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: 21, 41, 26, 47, 41, 34, 20.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is below 510?
μ = 500σ = 25x = 510
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.3, 0.1
σ =
4.ProbabilityHard
A game costs €2 to play. You win €4 with probability 0.2 and €40 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€4: p = 0.2€40: p = 0.05
E(G) = €
5.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 95.5 and 118 points? Give P as a decimal.
μ = 100σ = 15x1 = 95.5x2 = 118
P =
6.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 2.1. What is the margin of error E₂ of a 90% interval based on 256 observations?
99%n₁ = 64E₁ = 2.190%n₂ = 256
E₂ =
8.Confidence intervalsHard
The population standard deviation is 19. 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 = 19E = 295%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 122.3 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 122.3s = 12n = 81
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 118.8. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 118.8σ = 4n = 8199%
L =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. 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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (26.5, 132) is made. Find its residual e = y − ŷ.
n = 5(26.5, 132)
e =
3Solutions
1.Descriptive statistics
Answer: s = 10.7
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 = 230/7 = 32.86
Squared deviations
Σ(x − x̄)² = (−11.86)² + 8.143² + (−6.857)² + 14.14² + 8.143² + 1.143² + (−12.86)² = 686.9
Standard deviation
s = √(686.9/6) = 10.7
Answer
s = 10.7
2.Descriptive statistics
Answer: P = 0.6554
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
3.Probability
Answer: σ = 9.421
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.
Mean
μ = 10·0.25 + 20·0.35 + 30·0.3 + 40·0.1 = 22.5
Mean of the squares
E(X²) = 100·0.25 + 400·0.35 + 900·0.3 + 1600·0.1 = 595
Variance
Var(X) = 595 − 22.5² = 88.75
σ = √88.75 = 9.421
Answer
σ = 9.421
4.Probability
Answer: E(G) = 0.8 €
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.
Net outcomes
G: 2 (p = 0.2), 38 (p = 0.05), −2 (p = 0.75)
Expected value
E(G) = 2·0.2 + 38·0.05 − 2·0.75 = 0.8 €
Answer
E(G) = 0.8 €
A positive value means you win money on average.
5.Normal distribution
Answer: P = 0.5028
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 = (95.5 − 100)/15 = −0.3
z₂ = (118 − 100)/15 = 1.2
Area between
P = Φ(1.2) − Φ(−0.3) = 0.8849 − 0.3821 = 0.5028
Answer
P = 0.5028
6.Normal distribution
Answer: P = 0.8413
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 difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
7.Confidence intervals
Answer: E₂ = 0.6705
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.
Ratio of margins
E₂/E₁ = (1.645/2.576)·√(64/256) = 0.3193
New margin
E₂ = 2.1·0.3193 = 0.6705
Answer
E₂ = 0.6705
8.Confidence intervals
Answer: n = 347
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 = (1.96·19/2)² = 346.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 347
Answer
n = 347
9.Hypothesis tests & confidence intervals
Answer: t = 1.725
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 = (122.3 − 120)/1.333 = 1.725
Answer
t = 1.725
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: L = 117.7
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).
Margin of error
E = 2.576·4/√81 = 1.145
Interval limit
L = 118.8 − 1.145 = 117.7
Answer
L = 117.7
11.Linear regression & correlation
Answer: r² = 0.8335
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 = 36/6 = 6
ȳ = 358/6 = 59.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 170
Syy = Σ(y − ȳ)² = 495.3
Coefficient of determination
r² = 170²/(70·495.3) = 0.8335
Answer
r² = 0.8335
About 83.35% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 2.3
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 = 110/5 = 22
ȳ = 527/5 = 105.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 486
Regression line
b = 486/90 = 5.4
a = 105.4 − 5.4·22 = −13.4
ŷ = −13.4 + 5.4·x
Predicted value
ŷ = −13.4 + 5.4·26.5 = 129.7
Residual
e = 132 − 129.7 = 2.3
Answer
e = 2.3
The point lies above the line.
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