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: 54, 29, 21, 27, 42, 51.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 210?
μ = 170σ = 20x = 210
P =
3.ProbabilityMedium
A bag holds 8 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 8
P =
4.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?
n = 6p = 0.4k = 1
P =
5.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.4647. Find z to two decimals.
A = 0.4647
z =
6.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 32σ = 6P = 97.5%
x = min
7.Confidence intervalsMedium
The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 1.3 and standard deviation 4.1. Find the margin of error E of the 99% confidence interval for the mean difference.
n = 8d̄ = 1.3sd = 4.199%t(7) = 3.499
E = mmHg
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 9 plots, mean yield 81.5 kg, standard deviation 4 kg. Plot group 2: 7 plots, mean 82 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
n₁ = 9x̄₁ = 81.5s₁ = 4n₂ = 7x̄₂ = 82s₂ = 390%t(14) = 1.761
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 124.8 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 124.8s = 15n = 81
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 199. The population standard deviation is 4. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 199σ = 4n = 4995%
U =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (8, 61) is made. Find its residual e = y − ŷ.
n = 5(8, 61)
e =
3Solutions
1.Descriptive statistics
Answer: s = 13.63
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 = 224/6 = 37.33
Squared deviations
Σ(x − x̄)² = 16.67² + (−8.333)² + (−16.33)² + (−10.33)² + 4.667² + 13.67² = 929.3
Standard deviation
s = √(929.3/5) = 13.63
Answer
s = 13.63
2.Descriptive statistics
Answer: P = 0.9772
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
3.Probability
Answer: P = 0.2333
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 = 8/16 · 7/15 = 56/240 = 0.2333
Answer
P = 0.2333
4.Probability
Answer: P = 0.1866
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 = 1) = 6·0.41·0.65 = 6·0.4·0.07776 = 0.1866
Answer
P = 0.1866
5.Normal distribution
Answer: z = 0.62
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 + 0.4647)/2 = 0.7324
Read the z-table backwards
Find 0.7324 inside the table and read off z.
z = 0.62
Answer
z = 0.62
6.Normal distribution
Answer: x = 43.76 min
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 for the percentile
Φ(z) = 0.975 → z = 1.96
Back to x
x = 32 + 1.96·6 = 43.76 min
Answer
x = 43.76 min
7.Confidence intervals
Answer: E = 5.072 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 = 7, t = 3.499
Margin of error
E = 3.499·4.1/√8 = 5.072
Answer
E = 5.072 mmHg
8.Confidence intervals
Answer: upper limit = 2.7 kg
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.
Pooled standard deviation
sp² = (8·4² + 6·3²)/14 = 13
sp = 3.606
Margin of error
E = 1.761·3.606·√(1/9 + 1/7) = 3.2
Interval
−0.5 ± 3.2 → [−3.7, 2.7]
Answer
upper limit = 2.7 kg
9.Hypothesis tests & confidence intervals
Answer: t = 2.88
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/√81 = 15/9 = 1.667
Test statistic
t = (124.8 − 120)/1.667 = 2.88
Answer
t = 2.88
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 200.1
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 = 1.96·4/√49 = 1.12
Interval limit
U = 199 + 1.12 = 200.1
Answer
U = 200.1
11.Linear regression & correlation
Answer: r = 0.9832
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 = 461/6 = 76.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 433
Syy = Σ(y − ȳ)² = 2771
Correlation coefficient
r = 433/√(70·2771) = 0.9832
Answer
r = 0.9832
12.Linear regression & correlation
Answer: e = −3.35
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 = 25/5 = 5
ȳ = 279/5 = 55.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 114
Regression line
b = 114/40 = 2.85
a = 55.8 − 2.85·5 = 41.55
ŷ = 41.55 + 2.85·x
Predicted value
ŷ = 41.55 + 2.85·8 = 64.35
Residual
e = 61 − 64.35 = −3.35
Answer
e = −3.35
The point lies below the line.
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