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: 49, 46, 29, 13, 40.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is below 75?
μ = 50σ = 20x = 75
P =
3.ProbabilityMedium
A bag holds 5 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 4
P =
4.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.05, 0.2, 0.25, 0.25Y = 0.5X + 1
σY =
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 260.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 244.25x2 = 260.25
P =
6.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
7.Confidence intervalsMedium
A sample of 6 items has mean 104.5 and standard deviation s = 11.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 6x̄ = 104.5s = 11.599%t(5) = 4.032
upper limit =
8.Confidence intervalsHard
The population standard deviation is 31. 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 = 31E = 2.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 121.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 121.4s = 8n = 100
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 1.590%
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 48Σy = 456Σx² = 454Σy² = 35604Σxy = 3902
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 27) is made. Find its residual e = y − ŷ.
n = 6(2.5, 27)
e =
3Solutions
1.Descriptive statistics
Answer: s = 14.67
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 = 177/5 = 35.4
Squared deviations
Σ(x − x̄)² = 13.6² + 10.6² + (−6.4)² + (−22.4)² + 4.6² = 861.2
Standard deviation
s = √(861.2/4) = 14.67
Answer
s = 14.67
2.Descriptive statistics
Answer: P = 0.8944
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
3.Probability
Answer: P = 0.2778
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 = 5/9 · 4/8 = 20/72 = 0.2778
Answer
P = 0.2778
4.Probability
Answer: σY = 0.7517
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 = 2.2, Var(X) = 2.26, σX = 1.503
Linear change
σY = |0.5|·1.503 = 0.7517
Answer
σY = 0.7517
The mean does move: E(Y) = 0.5·2.2 + 1 = 2.1.
5.Normal distribution
Answer: P = 0.8547
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 = (260.25 − 250)/5 = 2.05
Area between
P = Φ(2.05) − Φ(−1.15) = 0.9798 − 0.1251 = 0.8547
Answer
P = 0.8547
6.Normal distribution
Answer: P = 0.7389
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 = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
7.Confidence intervals
Answer: upper limit = 123.4
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 = 4.032
Margin of error
E = 4.032·11.5/√6 = 18.93
Interval
104.5 ± 18.93 → [85.57, 123.4]
Answer
upper limit = 123.4
8.Confidence intervals
Answer: n = 591
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 = (1.96·31/2.5)² = 590.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 591
Answer
n = 591
9.Hypothesis tests & confidence intervals
Answer: t = 1.75
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 = 8/√100 = 8/10 = 0.8
Test statistic
t = (121.4 − 120)/0.8 = 1.75
Answer
t = 1.75
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: n = 482
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).
Solve for n
n = (1.645·20/1.5)² = 481.1
Round up
n = 482
Answer
n = 482
11.Linear regression & correlation
Answer: r = 0.986
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 = 3902 − 48·456/6 = 254
Sxx = 454 − 48²/6 = 70
Syy = 35604 − 456²/6 = 948
Correlation coefficient
r = 254/√(70·948) = 0.986
Answer
r = 0.986
12.Linear regression & correlation
Answer: e = 3.762
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 = 124/6 = 20.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −45
Regression line
b = −45/17.5 = −2.571
a = 20.67 − (−2.571)·3.5 = 29.67
ŷ = 29.67 − 2.571·x
Predicted value
ŷ = 29.67 + (−2.571)·2.5 = 23.24
Residual
e = 27 − 23.24 = 3.762
Answer
e = 3.762
The point lies above the line.
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