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: 19, 28, 16, 45, 53.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 38?
μ = 50σ = 10x = 38
P =
3.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 6? The order does not matter.
n = 6k = 3
N =
4.ProbabilityHard
A game costs €5 to play. You win €20 with probability 0.15 and €250 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€20: p = 0.15€250: p = 0.05
E(G) = €
5.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 9 of them exceeds 226.5 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 9> 226.5 kg
P =
6.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 80% of them lie (the 80th percentile).
μ = 500σ = 40P = 80%
x = h
7.Confidence intervalsMedium
A sample of 6 items has mean 72.5 and standard deviation s = 8.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 6x̄ = 72.5s = 8.590%t(5) = 2.015
upper limit =
8.Confidence intervalsHard
The population standard deviation is 33. 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 2.5?
σ = 33E = 2.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 496.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 496.4s = 20n = 100
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 101.8. The population standard deviation is 8. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 101.8σ = 8n = 10090%
U =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5, 69) is made. Find its residual e = y − ŷ.
n = 6(5, 69)
e =
3Solutions
1.Descriptive statistics
Answer: s = 16.21
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 = 161/5 = 32.2
Squared deviations
Σ(x − x̄)² = (−13.2)² + (−4.2)² + (−16.2)² + 12.8² + 20.8² = 1051
Standard deviation
s = √(1051/4) = 16.21
Answer
s = 16.21
2.Descriptive statistics
Answer: P = 0.8849
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
3.Probability
Answer: N = 20
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(6, 3) = (6·5·4)/(3·2·1) = 20
Answer
N = 20
4.Probability
Answer: E(G) = 10.5 €
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: 15 (p = 0.15), 245 (p = 0.05), −5 (p = 0.8)
Expected value
E(G) = 15·0.15 + 245·0.05 − 5·0.8 = 10.5 €
Answer
E(G) = 10.5 €
A positive value means you win money on average.
5.Normal distribution
Answer: P = 0.1587
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 = 225 kg, σT = 0.5·√9 = 1.5 kg
Standardise
z = (226.5 − 225)/1.5 = 1
Probability
P = 1 − Φ(1) = 0.1587
Answer
P = 0.1587
6.Normal distribution
Answer: x = 533.7 h
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 for the percentile
Φ(z) = 0.8 → z = 0.8416
Back to x
x = 500 + 0.8416·40 = 533.66 h
Answer
x = 533.66 h
7.Confidence intervals
Answer: upper limit = 79.49
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 = 5, t = 2.015
Margin of error
E = 2.015·8.5/√6 = 6.992
Interval
72.5 ± 6.992 → [65.51, 79.49]
Answer
upper limit = 79.49
8.Confidence intervals
Answer: n = 670
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 = (1.96·33/2.5)² = 669.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 670
Answer
n = 670
9.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 = 20/√100 = 20/10 = 2
Test statistic
t = (496.4 − 500)/2 = −1.8
Answer
t = −1.8
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 103.1
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 = 1.645·8/√100 = 1.316
Interval limit
U = 101.8 + 1.316 = 103.1
Answer
U = 103.1
11.Linear regression & correlation
Answer: r² = 0.8287
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 = 25/5 = 5
ȳ = 233/5 = 46.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 37
Syy = Σ(y − ȳ)² = 165.2
Coefficient of determination
r² = 37²/(10·165.2) = 0.8287
Answer
r² = 0.8287
About 82.87% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 4.567
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 = 455/6 = 75.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 399
Regression line
b = 399/70 = 5.7
a = 75.83 − 5.7·7 = 35.93
ŷ = 35.93 + 5.7·x
Predicted value
ŷ = 35.93 + 5.7·5 = 64.43
Residual
e = 69 − 64.43 = 4.567
Answer
e = 4.567
The point lies above the line.
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