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: 55, 18, 53, 48, 14.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is below 590?
μ = 500σ = 60x = 590
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.1, 0.15, 0.25, 0.2, 0.3
σ =
4.ProbabilityHard
A game costs €10 to play. You win €20 with probability 0.2 and €500 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€20: p = 0.2€500: p = 0.05
E(G) = €
5.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 176.25 and 181.85 cm? Give P as a decimal.
μ = 178σ = 7x1 = 176.25x2 = 181.85
P =
6.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 60% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g60%
x = g
7.Confidence intervalsMedium
A sample of 21 items has mean 60.5 and standard deviation s = 1.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 21x̄ = 60.5s = 1.599%t(20) = 2.845
upper limit =
8.Confidence intervalsHard
The population standard deviation is 15. 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 5?
σ = 15E = 599%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 495.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 495.5s = 15n = 36
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 99.8. The population standard deviation is 20. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 99.8σ = 20n = 6490%
L =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. For which value x does the line predict ŷ = 21?
n = 6ŷ = 21
x =
3Solutions
1.Descriptive statistics
Answer: s = 19.93
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 = 188/5 = 37.6
Squared deviations
Σ(x − x̄)² = 17.4² + (−19.6)² + 15.4² + 10.4² + (−23.6)² = 1589
Standard deviation
s = √(1589/4) = 19.93
Answer
s = 19.93
2.Descriptive statistics
Answer: P = 0.9332
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
3.Probability
Answer: σ = 1.322
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 = 0·0.1 + 1·0.15 + 2·0.25 + 3·0.2 + 4·0.3 = 2.45
Mean of the squares
E(X²) = 0·0.1 + 1·0.15 + 4·0.25 + 9·0.2 + 16·0.3 = 7.75
Variance
Var(X) = 7.75 − 2.45² = 1.747
σ = √1.747 = 1.322
Answer
σ = 1.322
4.Probability
Answer: E(G) = 19 €
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: 10 (p = 0.2), 490 (p = 0.05), −10 (p = 0.75)
Expected value
E(G) = 10·0.2 + 490·0.05 − 10·0.75 = 19 €
Answer
E(G) = 19 €
A positive value means you win money on average.
5.Normal distribution
Answer: P = 0.3075
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 = (176.25 − 178)/7 = −0.25
z₂ = (181.85 − 178)/7 = 0.55
Area between
P = Φ(0.55) − Φ(−0.25) = 0.7088 − 0.4013 = 0.3075
Answer
P = 0.3075
6.Normal distribution
Answer: x = 251.3 g
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 250 + 0.25·5 = 251.3 g
Answer
x = 251.3 g
7.Confidence intervals
Answer: upper limit = 61.43
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 = 20, t = 2.845
Margin of error
E = 2.845·1.5/√21 = 0.9312
Interval
60.5 ± 0.9312 → [59.57, 61.43]
Answer
upper limit = 61.43
8.Confidence intervals
Answer: n = 60
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 = (2.576·15/5)² = 59.72
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 60
Answer
n = 60
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 = 15/√36 = 15/6 = 2.5
Test statistic
t = (495.5 − 500)/2.5 = −1.8
Answer
t = −1.8
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: L = 95.69
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 = 1.645·20/√64 = 4.112
Interval limit
L = 99.8 − 4.112 = 95.69
Answer
L = 95.69
11.Linear regression & correlation
Answer: r² = 0.9046
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 = 25/5 = 5
ȳ = 222/5 = 44.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 64
Syy = Σ(y − ȳ)² = 113.2
Coefficient of determination
r² = 64²/(40·113.2) = 0.9046
Answer
r² = 0.9046
About 90.46% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 4.956
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 = 174/6 = 29
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 274
Regression line
b = 274/70 = 3.914
a = 29 − 3.914·7 = 1.6
ŷ = 1.6 + 3.914·x
Solve for x
x = (21 − 1.6)/3.914 = 4.956
Answer
x = 4.956
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