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: 42, 49, 57, 30, 33, 38.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 55?
μ = 50σ = 5x = 55
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.15, 0.35, 0.05, 0.15
σ =
4.ProbabilityHard
X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.
x: 10, 20, 30, 40p(x): 0.2, 0.35, 0.2, 0.25Y = 3X + 10
σY =
5.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 4 of them is below 316 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 4< 316 kg
P =
6.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 25% of them lie (the 25th percentile).
μ = 178σ = 7P = 25%
x = cm
7.Confidence intervalsMedium
A sample of 15 items has mean 74.5 and standard deviation s = 14.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 15x̄ = 74.5s = 14.590%t(14) = 1.761
lower limit =
8.Confidence intervalsHard
The population standard deviation is 9. 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 = 9E = 2.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 49.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 49.5s = 5n = 81
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 299%
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (11, 91) is made. Find its residual e = y − ŷ.
n = 6(11, 91)
e =
3Solutions
1.Descriptive statistics
Answer: s = 10.13
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 = 249/6 = 41.5
Squared deviations
Σ(x − x̄)² = 0.5² + 7.5² + 15.5² + (−11.5)² + (−8.5)² + (−3.5)² = 513.5
Standard deviation
s = √(513.5/5) = 10.13
Answer
s = 10.13
2.Descriptive statistics
Answer: P = 0.8413
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) = 0.8413
P = 0.8413
Answer
P = 0.8413
3.Probability
Answer: σ = 1.356
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 = 1·0.3 + 2·0.15 + 3·0.35 + 4·0.05 + 5·0.15 = 2.6
Mean of the squares
E(X²) = 1·0.3 + 4·0.15 + 9·0.35 + 16·0.05 + 25·0.15 = 8.6
Variance
Var(X) = 8.6 − 2.6² = 1.84
σ = √1.84 = 1.356
Answer
σ = 1.356
4.Probability
Answer: σY = 32.17
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 = 25, Var(X) = 115, σX = 10.72
Linear change
σY = |3|·10.72 = 32.17
Answer
σY = 32.17
The mean does move: E(Y) = 3·25 + 10 = 85.
5.Normal distribution
Answer: P = 0.1587
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 total
μT = 340 kg, σT = 12·√4 = 24 kg
Standardise
z = (316 − 340)/24 = −1
Probability
P = Φ(−1) = 0.1587
Answer
P = 0.1587
6.Normal distribution
Answer: x = 173.3 cm
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) = 0.25 → z = −0.6745
Back to x
x = 178 + −0.6745·7 = 173.28 cm
Answer
x = 173.28 cm
7.Confidence intervals
Answer: lower limit = 67.91
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 = 14, t = 1.761
Margin of error
E = 1.761·14.5/√15 = 6.593
Interval
74.5 ± 6.593 → [67.91, 81.09]
Answer
lower limit = 67.91
8.Confidence intervals
Answer: n = 50
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 = (1.96·9/2.5)² = 49.79
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 50
Answer
n = 50
9.Hypothesis tests & confidence intervals
Answer: t = −0.9
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 = 5/√81 = 5/9 = 0.5556
Test statistic
t = (49.5 − 50)/0.5556 = −0.9
Answer
t = −0.9
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 166
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 = (2.576·10/2)² = 165.9
Round up
n = 166
Answer
n = 166
11.Linear regression & correlation
Answer: r² = 0.9835
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 = 103/6 = 17.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 175
Syy = Σ(y − ȳ)² = 444.8
Coefficient of determination
r² = 175²/(70·444.8) = 0.9835
Answer
r² = 0.9835
About 98.35% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 3.162
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 = 370/6 = 61.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 458
Regression line
b = 458/70 = 6.543
a = 61.67 − 6.543·7 = 15.87
ŷ = 15.87 + 6.543·x
Predicted value
ŷ = 15.87 + 6.543·11 = 87.84
Residual
e = 91 − 87.84 = 3.162
Answer
e = 3.162
The point lies above the line.
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