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, 57, 12, 27, 46, 53.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 174.8?
μ = 170σ = 12x = 174.8
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.35, 0.15
Var(X) =
4.ProbabilityHard
A game costs €10 to play. You win €50 with probability 0.2 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.2€200: 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 244.25 and 259.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 244.25x2 = 259.25
P =
6.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 20% of them lie (the 20th percentile).
μ = 65σ = 10P = 20%
x = points
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 146.5 h, known standard deviation 15 h. Model 2: 50 phones, mean 134.5 h, known standard deviation 16 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
n₁ = 50x̄₁ = 146.5 hσ₁ = 15 hn₂ = 50x̄₂ = 134.5 hσ₂ = 16 h95%
E = h
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 9 plots, mean yield 46.5 kg, standard deviation 9.5 kg. Plot group 2: 8 plots, mean 35 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
n₁ = 9x̄₁ = 46.5s₁ = 9.5n₂ = 8x̄₂ = 35s₂ = 399%t(15) = 2.947
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 77.91 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 77.91s = 12n = 49
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 71.3. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 71.3σ = 15n = 6490%
L =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
r =
12.Linear regression & correlationHard
The table gives 6 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, 7) is made. Find its residual e = y − ŷ.
n = 6(5, 7)
e =
3Solutions
1.Descriptive statistics
Answer: s = 18.84
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 = 214/6 = 35.67
Squared deviations
Σ(x − x̄)² = (−16.67)² + 21.33² + (−23.67)² + (−8.667)² + 10.33² + 17.33² = 1775
Standard deviation
s = √(1775/5) = 18.84
Answer
s = 18.84
2.Descriptive statistics
Answer: P = 0.6554
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
3.Probability
Answer: Var(X) = 0.85
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 + 2·0.35 + 3·0.35 + 4·0.15 = 2.5
Mean of the squares
E(X²) = 1·0.15 + 4·0.35 + 9·0.35 + 16·0.15 = 7.1
Variance
Var(X) = 7.1 − 2.5² = 0.85
Answer
Var(X) = 0.85
4.Probability
Answer: E(G) = 4 €
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: 40 (p = 0.2), 190 (p = 0.02), −10 (p = 0.78)
Expected value
E(G) = 40·0.2 + 190·0.02 − 10·0.78 = 4 €
Answer
E(G) = 4 €
A positive value means you win money on average.
5.Normal distribution
Answer: P = 0.8428
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 = (244.25 − 250)/5 = −1.15
z₂ = (259.25 − 250)/5 = 1.85
Area between
P = Φ(1.85) − Φ(−1.15) = 0.9678 − 0.1251 = 0.8428
Answer
P = 0.8428
6.Normal distribution
Answer: x = 56.58 points
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) = 0.2 → z = −0.8416
Back to x
x = 65 + −0.8416·10 = 56.58 points
Answer
x = 56.58 points
7.Confidence intervals
Answer: E = 6.079 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 = 146.5 − 134.5 = 12
Standard error
SE = √(15²/50 + 16²/50) = 3.102
Margin of error
E = 1.96·3.102 = 6.079
Answer
E = 6.079 h
The interval does not contain 0, so the means differ significantly.
8.Confidence intervals
Answer: lower limit = 1.141 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 = (8·9.5² + 7·3²)/15 = 52.33
sp = 7.234
Margin of error
E = 2.947·7.234·√(1/9 + 1/8) = 10.36
Interval
11.5 ± 10.36 → [1.141, 21.86]
Answer
lower limit = 1.141 kg
9.Hypothesis tests & confidence intervals
Answer: t = 1.697
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 = 12/√49 = 12/7 = 1.714
Test statistic
t = (77.91 − 75)/1.714 = 1.697
Answer
t = 1.697
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: L = 68.22
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 = 1.645·15/√64 = 3.084
Interval limit
L = 71.3 − 3.084 = 68.22
Answer
L = 68.22
11.Linear regression & correlation
Answer: r = 0.9929
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 = 132/6 = 22
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 232
Syy = Σ(y − ȳ)² = 780
Correlation coefficient
r = 232/√(70·780) = 0.9929
Answer
r = 0.9929
12.Linear regression & correlation
Answer: e = −3.629
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 = 84/6 = 14
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 118
Regression line
b = 118/70 = 1.686
a = 14 − 1.686·7 = 2.2
ŷ = 2.2 + 1.686·x
Predicted value
ŷ = 2.2 + 1.686·5 = 10.63
Residual
e = 7 − 10.63 = −3.629
Answer
e = −3.629
The point lies below the line.
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