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: 57, 39, 43, 51, 13.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 10. What proportion P of values is below 82.5?
μ = 70σ = 10x = 82.5
P =
3.ProbabilityMedium
A bag holds 4 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 6
P =
4.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 2 correct answers?
n = 7p = 0.6k = 2
P =
5.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7764. Find z to two decimals.
A = 0.7764
z =
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 75% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h75%
x = h
7.Confidence intervalsMedium
A sample of 15 items has mean 25.5 and standard deviation s = 19.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 15x̄ = 25.5s = 19.595%t(14) = 2.145
upper 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 95% confidence interval for the proportion has a margin of error of at most 0.01?
p = 0.5E = 0.0195%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 495.3 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 495.3s = 15n = 49
t =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 1 units with a power of 95%. The standard deviation is 8. Find the minimum sample size n.
α = 0.05power = 95%δ = 1σ = 8zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. 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 = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 16) is made. Find its residual e = y − ŷ.
n = 6(4.5, 16)
e =
3Solutions
1.Descriptive statistics
Answer: s = 16.94
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 = 203/5 = 40.6
Squared deviations
Σ(x − x̄)² = 16.4² + (−1.6)² + 2.4² + 10.4² + (−27.6)² = 1147
Standard deviation
s = √(1147/4) = 16.94
Answer
s = 16.94
2.Descriptive statistics
Answer: P = 0.8944
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
3.Probability
Answer: P = 0.1333
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 = 4/10 · 3/9 = 12/90 = 0.1333
Answer
P = 0.1333
4.Probability
Answer: P = 0.07741
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.
Binomial formula
P(X = 2) = 21·0.62·0.45 = 21·0.36·0.01024 = 0.07741
Answer
P = 0.07741
5.Normal distribution
Answer: z = −0.76
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 − 0.7764 = 0.2236
Read the z-table backwards
Find 0.2236 inside the table and read off z.
z = −0.76
Answer
z = −0.76
6.Normal distribution
Answer: x = 527 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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 500 + 0.67·40 = 527 h
Answer
x = 527 h
7.Confidence intervals
Answer: upper limit = 36.3
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 = 14, t = 2.145
Margin of error
E = 2.145·19.5/√15 = 10.8
Interval
25.5 ± 10.8 → [14.7, 36.3]
Answer
upper limit = 36.3
8.Confidence intervals
Answer: n = 9604
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 = 1.96²·0.5·0.5/0.01² = 9604
Round up
n = 9604
Answer
n = 9604
9.Hypothesis tests & confidence intervals
Answer: t = −2.193
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/√49 = 15/7 = 2.143
Test statistic
t = (495.3 − 500)/2.143 = −2.193
Answer
t = −2.193
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: n = 693
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 = ((1.645 + 1.645)·8/1)² = 692.7
Round up
n = 693
Answer
n = 693
11.Linear regression & correlation
Answer: r² = 0.9935
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 = 121/6 = 20.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −58.5
Syy = Σ(y − ȳ)² = 196.8
Coefficient of determination
r² = −58.5²/(17.5·196.8) = 0.9935
Answer
r² = 0.9935
About 99.35% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = −3.405
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 = 131/6 = 21.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −42.5
Regression line
b = −42.5/17.5 = −2.429
a = 21.83 − (−2.429)·3.5 = 30.33
ŷ = 30.33 − 2.429·x
Predicted value
ŷ = 30.33 + (−2.429)·4.5 = 19.4
Residual
e = 16 − 19.4 = −3.405
Answer
e = −3.405
The point lies below the line.
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