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: 22, 40, 31, 16, 42.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 425?
μ = 500σ = 75x = 425
P =
3.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 6? The order does not matter.
n = 6k = 4
N =
4.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?
n = 12p = 0.3k = 4
P =
5.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 12 of them is below 1041 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 12< 1041 kg
P =
6.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points95%
x = points
7.Confidence intervalsMedium
A sample of 25 items has mean 30.5 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 25x̄ = 30.5s = 9.599%t(24) = 2.797
upper limit =
8.Confidence intervalsHard
The population standard deviation is 10. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 0.5?
σ = 10E = 0.595%
n =
9.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 74.5 mm, and the line's known standard deviation is 14 mm. Line 2: a sample of 60 parts has mean 71.1 mm, known standard deviation 11 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
n₁ = 50x̄₁ = 74.5 mmσ₁ = 14 mmn₂ = 60x̄₂ = 71.1 mmσ₂ = 11 mm
z =
10.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 119.6. The population standard deviation is 8. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 119.6σ = 8n = 8190%
U =
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 = 77Σx² = 220Σy² = 1381Σxy = 550
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 11) is made. Find its residual e = y − ŷ.
n = 6(3, 11)
e =
3Solutions
1.Descriptive statistics
Answer: s = 11.23
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 = 151/5 = 30.2
Squared deviations
Σ(x − x̄)² = (−8.2)² + 9.8² + 0.8² + (−14.2)² + 11.8² = 504.8
Standard deviation
s = √(504.8/4) = 11.23
Answer
s = 11.23
2.Descriptive statistics
Answer: P = 0.1587
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
3.Probability
Answer: N = 15
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(6, 4) = (6·5·4·3)/(4·3·2·1) = 15
Answer
N = 15
4.Probability
Answer: P = 0.2311
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) = 495·0.34·0.78 = 495·0.0081·0.05765 = 0.2311
Answer
P = 0.2311
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 = 1020 kg, σT = 12·√12 = 41.57 kg
Standardise
z = (1041 − 1020)/41.57 = 0.5
Probability
P = Φ(0.5) = 0.6915
Answer
P = 0.6915
6.Normal distribution
Answer: x = 75.4 points
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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 100 − 1.64·15 = 75.4 points
Answer
x = 75.4 points
7.Confidence intervals
Answer: upper limit = 35.81
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 = 24, t = 2.797
Margin of error
E = 2.797·9.5/√25 = 5.314
Interval
30.5 ± 5.314 → [25.19, 35.81]
Answer
upper limit = 35.81
8.Confidence intervals
Answer: n = 1537
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 = (1.96·10/0.5)² = 1537
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1537
Answer
n = 1537
9.Hypothesis tests & confidence intervals
Answer: z = 1.395
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 = √(14²/50 + 11²/60) = √5.937 = 2.437
Test statistic
z = (74.5 − 71.1)/2.437 = 1.395
Answer
z = 1.395
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 121.1
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).
Margin of error
E = 1.645·8/√81 = 1.462
Interval limit
U = 119.6 + 1.462 = 121.1
Answer
U = 121.1
11.Linear regression & correlation
Answer: r = 0.9959
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 = 550 − 30·77/5 = 88
Sxx = 220 − 30²/5 = 40
Syy = 1381 − 77²/5 = 195.2
Correlation coefficient
r = 88/√(40·195.2) = 0.9959
Answer
r = 0.9959
12.Linear regression & correlation
Answer: e = 3.952
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 = 42/6 = 7
ȳ = 104/6 = 17.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 180
Regression line
b = 180/70 = 2.571
a = 17.33 − 2.571·7 = −0.6667
ŷ = −0.6667 + 2.571·x
Predicted value
ŷ = −0.6667 + 2.571·3 = 7.048
Residual
e = 11 − 7.048 = 3.952
Answer
e = 3.952
The point lies above the line.
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