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: 12, 47, 49, 36, 11, 33.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 162.5?
μ = 170σ = 15x = 162.5
P =
3.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
4.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 4 correct answers?
n = 9p = 0.6k = 4
P =
5.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 4 of them is below 102 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 4< 102 kg
P =
6.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g70%
x = g
7.Confidence intervalsMedium
A sample of 9 items has mean 117 and standard deviation s = 3.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 9x̄ = 117s = 3.590%t(8) = 1.86
lower limit =
8.Confidence intervalsHard
The population standard deviation is 27. 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 0.5?
σ = 27E = 0.590%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 121.7 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 121.7s = 5n = 25
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 60 of 150 patients on treatment A recover, and 55 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 60n₁ = 150x₂ = 55n₂ = 100
z =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 137Σx² = 91Σy² = 3259Σxy = 433
r =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 11?
n = 5ŷ = 11
x =
3Solutions
1.Descriptive statistics
Answer: s = 16.55
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 = 188/6 = 31.33
Squared deviations
Σ(x − x̄)² = (−19.33)² + 15.67² + 17.67² + 4.667² + (−20.33)² + 1.667² = 1369
Standard deviation
s = √(1369/5) = 16.55
Answer
s = 16.55
2.Descriptive statistics
Answer: P = 0.6915
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
3.Probability
Answer: P = 0.6651
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)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
4.Probability
Answer: P = 0.1672
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 = 4) = 126·0.64·0.45 = 126·0.1296·0.01024 = 0.1672
Answer
P = 0.1672
5.Normal distribution
Answer: P = 0.9772
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 = 100 kg, σT = 0.5·√4 = 1 kg
Standardise
z = (102 − 100)/1 = 2
Probability
P = Φ(2) = 0.9772
Answer
P = 0.9772
6.Normal distribution
Answer: x = 247.4 g
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 250 − 0.52·5 = 247.4 g
Answer
x = 247.4 g
7.Confidence intervals
Answer: lower limit = 114.8
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 = 8, t = 1.86
Margin of error
E = 1.86·3.5/√9 = 2.17
Interval
117 ± 2.17 → [114.8, 119.2]
Answer
lower limit = 114.8
8.Confidence intervals
Answer: n = 7891
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 = (1.645·27/0.5)² = 7891
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 7891
Answer
n = 7891
9.Hypothesis tests & confidence intervals
Answer: t = 1.7
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 = 5/√25 = 5/5 = 1
Test statistic
t = (121.7 − 120)/1 = 1.7
Answer
t = 1.7
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: z = −2.331
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.
Sample proportions
p̂₁ = 60/150 = 0.4
p̂₂ = 55/100 = 0.55
p̂ = 115/250 = 0.46
Standard error under H₀
SE = √(0.46·0.54·(1/150 + 1/100)) = 0.06434
Test statistic
z = (0.4 − 0.55)/0.06434 = −2.331
Answer
z = −2.331
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r = −0.9718
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 = 433 − 21·137/6 = −46.5
Sxx = 91 − 21²/6 = 17.5
Syy = 3259 − 137²/6 = 130.8
Correlation coefficient
r = −46.5/√(17.5·130.8) = −0.9718
Answer
r = −0.9718
12.Linear regression & correlation
Answer: x = 2.889
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 = 118/5 = 23.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 162
Regression line
b = 162/40 = 4.05
a = 23.6 − 4.05·6 = −0.7
ŷ = −0.7 + 4.05·x
Solve for x
x = (11 − (−0.7))/4.05 = 2.889
Answer
x = 2.889
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