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: 41, 20, 52, 25, 53, 37, 42.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is below 77.5?
μ = 100σ = 15x = 77.5
P =
3.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 14? The order does not matter.
n = 14k = 5
N =
4.ProbabilityHard
5 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 5p = 0.1
P =
5.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 30.5 and 38 min? Give P as a decimal.
μ = 32σ = 6x1 = 30.5x2 = 38
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 top 15% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h15%
x = h
7.Confidence intervalsMedium
The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.7 and standard deviation 1.1. Find the upper limit of the 99% confidence interval for the mean difference.
n = 9d̄ = 8.7sd = 1.199%t(8) = 3.355
upper limit = mmHg
8.Confidence intervalsHard
A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.6E = 0.0495%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 503.4 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 503.4s = 10n = 25
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 198. The population standard deviation is 10. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 198σ = 10n = 6499%
U =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
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. For which value x does the line predict ŷ = 20?
n = 6ŷ = 20
x =
3Solutions
1.Descriptive statistics
Answer: s = 12.5
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 = 270/7 = 38.57
Squared deviations
Σ(x − x̄)² = 2.429² + (−18.57)² + 13.43² + (−13.57)² + 14.43² + (−1.571)² + 3.429² = 937.7
Standard deviation
s = √(937.7/6) = 12.5
Answer
s = 12.5
2.Descriptive statistics
Answer: P = 0.06681
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 0.06681
Answer
P = 0.06681
3.Probability
Answer: N = 2002
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.
Combinations
C(14, 5) = (14·13·12·11·10)/(5·4·3·2·1) = 2002
Answer
N = 2002
4.Probability
Answer: P = 0.9185
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) = 0.95 = 0.5905
P(X = 1) = 5·0.1·0.94 = 0.3281
Add them
P = 0.5905 + 0.3281 = 0.9185
Answer
P = 0.9185
5.Normal distribution
Answer: P = 0.4401
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 = (30.5 − 32)/6 = −0.25
z₂ = (38 − 32)/6 = 1
Area between
P = Φ(1) − Φ(−0.25) = 0.8413 − 0.4013 = 0.4401
Answer
P = 0.4401
6.Normal distribution
Answer: x = 542 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) = 0.85
z from the table
z ≈ 1.04
Back to the original units
x = 500 + 1.04·40 = 542 h
Answer
x = 542 h
7.Confidence intervals
Answer: upper limit = 9.93 mmHg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Critical value
df = 8, t = 3.355
Margin of error
E = 3.355·1.1/√9 = 1.23
Interval
8.7 ± 1.23 → [7.47, 9.93]
Answer
upper limit = 9.93 mmHg
8.Confidence intervals
Answer: n = 577
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 = 1.96²·0.6·0.4/0.04² = 576.2
Round up
n = 577
Answer
n = 577
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 = 10/√25 = 10/5 = 2
Test statistic
t = (503.4 − 500)/2 = 1.7
Answer
t = 1.7
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 201.2
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).
Margin of error
E = 2.576·10/√64 = 3.22
Interval limit
U = 198 + 3.22 = 201.2
Answer
U = 201.2
11.Linear regression & correlation
Answer: r = 0.9978
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 = 127/6 = 21.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 207
Syy = Σ(y − ȳ)² = 614.8
Correlation coefficient
r = 207/√(70·614.8) = 0.9978
Answer
r = 0.9978
12.Linear regression & correlation
Answer: x = 2.667
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 = 114/6 = 19
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −21
Regression line
b = −21/17.5 = −1.2
a = 19 − (−1.2)·3.5 = 23.2
ŷ = 23.2 − 1.2·x
Solve for x
x = (20 − 23.2)/(−1.2) = 2.667
Answer
x = 2.667
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