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: 45, 36, 22, 48, 33.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 57.5?
μ = 50σ = 5x = 57.5
P =
3.ProbabilityMedium
A fair die is rolled 7 times. What is the probability P of getting at least one six?
n = 7
P =
4.ProbabilityHard
X has the distribution shown, and Y = 5X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 5X + 1
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 exceeds 364 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 4> 364 kg
P =
6.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 1.8. What is the margin of error E₂ of a 99% interval from the same data?
90%n₁ = 100E₁ = 1.899%
E₂ =
8.Confidence intervalsHard
The population standard deviation is 6. 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 1?
σ = 6E = 195%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 205.7 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 205.7s = 20n = 36
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 120.2. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 120.2σ = 10n = 8190%
U =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 142Σx² = 364Σy² = 3916Σxy = 1190
r =
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 (4, 52) is made. Find its residual e = y − ŷ.
n = 6(4, 52)
e =
3Solutions
1.Descriptive statistics
Answer: s = 10.33
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 = 184/5 = 36.8
Squared deviations
Σ(x − x̄)² = 8.2² + (−0.8)² + (−14.8)² + 11.2² + (−3.8)² = 426.8
Standard deviation
s = √(426.8/4) = 10.33
Answer
s = 10.33
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: P = 0.7209
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.
Complement
P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209
Answer
P = 0.7209
4.Probability
Answer: σY = 4.023
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 = 0.95, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |5|·0.8047 = 4.023
Answer
σY = 4.023
The mean does move: E(Y) = 5·0.95 + 1 = 5.75.
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 = (364 − 340)/24 = 1
Probability
P = 1 − Φ(1) = 0.1587
Answer
P = 0.1587
6.Normal distribution
Answer: P = 0.7881
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 = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
7.Confidence intervals
Answer: E₂ = 2.819
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.
Ratio of margins
E₂/E₁ = (2.576/1.645)·√(100/100) = 1.566
New margin
E₂ = 1.8·1.566 = 2.819
Answer
E₂ = 2.819
8.Confidence intervals
Answer: n = 139
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 = (1.96·6/1)² = 138.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 139
Answer
n = 139
9.Hypothesis tests & confidence intervals
Answer: t = 1.71
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/√36 = 20/6 = 3.333
Test statistic
t = (205.7 − 200)/3.333 = 1.71
Answer
t = 1.71
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: U = 122
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 = 1.645·10/√81 = 1.828
Interval limit
U = 120.2 + 1.828 = 122
Answer
U = 122
11.Linear regression & correlation
Answer: r = 0.9941
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 = 1190 − 42·142/6 = 196
Sxx = 364 − 42²/6 = 70
Syy = 3916 − 142²/6 = 555.3
Correlation coefficient
r = 196/√(70·555.3) = 0.9941
Answer
r = 0.9941
12.Linear regression & correlation
Answer: e = 4.933
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 = 36/6 = 6
ȳ = 340/6 = 56.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 336
Regression line
b = 336/70 = 4.8
a = 56.67 − 4.8·6 = 27.87
ŷ = 27.87 + 4.8·x
Predicted value
ŷ = 27.87 + 4.8·4 = 47.07
Residual
e = 52 − 47.07 = 4.933
Answer
e = 4.933
The point lies above the line.
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