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: 35, 51, 32, 20, 49, 57.
n = 6
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 27.5.
μ = 50σ = 15x = 27.5
z =
3.ProbabilityMedium
A bag holds 6 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 9
P =
4.ProbabilityHard
X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.
x: 1, 2, 3, 4, 5p(x): 0.25, 0.15, 0.3, 0.2, 0.1Y = 2X + 4
σY =
5.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 28.3 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 28.3 min
P =
6.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 100 phones, mean 94 h, known standard deviation 6 h. Model 2: 60 phones, mean 95 h, known standard deviation 10 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
n₁ = 100x̄₁ = 94 hσ₁ = 6 hn₂ = 60x̄₂ = 95 hσ₂ = 10 h90%
E = h
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 13 plots, mean yield 50.5 kg, standard deviation 1.5 kg. Plot group 2: 8 plots, mean 57.5 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
n₁ = 13x̄₁ = 50.5s₁ = 1.5n₂ = 8x̄₂ = 57.5s₂ = 8.595%t(19) = 2.093
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 491 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 491s = 20n = 16
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 74.5. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 74.5σ = 8n = 2595%
U =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
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. For which value x does the line predict ŷ = 27?
n = 5ŷ = 27
x =
3Solutions
1.Descriptive statistics
Answer: s = 13.98
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 = (−5.667)² + 10.33² + (−8.667)² + (−20.67)² + 8.333² + 16.33² = 977.3
Standard deviation
s = √(977.3/5) = 13.98
Answer
s = 13.98
2.Descriptive statistics
Answer: z = −1.5
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 = (27.5 − 50)/15 = −1.5
Answer
z = −1.5
3.Probability
Answer: P = 0.1429
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.
First and second draw
P = 6/15 · 5/14 = 30/210 = 0.1429
Answer
P = 0.1429
4.Probability
Answer: σY = 2.598
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.
Spread of X
μ = 2.75, Var(X) = 1.688, σX = 1.299
Linear change
σY = |2|·1.299 = 2.598
Answer
σY = 2.598
The mean does move: E(Y) = 2·2.75 + 4 = 9.5.
5.Normal distribution
Answer: P = 0.9306
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 = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (28.3 − 32)/2.5 = −1.48
Upper tail
P = 1 − Φ(−1.48) = 1 − 0.06944 = 0.9306
Answer
P = 0.9306
6.Normal distribution
Answer: P = 0.9515
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 = 25.03 − 25 = 0.03 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.03)/0.01803 = −1.66
Probability
P = 1 − Φ(−1.66) = 0.9515
Answer
P = 0.9515
7.Confidence intervals
Answer: E = 2.342 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 = 94 − 95 = −1
Standard error
SE = √(6²/100 + 10²/60) = 1.424
Margin of error
E = 1.645·1.424 = 2.342
Answer
E = 2.342 h
The interval contains 0, so the data do not show a significant difference.
8.Confidence intervals
Answer: lower limit = −11.98 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 = (12·1.5² + 7·8.5²)/19 = 28.04
sp = 5.295
Margin of error
E = 2.093·5.295·√(1/13 + 1/8) = 4.98
Interval
−7 ± 4.98 → [−11.98, −2.02]
Answer
lower limit = −11.98 kg
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/√16 = 20/4 = 5
Test statistic
t = (491 − 500)/5 = −1.8
Answer
t = −1.8
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 77.64
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 = 1.96·8/√25 = 3.136
Interval limit
U = 74.5 + 3.136 = 77.64
Answer
U = 77.64
11.Linear regression & correlation
Answer: r² = 0.8719
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 = 33/6 = 5.5
ȳ = 445/6 = 74.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 99.5
Syy = Σ(y − ȳ)² = 648.8
Coefficient of determination
r² = 99.5²/(17.5·648.8) = 0.8719
Answer
r² = 0.8719
About 87.19% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 7
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 = 118/5 = 23.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 136
Regression line
b = 136/40 = 3.4
a = 23.6 − 3.4·6 = 3.2
ŷ = 3.2 + 3.4·x
Solve for x
x = (27 − 3.2)/3.4 = 7
Answer
x = 7
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