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: 59, 24, 29, 42, 20, 48.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 80?
μ = 60σ = 10x = 80
P =
3.ProbabilityMedium
A bag holds 4 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 8
P =
4.ProbabilityHard
A test has 6 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 = 6p = 0.5k = 2
P =
5.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85.75 and 103.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 85.75x2 = 103.75
P =
6.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 5% of them lie (the 5th percentile).
μ = 100σ = 15P = 5%
x = points
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 76 h, known standard deviation 10 h. Model 2: 50 phones, mean 62.5 h, known standard deviation 3 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
n₁ = 40x̄₁ = 76 hσ₁ = 10 hn₂ = 50x̄₂ = 62.5 hσ₂ = 3 h95%
lower limit = h
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 8 plots, mean yield 79.5 kg, standard deviation 3.5 kg. Plot group 2: 8 plots, mean 85 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
n₁ = 8x̄₁ = 79.5s₁ = 3.5n₂ = 8x̄₂ = 85s₂ = 1.595%t(14) = 2.145
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 9 students, mean score 24.5, standard deviation 3.5. Group 2: 8 students, mean score 30.1, standard deviation 8. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 9x̄₁ = 24.5s₁ = 3.5n₂ = 8x̄₂ = 30.1s₂ = 8
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 498.9. The population standard deviation is 4. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 498.9σ = 4n = 4990%
L =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 20Σy = 279Σx² = 90Σy² = 15827Σxy = 1165
r =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 77?
n = 5ŷ = 77
x =
3Solutions
1.Descriptive statistics
Answer: s = 15.18
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 = 222/6 = 37
Squared deviations
Σ(x − x̄)² = 22² + (−13)² + (−8)² + 5² + (−17)² + 11² = 1152
Standard deviation
s = √(1152/5) = 15.18
Answer
s = 15.18
2.Descriptive statistics
Answer: P = 0.02275
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
3.Probability
Answer: P = 0.09091
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 = 4/12 · 3/11 = 12/132 = 0.09091
Answer
P = 0.09091
4.Probability
Answer: P = 0.2344
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) = 15·0.52·0.54 = 15·0.25·0.0625 = 0.2344
Answer
P = 0.2344
5.Normal distribution
Answer: P = 0.4277
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 = (85.75 − 100)/15 = −0.95
z₂ = (103.75 − 100)/15 = 0.25
Area between
P = Φ(0.25) − Φ(−0.95) = 0.5987 − 0.1711 = 0.4277
Answer
P = 0.4277
6.Normal distribution
Answer: x = 75.33 points
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) = 0.05 → z = −1.645
Back to x
x = 100 + −1.645·15 = 75.33 points
Answer
x = 75.33 points
7.Confidence intervals
Answer: lower limit = 10.29 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 = 76 − 62.5 = 13.5
Standard error
SE = √(10²/40 + 3²/50) = 1.637
Margin of error
E = 1.96·1.637 = 3.209
Interval
13.5 ± 3.209 → [10.29, 16.71]
Answer
lower limit = 10.29 h
The interval does not contain 0, so the means differ significantly.
8.Confidence intervals
Answer: upper limit = −2.612 kg
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.
Pooled standard deviation
sp² = (7·3.5² + 7·1.5²)/14 = 7.25
sp = 2.693
Margin of error
E = 2.145·2.693·√(1/8 + 1/8) = 2.888
Interval
−5.5 ± 2.888 → [−8.388, −2.612]
Answer
upper limit = −2.612 kg
9.Hypothesis tests & confidence intervals
Answer: t = −1.91
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.
Pooled variance
sp² = (8·3.5² + 7·8²)/15 = 36.4
sp = 6.033
Standard error
SE = 6.033·√(1/9 + 1/8) = 2.932
Test statistic
t = (24.5 − 30.1)/2.932 = −1.91
Answer
t = −1.91
With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.
10.Hypothesis tests & confidence intervals
Answer: L = 498
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 = 1.645·4/√49 = 0.94
Interval limit
L = 498.9 − 0.94 = 498
Answer
L = 498
11.Linear regression & correlation
Answer: r = 0.9632
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 = 1165 − 20·279/5 = 49
Sxx = 90 − 20²/5 = 10
Syy = 15827 − 279²/5 = 258.8
Correlation coefficient
r = 49/√(10·258.8) = 0.9632
Answer
r = 0.9632
12.Linear regression & correlation
Answer: x = 5.95
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 = 361/5 = 72.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 202
Regression line
b = 202/40 = 5.05
a = 72.2 − 5.05·5 = 46.95
ŷ = 46.95 + 5.05·x
Solve for x
x = (77 − 46.95)/5.05 = 5.95
Answer
x = 5.95
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