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: 47, 31, 32, 32, 37, 28, 27.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is below 88?
μ = 100σ = 15x = 88
P =
3.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 15? The order does not matter.
n = 15k = 3
N =
4.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 5 correct answers?
n = 8p = 0.25k = 5
P =
5.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 422 and 522 h? Give P as a decimal.
μ = 500σ = 40x1 = 422x2 = 522
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 sample of 21 items has mean 14 and standard deviation s = 9. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 21x̄ = 14s = 990%t(20) = 1.725
lower limit =
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 11 plots, mean yield 12 kg, standard deviation 2.5 kg. Plot group 2: 6 plots, mean 2.5 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
n₁ = 11x̄₁ = 12s₁ = 2.5n₂ = 6x̄₂ = 2.5s₂ = 8.595%t(15) = 2.131
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 124.2 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 124.2s = 20n = 100
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 76.8. The population standard deviation is 10. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 76.8σ = 10n = 1695%
L =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. 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. A new observation (3.5, 19) is made. Find its residual e = y − ŷ.
n = 6(3.5, 19)
e =
3Solutions
1.Descriptive statistics
Answer: s = 6.803
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 = 234/7 = 33.43
Squared deviations
Σ(x − x̄)² = 13.57² + (−2.429)² + (−1.429)² + (−1.429)² + 3.571² + (−5.429)² + (−6.429)² = 277.7
Standard deviation
s = √(277.7/6) = 6.803
Answer
s = 6.803
2.Descriptive statistics
Answer: P = 0.2119
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
3.Probability
Answer: N = 455
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.
Combinations
C(15, 3) = (15·14·13)/(3·2·1) = 455
Answer
N = 455
4.Probability
Answer: P = 0.02307
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 = 5) = 56·0.255·0.753 = 56·0.0009766·0.4219 = 0.02307
Answer
P = 0.02307
5.Normal distribution
Answer: P = 0.6833
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 = (422 − 500)/40 = −1.95
z₂ = (522 − 500)/40 = 0.55
Area between
P = Φ(0.55) − Φ(−1.95) = 0.7088 − 0.02559 = 0.6833
Answer
P = 0.6833
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: lower limit = 10.61
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 = 20, t = 1.725
Margin of error
E = 1.725·9/√21 = 3.388
Interval
14 ± 3.388 → [10.61, 17.39]
Answer
lower limit = 10.61
8.Confidence intervals
Answer: lower limit = 3.752 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 = (10·2.5² + 5·8.5²)/15 = 28.25
sp = 5.315
Margin of error
E = 2.131·5.315·√(1/11 + 1/6) = 5.748
Interval
9.5 ± 5.748 → [3.752, 15.25]
Answer
lower limit = 3.752 kg
9.Hypothesis tests & confidence intervals
Answer: t = 2.1
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/√100 = 20/10 = 2
Test statistic
t = (124.2 − 120)/2 = 2.1
Answer
t = 2.1
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: L = 71.9
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).
Margin of error
E = 1.96·10/√16 = 4.9
Interval limit
L = 76.8 − 4.9 = 71.9
Answer
L = 71.9
11.Linear regression & correlation
Answer: r² = 0.9949
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 = 30/5 = 6
ȳ = 117/5 = 23.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134
Syy = Σ(y − ȳ)² = 451.2
Coefficient of determination
r² = 134²/(40·451.2) = 0.9949
Answer
r² = 0.9949
About 99.49% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = −3
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 = 132/6 = 22
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −22
Regression line
b = −22/17.5 = −1.257
a = 22 − (−1.257)·3.5 = 26.4
ŷ = 26.4 − 1.257·x
Predicted value
ŷ = 26.4 + (−1.257)·3.5 = 22
Residual
e = 19 − 22 = −3
Answer
e = −3
The point lies below the line.
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