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: 56, 35, 20, 36, 29, 12, 29.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 34?
μ = 50σ = 8x = 34
P =
3.ProbabilityMedium
A bag holds 7 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 7
P =
4.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 4 correct answers?
n = 12p = 0.25k = 4
P =
5.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 480 and 550 h? Give P as a decimal.
μ = 500σ = 40x1 = 480x2 = 550
P =
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 battery life of two phone models is compared. Model 1: 100 phones, mean 106 h, known standard deviation 20 h. Model 2: 30 phones, mean 99 h, known standard deviation 13 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
n₁ = 100x̄₁ = 106 hσ₁ = 20 hn₂ = 30x̄₂ = 99 hσ₂ = 13 h99%
upper limit = h
8.Confidence intervalsHard
The diameters of 25 machined shafts have a sample standard deviation of 2.4 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.
n = 25s = 2.4 µmχ² = 12.401, 39.364
σ = µm
9.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 60 parts has mean length 125.5 mm, and the line's known standard deviation is 13 mm. Line 2: a sample of 60 parts has mean 131.4 mm, known standard deviation 8 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
n₁ = 60x̄₁ = 125.5 mmσ₁ = 13 mmn₂ = 60x̄₂ = 131.4 mmσ₂ = 8 mm
z =
10.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 120.1. The population standard deviation is 5. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 120.1σ = 5n = 4990%
U =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. 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 = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 33?
n = 6ŷ = 33
x =
3Solutions
1.Descriptive statistics
Answer: s = 13.88
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 = 217/7 = 31
Squared deviations
Σ(x − x̄)² = 25² + 4² + (−11)² + 5² + (−2)² + (−19)² + (−2)² = 1156
Standard deviation
s = √(1156/6) = 13.88
Answer
s = 13.88
2.Descriptive statistics
Answer: P = 0.02275
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) = 0.02275
P = 0.02275
Answer
P = 0.02275
3.Probability
Answer: P = 0.2308
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 = 7/14 · 6/13 = 42/182 = 0.2308
Answer
P = 0.2308
4.Probability
Answer: P = 0.1936
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 = 4) = 495·0.254·0.758 = 495·0.003906·0.1001 = 0.1936
Answer
P = 0.1936
5.Normal distribution
Answer: P = 0.5858
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 = (480 − 500)/40 = −0.5
z₂ = (550 − 500)/40 = 1.25
Area between
P = Φ(1.25) − Φ(−0.5) = 0.8944 − 0.3085 = 0.5858
Answer
P = 0.5858
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: upper limit = 15 h
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.
Point estimate
x̄₁ − x̄₂ = 106 − 99 = 7
Standard error
SE = √(20²/100 + 13²/30) = 3.104
Margin of error
E = 2.576·3.104 = 7.995
Interval
7 ± 7.995 → [−0.9953, 15]
Answer
upper limit = 15 h
The interval contains 0, so the data do not show a significant difference.
8.Confidence intervals
Answer: σ = 1.874 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 24·2.4² = 138.2
Limits for σ²
138.2/39.364 = 3.512
138.2/12.401 = 11.15
Limit for σ
σ = √3.512 = 1.874 µm
Answer
σ = 1.874 µm
95% interval for σ: [1.874, 3.339] µm.
9.Hypothesis tests & confidence intervals
Answer: z = −2.994
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Standard error of the difference
SE = √(13²/60 + 8²/60) = √3.883 = 1.971
Test statistic
z = (125.5 − 131.4)/1.971 = −2.994
Answer
z = −2.994
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 121.3
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 = 1.645·5/√49 = 1.175
Interval limit
U = 120.1 + 1.175 = 121.3
Answer
U = 121.3
11.Linear regression & correlation
Answer: r² = 0.9833
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 = 147/6 = 24.5
ȳ = 789/6 = 131.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 748.5
Syy = Σ(y − ȳ)² = 3618
Coefficient of determination
r² = 748.5²/(157.5·3618) = 0.9833
Answer
r² = 0.9833
About 98.33% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 2.023
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 = 36/6 = 6
ȳ = 348/6 = 58
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 440
Regression line
b = 440/70 = 6.286
a = 58 − 6.286·6 = 20.29
ŷ = 20.29 + 6.286·x
Solve for x
x = (33 − 20.29)/6.286 = 2.023
Answer
x = 2.023
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