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: 54, 34, 44, 60, 25.
n = 5
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
A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 5
P =
4.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 0.5X + 1
σY =
5.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 422 and 550 h? Give P as a decimal.
μ = 500σ = 40x1 = 422x2 = 550
P =
6.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 100σ = 15P = 10%
x = points
7.Confidence intervalsMedium
A sample of 25 items has mean 96 and standard deviation s = 6.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 25x̄ = 96s = 6.595%t(24) = 2.064
lower limit =
8.Confidence intervalsHard
The population standard deviation is 34. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2?
σ = 34E = 299%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 123.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 123.4s = 20n = 100
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 4. Find the smallest sample size n.
σ = 4E = 395%
n =
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 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.
n = 6x = 9
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 14.28
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 = 217/5 = 43.4
Squared deviations
Σ(x − x̄)² = 10.6² + (−9.4)² + 0.6² + 16.6² + (−18.4)² = 815.2
Standard deviation
s = √(815.2/4) = 14.28
Answer
s = 14.28
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: P = 0.359
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 = 8/13 · 7/12 = 56/156 = 0.359
Answer
P = 0.359
4.Probability
Answer: σY = 0.4023
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 = 0.95, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |0.5|·0.8047 = 0.4023
Answer
σY = 0.4023
The mean does move: E(Y) = 0.5·0.95 + 1 = 1.475.
5.Normal distribution
Answer: P = 0.8688
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 = (550 − 500)/40 = 1.25
Area between
P = Φ(1.25) − Φ(−1.95) = 0.8944 − 0.02559 = 0.8688
Answer
P = 0.8688
6.Normal distribution
Answer: x = 80.78 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 = 100 + −1.282·15 = 80.78 points
Answer
x = 80.78 points
7.Confidence intervals
Answer: lower limit = 93.32
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 = 2.064
Margin of error
E = 2.064·6.5/√25 = 2.683
Interval
96 ± 2.683 → [93.32, 98.68]
Answer
lower limit = 93.32
8.Confidence intervals
Answer: n = 1918
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 = (2.576·34/2)² = 1918
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1918
Answer
n = 1918
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 = 20/√100 = 20/10 = 2
Test statistic
t = (123.4 − 120)/2 = 1.7
Answer
t = 1.7
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 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).
Solve for n
n = (1.96·4/3)² = 6.83
Round up
n = 7
Answer
n = 7
11.Linear regression & correlation
Answer: r² = 0.9834
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 = 105/6 = 17.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 165
Syy = Σ(y − ȳ)² = 395.5
Coefficient of determination
r² = 165²/(70·395.5) = 0.9834
Answer
r² = 0.9834
About 98.34% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: ŷ = 78.07
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 = 430/6 = 71.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 224
Regression line
b = 224/70 = 3.2
a = 71.67 − 3.2·7 = 49.27
ŷ = 49.27 + 3.2·x
Prediction
ŷ = 49.27 + 3.2·9 = 78.07
Answer
ŷ = 78.07
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