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: 30, 13, 42, 48, 26, 51, 13.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 15. Find the z-score z of a score of 48.
μ = 60σ = 15x = 48
z =
3.ProbabilityMedium
A bag holds 3 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 5
P =
4.ProbabilityHard
12 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?
n = 12p = 0.5
P =
5.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 78.25 and 103.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 78.25x2 = 103.75
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 bottom 75% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 100 phones, mean 49 h, known standard deviation 18 h. Model 2: 100 phones, mean 41.5 h, known standard deviation 4 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
n₁ = 100x̄₁ = 49 hσ₁ = 18 hn₂ = 100x̄₂ = 41.5 hσ₂ = 4 h99%
lower limit = h
8.Confidence intervalsHard
The population standard deviation is 27. 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 3?
σ = 27E = 395%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 198.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 198.8s = 12n = 81
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 119.8. The population standard deviation is 5. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 119.8σ = 5n = 6495%
U =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. Predict ŷ for x = 9.
n = 5x = 9
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 15.7
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 = 223/7 = 31.86
Squared deviations
Σ(x − x̄)² = (−1.857)² + (−18.86)² + 10.14² + 16.14² + (−5.857)² + 19.14² + (−18.86)² = 1479
Standard deviation
s = √(1479/6) = 15.7
Answer
s = 15.7
2.Descriptive statistics
Answer: z = −0.8
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 = (48 − 60)/15 = −0.8
Answer
z = −0.8
3.Probability
Answer: P = 0.1071
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 = 3/8 · 2/7 = 6/56 = 0.1071
Answer
P = 0.1071
4.Probability
Answer: P = 0.003174
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.
Zero and one failure
P(X = 0) = 0.512 = 0.0002441
P(X = 1) = 12·0.5·0.511 = 0.00293
Add them
P = 0.0002441 + 0.00293 = 0.003174
Answer
P = 0.003174
5.Normal distribution
Answer: P = 0.5252
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 = (78.25 − 100)/15 = −1.45
z₂ = (103.75 − 100)/15 = 0.25
Area between
P = Φ(0.25) − Φ(−1.45) = 0.5987 − 0.07353 = 0.5252
Answer
P = 0.5252
6.Normal distribution
Answer: x = 253.4 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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 250 + 0.67·5 = 253.4 g
Answer
x = 253.4 g
7.Confidence intervals
Answer: lower limit = 2.75 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 = 49 − 41.5 = 7.5
Standard error
SE = √(18²/100 + 4²/100) = 1.844
Margin of error
E = 2.576·1.844 = 4.75
Interval
7.5 ± 4.75 → [2.75, 12.25]
Answer
lower limit = 2.75 h
The interval does not contain 0, so the means differ significantly.
8.Confidence intervals
Answer: n = 312
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 = (1.96·27/3)² = 311.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 312
Answer
n = 312
9.Hypothesis tests & confidence intervals
Answer: t = −0.9
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).
Standard error
SE = 12/√81 = 12/9 = 1.333
Test statistic
t = (198.8 − 200)/1.333 = −0.9
Answer
t = −0.9
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 121
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 = 1.96·5/√64 = 1.225
Interval limit
U = 119.8 + 1.225 = 121
Answer
U = 121
11.Linear regression & correlation
Answer: r² = 0.9389
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 = 15/5 = 3
ȳ = 125/5 = 25
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −13
Syy = Σ(y − ȳ)² = 18
Coefficient of determination
r² = −13²/(10·18) = 0.9389
Answer
r² = 0.9389
About 93.89% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: ŷ = 33.85
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 = 113/5 = 22.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 150
Regression line
b = 150/40 = 3.75
a = 22.6 − 3.75·6 = 0.1
ŷ = 0.1 + 3.75·x
Prediction
ŷ = 0.1 + 3.75·9 = 33.85
Answer
ŷ = 33.85
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