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: 14, 24, 60, 20, 38, 20.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is below 440?
μ = 500σ = 40x = 440
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.3, 0.15, 0.35, 0.2
σ =
4.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 2 correct answers?
n = 10p = 0.5k = 2
P =
5.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 9 of them is below 109.2 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 9< 109.2 kg
P =
6.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 140 h, known standard deviation 13 h. Model 2: 100 phones, mean 142.5 h, known standard deviation 7 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
n₁ = 40x̄₁ = 140 hσ₁ = 13 hn₂ = 100x̄₂ = 142.5 hσ₂ = 7 h90%
E = h
8.Confidence intervalsHard
The population standard deviation is 30. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 3?
σ = 30E = 390%
n =
9.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 149.5 mm, and the line's known standard deviation is 10 mm. Line 2: a sample of 60 parts has mean 145.3 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
n₁ = 50x̄₁ = 149.5 mmσ₁ = 10 mmn₂ = 60x̄₂ = 145.3 mmσ₂ = 5 mm
z =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 500 against H₁: μ < 500 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 3. In reality the mean is 499.2. Find the probability of a Type II error β in percent.
μ₀ = 500μ = 499.2σ = 3n = 36α = 0.10
β = %
11.Linear regression & correlationMedium
The table gives 5 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 = 5Σx = 30Σy = 49Σx² = 220Σy² = 573Σxy = 354
r =
12.Linear regression & correlationHard
The table gives 6 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 = 104?
n = 6ŷ = 104
x =
3Solutions
1.Descriptive statistics
Answer: s = 17.05
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 = 176/6 = 29.33
Squared deviations
Σ(x − x̄)² = (−15.33)² + (−5.333)² + 30.67² + (−9.333)² + 8.667² + (−9.333)² = 1453
Standard deviation
s = √(1453/5) = 17.05
Answer
s = 17.05
2.Descriptive statistics
Answer: P = 0.06681
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) = 0.06681
P = 0.06681
Answer
P = 0.06681
3.Probability
Answer: σ = 1.117
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 + 4·0.2 = 2.45
Mean of the squares
E(X²) = 1·0.3 + 4·0.15 + 9·0.35 + 16·0.2 = 7.25
Variance
Var(X) = 7.25 − 2.45² = 1.247
σ = √1.247 = 1.117
Answer
σ = 1.117
4.Probability
Answer: P = 0.04395
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.
Binomial formula
P(X = 2) = 45·0.52·0.58 = 45·0.25·0.003906 = 0.04395
Answer
P = 0.04395
5.Normal distribution
Answer: P = 0.6915
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 = 108 kg, σT = 0.8·√9 = 2.4 kg
Standardise
z = (109.2 − 108)/2.4 = 0.5
Probability
P = Φ(0.5) = 0.6915
Answer
P = 0.6915
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: E = 3.572 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 140 − 142.5 = −2.5
Standard error
SE = √(13²/40 + 7²/100) = 2.171
Margin of error
E = 1.645·2.171 = 3.572
Answer
E = 3.572 h
The interval contains 0, so the data do not show a significant difference.
8.Confidence intervals
Answer: n = 271
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 = (1.645·30/3)² = 270.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 271
Answer
n = 271
9.Hypothesis tests & confidence intervals
Answer: z = 2.702
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.
Standard error of the difference
SE = √(10²/50 + 5²/60) = √2.417 = 1.555
Test statistic
z = (149.5 − 145.3)/1.555 = 2.702
Answer
z = 2.702
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: β = 37.52 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 3/√36 = 0.5
x̄c = 500 − 1.282·0.5 = 499.4
z under the true mean
z = (499.4 − 499.2)/0.5 = 0.318
Type II error
power = P(Z < 0.318) = 62.48%
β = 100% − 62.48% = 37.52%
Answer
β = 37.52%
11.Linear regression & correlation
Answer: r = 0.9848
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 = 354 − 30·49/5 = 60
Sxx = 220 − 30²/5 = 40
Syy = 573 − 49²/5 = 92.8
Correlation coefficient
r = 60/√(40·92.8) = 0.9848
Answer
r = 0.9848
12.Linear regression & correlation
Answer: x = 24.94
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 = 138/6 = 23
ȳ = 563/6 = 93.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 367
Regression line
b = 367/70 = 5.243
a = 93.83 − 5.243·23 = −26.75
ŷ = −26.75 + 5.243·x
Solve for x
x = (104 − (−26.75))/5.243 = 24.94
Answer
x = 24.94
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