77.8 points
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) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 65 + 1.28·10 = 77.8 points
Answer
x = 77.8 points
7.Confidence intervals
Answer: lower limit = 46.9
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 = 29, t = 2.045
Margin of error
E = 2.045·15/√30 = 5.6
Interval
52.5 ± 5.6 → [46.9, 58.1]
Answer
lower limit = 46.9
8.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
9.Hypothesis tests & confidence intervals
Answer: t = −1.4
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 = 4/√25 = 4/5 = 0.8
Test statistic
t = (73.88 − 75)/0.8 = −1.4
Answer
t = −1.4
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 664
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 = (2.576·20/2)² = 663.6
Round up
n = 664
Answer
n = 664
11.Linear regression & correlation
Answer: r² = 0.9982
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 = 42/6 = 7
ȳ = 166/6 = 27.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 254
Syy = Σ(y − ȳ)² = 923.3
Coefficient of determination
r² = 254²/(70·923.3) = 0.9982
Answer
r² = 0.9982
About 99.82% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 5.356
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 = 21/6 = 3.5
ȳ = 140/6 = 23.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −22
Regression line
b = −22/17.5 = −1.257
a = 23.33 − (−1.257)·3.5 = 27.73
ŷ = 27.73 − 1.257·x
Solve for x
x = (21 − 27.73)/(−1.257) = 5.356
Answer
x = 5.356
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