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, 22, 24, 47, 58.
n = 5
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 545.
μ = 500σ = 75x = 545
z =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
4.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 20. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.3, 0.2, 0.2, 0.3Y = 0.5X + 20
σY =
5.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242 and 259 g? Give P as a decimal.
μ = 250σ = 5x1 = 242x2 = 259
P =
6.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 95% of them lie (the 95th percentile).
μ = 32σ = 6P = 95%
x = min
7.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.5. What is the margin of error E₂ if the same study uses 450 observations instead?
99%n₁ = 50E₁ = 4.5n₂ = 450
E₂ =
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 74 kg, standard deviation 5.5 kg. Plot group 2: 5 plots, mean 82 kg, standard deviation 9 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
n₁ = 7x̄₁ = 74s₁ = 5.5n₂ = 5x̄₂ = 82s₂ = 990%t(10) = 1.812
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 69.43 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 69.43s = 15n = 49
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 4. Find the smallest sample size n.
σ = 4E = 1.590%
n =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. 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 = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.
n = 6x = 5.5
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 15.61
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 = 196/5 = 39.2
Squared deviations
Σ(x − x̄)² = 5.8² + (−17.2)² + (−15.2)² + 7.8² + 18.8² = 974.8
Standard deviation
s = √(974.8/4) = 15.61
Answer
s = 15.61
2.Descriptive statistics
Answer: z = 0.6
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 = (545 − 500)/75 = 0.6
Answer
z = 0.6
3.Probability
Answer: σ = 0.8047
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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05
Mean of the squares
E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75
Variance
Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
4.Probability
Answer: σY = 1.204
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 = 3, Var(X) = 5.8, σX = 2.408
Linear change
σY = |0.5|·2.408 = 1.204
Answer
σY = 1.204
The mean does move: E(Y) = 0.5·3 + 20 = 21.5.
5.Normal distribution
Answer: P = 0.9093
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 = (242 − 250)/5 = −1.6
z₂ = (259 − 250)/5 = 1.8
Area between
P = Φ(1.8) − Φ(−1.6) = 0.9641 − 0.0548 = 0.9093
Answer
P = 0.9093
6.Normal distribution
Answer: x = 41.87 min
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) = 0.95 → z = 1.645
Back to x
x = 32 + 1.645·6 = 41.87 min
Answer
x = 41.87 min
7.Confidence intervals
Answer: E₂ = 1.5
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 = (2.576/2.576)·√(50/450) = 0.3333
New margin
E₂ = 4.5·0.3333 = 1.5
Answer
E₂ = 1.5
8.Confidence intervals
Answer: lower limit = −15.54 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 = (6·5.5² + 4·9²)/10 = 50.55
sp = 7.11
Margin of error
E = 1.812·7.11·√(1/7 + 1/5) = 7.544
Interval
−8 ± 7.544 → [−15.54, −0.4565]
Answer
lower limit = −15.54 kg
9.Hypothesis tests & confidence intervals
Answer: t = −2.599
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/√49 = 15/7 = 2.143
Test statistic
t = (69.43 − 75)/2.143 = −2.599
Answer
t = −2.599
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 20
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 = (1.645·4/1.5)² = 19.24
Round up
n = 20
Answer
n = 20
11.Linear regression & correlation
Answer: r² = 0.958
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 = 100/5 = 20
ȳ = 300/5 = 60
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 288
Syy = Σ(y − ȳ)² = 962
Coefficient of determination
r² = 288²/(90·962) = 0.958
Answer
r² = 0.958
About 95.8% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: ŷ = 17.26
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 = 21/6 = 3.5
ȳ = 120/6 = 20
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −24
Regression line
b = −24/17.5 = −1.371
a = 20 − (−1.371)·3.5 = 24.8
ŷ = 24.8 − 1.371·x
Prediction
ŷ = 24.8 + (−1.371)·5.5 = 17.26
Answer
ŷ = 17.26
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