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: 12, 12, 39, 26, 27.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is above 50?
μ = 60σ = 5x = 50
P =
3.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 6? The order does not matter.
n = 6k = 2
N =
4.ProbabilityHard
A game costs €1 to play. You win €4 with probability 0.3 and €10 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€4: p = 0.3€10: p = 0.01
E(G) = €
5.Normal distributionMedium
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
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 30 phones, mean 63.5 h, known standard deviation 16 h. Model 2: 60 phones, mean 59 h, known standard deviation 12 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
n₁ = 30x̄₁ = 63.5 hσ₁ = 16 hn₂ = 60x̄₂ = 59 hσ₂ = 12 h99%
upper limit = h
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 8 plots, mean yield 73.5 kg, standard deviation 8.5 kg. Plot group 2: 13 plots, mean 72.5 kg, standard deviation 8 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
n₁ = 8x̄₁ = 73.5s₁ = 8.5n₂ = 13x̄₂ = 72.5s₂ = 890%t(19) = 1.729
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 5 students, mean score 87.5, standard deviation 11. Group 2: 11 students, mean score 93.1, standard deviation 10. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 5x̄₁ = 87.5s₁ = 11n₂ = 11x̄₂ = 93.1s₂ = 10
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 165 of 300 patients on treatment A recover, and 96 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 165n₁ = 300x₂ = 96n₂ = 150
z =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 119Σx² = 91Σy² = 2491Σxy = 373
r =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (5, 20) is made. Find its residual e = y − ŷ.
n = 5(5, 20)
e =
3Solutions
1.Descriptive statistics
Answer: s = 11.43
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 = 116/5 = 23.2
Squared deviations
Σ(x − x̄)² = (−11.2)² + (−11.2)² + 15.8² + 2.8² + 3.8² = 522.8
Standard deviation
s = √(522.8/4) = 11.43
Answer
s = 11.43
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: N = 15
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, 2) = (6·5)/(2·1) = 15
Answer
N = 15
4.Probability
Answer: E(G) = 0.3 €
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: 3 (p = 0.3), 9 (p = 0.01), −1 (p = 0.69)
Expected value
E(G) = 3·0.3 + 9·0.01 − 1·0.69 = 0.3 €
Answer
E(G) = 0.3 €
A positive value means you win money on average.
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 = 0.7881
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 = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
7.Confidence intervals
Answer: upper limit = 13.02 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 = 63.5 − 59 = 4.5
Standard error
SE = √(16²/30 + 12²/60) = 3.307
Margin of error
E = 2.576·3.307 = 8.518
Interval
4.5 ± 8.518 → [−4.018, 13.02]
Answer
upper limit = 13.02 h
The interval contains 0, so the data do not show a significant difference.
8.Confidence intervals
Answer: lower limit = −5.361 kg
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.
Pooled standard deviation
sp² = (7·8.5² + 12·8²)/19 = 67.04
sp = 8.188
Margin of error
E = 1.729·8.188·√(1/8 + 1/13) = 6.361
Interval
1 ± 6.361 → [−5.361, 7.361]
Answer
lower limit = −5.361 kg
9.Hypothesis tests & confidence intervals
Answer: t = −1.008
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.
Pooled variance
sp² = (4·11² + 10·10²)/14 = 106
sp = 10.3
Standard error
SE = 10.3·√(1/5 + 1/11) = 5.553
Test statistic
t = (87.5 − 93.1)/5.553 = −1.008
Answer
t = −1.008
With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.
10.Hypothesis tests & confidence intervals
Answer: z = −1.823
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 = 165/300 = 0.55
p̂₂ = 96/150 = 0.64
p̂ = 261/450 = 0.58
Standard error under H₀
SE = √(0.58·0.42·(1/300 + 1/150)) = 0.04936
Test statistic
z = (0.55 − 0.64)/0.04936 = −1.823
Answer
z = −1.823
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r = −0.9091
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 = 373 − 21·119/6 = −43.5
Sxx = 91 − 21²/6 = 17.5
Syy = 2491 − 119²/6 = 130.8
Correlation coefficient
r = −43.5/√(17.5·130.8) = −0.9091
Answer
r = −0.9091
12.Linear regression & correlation
Answer: e = 3.35
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 = 99/5 = 19.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 126
Regression line
b = 126/40 = 3.15
a = 19.8 − 3.15·6 = 0.9
ŷ = 0.9 + 3.15·x
Predicted value
ŷ = 0.9 + 3.15·5 = 16.65
Residual
e = 20 − 16.65 = 3.35
Answer
e = 3.35
The point lies above the line.
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