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: 26, 38, 38, 26, 21, 57, 13.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is below 50?
μ = 70σ = 20x = 50
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.35, 0.05, 0.15, 0.15, 0.3
Var(X) =
4.ProbabilityHard
6 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 6p = 0.4
P =
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 34 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34 min
P =
6.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Confidence intervalsMedium
A sample of 13 items has mean 107.5 and standard deviation s = 13.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 13x̄ = 107.5s = 13.590%t(12) = 1.782
upper limit =
8.Confidence intervalsHard
The population standard deviation is 21. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2.5?
σ = 21E = 2.599%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 81.8 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 81.8s = 20n = 25
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 113 of 250 patients on treatment A recover, and 37 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 113n₁ = 250x₂ = 37n₂ = 100
z =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. 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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 86?
n = 5ŷ = 86
x =
3Solutions
1.Descriptive statistics
Answer: s = 14.42
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 = 219/7 = 31.29
Squared deviations
Σ(x − x̄)² = (−5.286)² + 6.714² + 6.714² + (−5.286)² + (−10.29)² + 25.71² + (−18.29)² = 1247
Standard deviation
s = √(1247/6) = 14.42
Answer
s = 14.42
2.Descriptive statistics
Answer: P = 0.1587
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
3.Probability
Answer: Var(X) = 2.8
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 = 1·0.35 + 2·0.05 + 3·0.15 + 4·0.15 + 5·0.3 = 3
Mean of the squares
E(X²) = 1·0.35 + 4·0.05 + 9·0.15 + 16·0.15 + 25·0.3 = 11.8
Variance
Var(X) = 11.8 − 3² = 2.8
Answer
Var(X) = 2.8
4.Probability
Answer: P = 0.2333
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.
Zero and one failure
P(X = 0) = 0.66 = 0.04666
P(X = 1) = 6·0.4·0.65 = 0.1866
Add them
P = 0.04666 + 0.1866 = 0.2333
Answer
P = 0.2333
5.Normal distribution
Answer: P = 0.2119
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 = (34 − 32)/2.5 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
6.Normal distribution
Answer: P = 0.9868
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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.04)/0.01803 = −2.22
Probability
P = 1 − Φ(−2.22) = 0.9868
Answer
P = 0.9868
7.Confidence intervals
Answer: upper limit = 114.2
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 = 12, t = 1.782
Margin of error
E = 1.782·13.5/√13 = 6.672
Interval
107.5 ± 6.672 → [100.8, 114.2]
Answer
upper limit = 114.2
8.Confidence intervals
Answer: n = 469
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 = (2.576·21/2.5)² = 468.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 469
Answer
n = 469
9.Hypothesis tests & confidence intervals
Answer: t = 1.7
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/√25 = 20/5 = 4
Test statistic
t = (81.8 − 75)/4 = 1.7
Answer
t = 1.7
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: z = 1.4
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 = 113/250 = 0.452
p̂₂ = 37/100 = 0.37
p̂ = 150/350 = 0.4286
Standard error under H₀
SE = √(0.4286·0.5714·(1/250 + 1/100)) = 0.05855
Test statistic
z = (0.452 − 0.37)/0.05855 = 1.4
Answer
z = 1.4
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r² = 0.9742
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 = 103/6 = 17.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −46.5
Syy = Σ(y − ȳ)² = 126.8
Coefficient of determination
r² = −46.5²/(17.5·126.8) = 0.9742
Answer
r² = 0.9742
About 97.42% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 22.43
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 = 105/5 = 21
ȳ = 401/5 = 80.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 366
Regression line
b = 366/90 = 4.067
a = 80.2 − 4.067·21 = −5.2
ŷ = −5.2 + 4.067·x
Solve for x
x = (86 − (−5.2))/4.067 = 22.43
Answer
x = 22.43
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