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: 22, 25, 28, 42, 54, 18.
n = 6
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 20. Find the z-score z of a score of 190.
μ = 170σ = 20x = 190
z =
3.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
4.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 3 correct answers?
n = 5p = 0.7k = 3
P =
5.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.85 and 186.4 cm? Give P as a decimal.
μ = 178σ = 7x1 = 167.85x2 = 186.4
P =
6.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 90% of them lie (the 90th percentile).
μ = 250σ = 5P = 90%
x = g
7.Confidence intervalsMedium
The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −7 and standard deviation 2.2. Find the upper limit of the 90% confidence interval for the mean difference.
n = 16d̄ = −7sd = 2.290%t(15) = 1.753
upper limit = mmHg
8.Confidence intervalsHard
The population standard deviation is 18. 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 2?
σ = 18E = 295%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 489.6 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 489.6s = 20n = 25
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 295%
n =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
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. For which value x does the line predict ŷ = 25?
n = 6ŷ = 25
x =
3Solutions
1.Descriptive statistics
Answer: s = 13.74
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 = 189/6 = 31.5
Squared deviations
Σ(x − x̄)² = (−9.5)² + (−6.5)² + (−3.5)² + 10.5² + 22.5² + (−13.5)² = 943.5
Standard deviation
s = √(943.5/5) = 13.74
Answer
s = 13.74
2.Descriptive statistics
Answer: z = 1
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 = (190 − 170)/20 = 1
Answer
z = 1
3.Probability
Answer: P = 0.3056
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)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
4.Probability
Answer: P = 0.3087
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 = 3) = 10·0.73·0.32 = 10·0.343·0.09 = 0.3087
Answer
P = 0.3087
5.Normal distribution
Answer: P = 0.8114
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 = (167.85 − 178)/7 = −1.45
z₂ = (186.4 − 178)/7 = 1.2
Area between
P = Φ(1.2) − Φ(−1.45) = 0.8849 − 0.07353 = 0.8114
Answer
P = 0.8114
6.Normal distribution
Answer: x = 256.4 g
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) = 0.9 → z = 1.282
Back to x
x = 250 + 1.282·5 = 256.41 g
Answer
x = 256.41 g
7.Confidence intervals
Answer: upper limit = −6.036 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 = 15, t = 1.753
Margin of error
E = 1.753·2.2/√16 = 0.9642
Interval
−7 ± 0.9642 → [−7.964, −6.036]
Answer
upper limit = −6.036 mmHg
8.Confidence intervals
Answer: n = 312
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 = (1.96·18/2)² = 311.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 312
Answer
n = 312
9.Hypothesis tests & confidence intervals
Answer: t = −2.6
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/√25 = 20/5 = 4
Test statistic
t = (489.6 − 500)/4 = −2.6
Answer
t = −2.6
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 385
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).
Solve for n
n = (1.96·20/2)² = 384.2
Round up
n = 385
Answer
n = 385
11.Linear regression & correlation
Answer: r² = 0.9898
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 = 30/5 = 6
ȳ = 81/5 = 16.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 86
Syy = Σ(y − ȳ)² = 186.8
Coefficient of determination
r² = 86²/(40·186.8) = 0.9898
Answer
r² = 0.9898
About 98.98% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 11.16
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 = 98/6 = 16.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 146
Regression line
b = 146/70 = 2.086
a = 16.33 − 2.086·7 = 1.733
ŷ = 1.733 + 2.086·x
Solve for x
x = (25 − 1.733)/2.086 = 11.16
Answer
x = 11.16
ExamProblems · examproblems.com/courses/statistics-1