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: 42, 58, 58, 55, 50, 17, 45.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 70.
μ = 60σ = 8x = 70
z =
3.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
4.ProbabilityHard
10 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 10p = 0.3
P =
5.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 12 of them is below 139.8 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 12< 139.8 kg
P =
6.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h80%
x = h
7.Confidence intervalsMedium
A sample of 11 items has mean 42.5 and standard deviation s = 1.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 11x̄ = 42.5s = 1.595%t(10) = 2.228
lower limit =
8.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?
p = 0.5E = 0.0599%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 70.6 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 70.6s = 20n = 100
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 51.6. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 51.6σ = 12n = 6490%
U =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 5
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.
n = 6x = 4.5
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 14.39
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 = 325/7 = 46.43
Squared deviations
Σ(x − x̄)² = (−4.429)² + 11.57² + 11.57² + 8.571² + 3.571² + (−29.43)² + (−1.429)² = 1242
Standard deviation
s = √(1242/6) = 14.39
Answer
s = 14.39
2.Descriptive statistics
Answer: z = 1.25
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 = (70 − 60)/8 = 1.25
Answer
z = 1.25
3.Probability
Answer: P = 0.5177
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
4.Probability
Answer: P = 0.1493
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) = 0.710 = 0.02825
P(X = 1) = 10·0.3·0.79 = 0.1211
Add them
P = 0.02825 + 0.1211 = 0.1493
Answer
P = 0.1493
5.Normal distribution
Answer: P = 0.06426
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 total
μT = 144 kg, σT = 0.8·√12 = 2.771 kg
Standardise
z = (139.8 − 144)/2.771 = −1.52
Probability
P = Φ(−1.52) = 0.06426
Answer
P = 0.06426
6.Normal distribution
Answer: x = 534 h
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) = 0.8
z from the table
z ≈ 0.84
Back to the original units
x = 500 + 0.84·40 = 534 h
Answer
x = 534 h
7.Confidence intervals
Answer: lower limit = 41.49
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 = 10, t = 2.228
Margin of error
E = 2.228·1.5/√11 = 1.008
Interval
42.5 ± 1.008 → [41.49, 43.51]
Answer
lower limit = 41.49
8.Confidence intervals
Answer: n = 664
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.5·0.5/0.05² = 663.6
Round up
n = 664
Answer
n = 664
9.Hypothesis tests & confidence intervals
Answer: t = −2.2
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 = (70.6 − 75)/2 = −2.2
Answer
t = −2.2
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 54.07
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 = 1.645·12/√64 = 2.468
Interval limit
U = 51.6 + 2.468 = 54.07
Answer
U = 54.07
11.Linear regression & correlation
Answer: r = −0.9138
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 = 114/5 = 22.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −18
Syy = Σ(y − ȳ)² = 38.8
Correlation coefficient
r = −18/√(10·38.8) = −0.9138
Answer
r = −0.9138
12.Linear regression & correlation
Answer: ŷ = 16.22
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 = 21/6 = 3.5
ȳ = 110/6 = 18.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −37
Regression line
b = −37/17.5 = −2.114
a = 18.33 − (−2.114)·3.5 = 25.73
ŷ = 25.73 − 2.114·x
Prediction
ŷ = 25.73 + (−2.114)·4.5 = 16.22
Answer
ŷ = 16.22
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