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: 35, 34, 58, 13, 51.
n = 5
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 575.
μ = 500σ = 50x = 575
z =
3.ProbabilityMedium
A fair die is rolled 3 times. What is the probability P of getting at least one six?
n = 3
P =
4.ProbabilityHard
X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.25, 0.15, 0.3, 0.3Y = 10X + 4
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 257.75 g? Give P as a decimal.
μ = 250σ = 5x1 = 242x2 = 257.75
P =
6.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 65σ = 10P = 10%
x = points
7.Confidence intervalsMedium
A sample of 9 items has mean 68 and standard deviation s = 4.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 9x̄ = 68s = 4.595%t(8) = 2.306
E =
8.Confidence intervalsHard
The population standard deviation is 37. 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 2?
σ = 37E = 295%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 44.69 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 44.69s = 12n = 49
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 49.7. The population standard deviation is 12. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 49.7σ = 12n = 6499%
L =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.
n = 5x = 5.5
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 17.46
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 = 191/5 = 38.2
Squared deviations
Σ(x − x̄)² = (−3.2)² + (−4.2)² + 19.8² + (−25.2)² + 12.8² = 1219
Standard deviation
s = √(1219/4) = 17.46
Answer
s = 17.46
2.Descriptive statistics
Answer: z = 1.5
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 = (575 − 500)/50 = 1.5
Answer
z = 1.5
3.Probability
Answer: P = 0.4213
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)3 = 0.5787
P = 1 − 0.5787 = 0.4213
Answer
P = 0.4213
4.Probability
Answer: σY = 11.52
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 = 2.65, Var(X) = 1.327, σX = 1.152
Linear change
σY = |10|·1.152 = 11.52
Answer
σY = 11.52
The mean does move: E(Y) = 10·2.65 + 4 = 30.5.
5.Normal distribution
Answer: P = 0.8846
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 = (257.75 − 250)/5 = 1.55
Area between
P = Φ(1.55) − Φ(−1.6) = 0.9394 − 0.0548 = 0.8846
Answer
P = 0.8846
6.Normal distribution
Answer: x = 52.18 points
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) = 0.1 → z = −1.282
Back to x
x = 65 + −1.282·10 = 52.18 points
Answer
x = 52.18 points
7.Confidence intervals
Answer: E = 3.459
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 = 2.306
Margin of error
E = 2.306·4.5/√9 = 3.459
Answer
E = 3.459
8.Confidence intervals
Answer: n = 1315
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 = (1.96·37/2)² = 1315
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1315
Answer
n = 1315
9.Hypothesis tests & confidence intervals
Answer: t = −3.098
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/√49 = 12/7 = 1.714
Test statistic
t = (44.69 − 50)/1.714 = −3.098
Answer
t = −3.098
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: L = 45.84
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 = 2.576·12/√64 = 3.864
Interval limit
L = 49.7 − 3.864 = 45.84
Answer
L = 45.84
11.Linear regression & correlation
Answer: r² = 0.954
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 = 85/6 = 14.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 145
Syy = Σ(y − ȳ)² = 314.8
Coefficient of determination
r² = 145²/(70·314.8) = 0.954
Answer
r² = 0.954
About 95.4% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: ŷ = 82.5
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 = 25/5 = 5
ȳ = 396/5 = 79.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 66
Regression line
b = 66/10 = 6.6
a = 79.2 − 6.6·5 = 46.2
ŷ = 46.2 + 6.6·x
Prediction
ŷ = 46.2 + 6.6·5.5 = 82.5
Answer
ŷ = 82.5
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