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: 52, 17, 24, 35, 27, 59, 46.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 84?
μ = 100σ = 20x = 84
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.1, 0.3, 0.35, 0.25
σ =
4.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 20. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.15, 0.2Y = 0.5X + 20
σY =
5.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9051. Find z to two decimals.
A = 0.9051
z =
6.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
A sample of 20 items has mean 50.5 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 20x̄ = 50.5s = 2090%t(19) = 1.729
lower limit =
8.Confidence intervalsHard
The population standard deviation is 17. 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 1.5?
σ = 17E = 1.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 115.1 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 115.1s = 15n = 64
t =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 1.5 units with a power of 90%. The standard deviation is 9. Find the minimum sample size n.
α = 0.01power = 90%δ = 1.5σ = 9zα = 2.326zβ = 1.282
n =
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 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 (9, 28) is made. Find its residual e = y − ŷ.
n = 6(9, 28)
e =
3Solutions
1.Descriptive statistics
Answer: s = 15.61
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 = 260/7 = 37.14
Squared deviations
Σ(x − x̄)² = 14.86² + (−20.14)² + (−13.14)² + (−2.143)² + (−10.14)² + 21.86² + 8.857² = 1463
Standard deviation
s = √(1463/6) = 15.61
Answer
s = 15.61
2.Descriptive statistics
Answer: P = 0.7881
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 = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
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 = 10·0.1 + 20·0.3 + 30·0.35 + 40·0.25 = 27.5
Mean of the squares
E(X²) = 100·0.1 + 400·0.3 + 900·0.35 + 1600·0.25 = 845
Variance
Var(X) = 845 − 27.5² = 88.75
σ = √88.75 = 9.421
Answer
σ = 9.421
4.Probability
Answer: σY = 0.5612
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 = 2.2, Var(X) = 1.26, σX = 1.122
Linear change
σY = |0.5|·1.122 = 0.5612
Answer
σY = 0.5612
The mean does move: E(Y) = 0.5·2.2 + 20 = 21.1.
5.Normal distribution
Answer: z = 1.67
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 of z
Φ(z) = (1 + 0.9051)/2 = 0.9525
Read the z-table backwards
Find 0.9525 inside the table and read off z.
z = 1.67
Answer
z = 1.67
6.Normal distribution
Answer: P = 0.7611
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 = 25.02 − 25 = 0.02 mm, σ = 0.02828 mm
Standardise
z = (0 − 0.02)/0.02828 = −0.71
Probability
P = 1 − Φ(−0.71) = 0.7611
Answer
P = 0.7611
7.Confidence intervals
Answer: lower limit = 42.77
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 = 19, t = 1.729
Margin of error
E = 1.729·20/√20 = 7.732
Interval
50.5 ± 7.732 → [42.77, 58.23]
Answer
lower limit = 42.77
8.Confidence intervals
Answer: n = 494
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 = (1.96·17/1.5)² = 493.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 494
Answer
n = 494
9.Hypothesis tests & confidence intervals
Answer: t = −2.613
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 = 15/√64 = 15/8 = 1.875
Test statistic
t = (115.1 − 120)/1.875 = −2.613
Answer
t = −2.613
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 469
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 1.282)·9/1.5)² = 468.6
Round up
n = 469
Answer
n = 469
11.Linear regression & correlation
Answer: r² = 0.9477
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 = 33/6 = 5.5
ȳ = 524/6 = 87.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 108
Syy = Σ(y − ȳ)² = 703.3
Coefficient of determination
r² = 108²/(17.5·703.3) = 0.9477
Answer
r² = 0.9477
About 94.77% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 3.043
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 = 117/6 = 19.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 191
Regression line
b = 191/70 = 2.729
a = 19.5 − 2.729·7 = 0.4
ŷ = 0.4 + 2.729·x
Predicted value
ŷ = 0.4 + 2.729·9 = 24.96
Residual
e = 28 − 24.96 = 3.043
Answer
e = 3.043
The point lies above the line.
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