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: 27, 21, 57, 11, 41, 26, 28.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 78.
μ = 70σ = 8x = 78
z =
3.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 13? The order does not matter.
n = 13k = 5
N =
4.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 2 correct answers?
n = 10p = 0.25k = 2
P =
5.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23 and 33.2 min? Give P as a decimal.
μ = 32σ = 6x1 = 23x2 = 33.2
P =
6.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g5%
x = g
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 100 phones, mean 121.5 h, known standard deviation 7 h. Model 2: 40 phones, mean 123.5 h, known standard deviation 11 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
n₁ = 100x̄₁ = 121.5 hσ₁ = 7 hn₂ = 40x̄₂ = 123.5 hσ₂ = 11 h99%
E = h
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 75 kg, standard deviation 2 kg. Plot group 2: 5 plots, mean 72 kg, standard deviation 2 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
n₁ = 7x̄₁ = 75s₁ = 2n₂ = 5x̄₂ = 72s₂ = 290%t(10) = 1.812
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
16 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −6.16 and the standard deviation of the differences is 7.7. Find the paired test statistic t for H₀: μd = 0.
n = 16d̄ = −6.16sd = 7.7
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 73.7. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 73.7σ = 4n = 6499%
L =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (16, 61) is made. Find its residual e = y − ŷ.
n = 5(16, 61)
e =
3Solutions
1.Descriptive statistics
Answer: s = 14.84
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 = 211/7 = 30.14
Squared deviations
Σ(x − x̄)² = (−3.143)² + (−9.143)² + 26.86² + (−19.14)² + 10.86² + (−4.143)² + (−2.143)² = 1321
Standard deviation
s = √(1321/6) = 14.84
Answer
s = 14.84
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 = (78 − 70)/8 = 1
Answer
z = 1
3.Probability
Answer: N = 1287
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(13, 5) = (13·12·11·10·9)/(5·4·3·2·1) = 1287
Answer
N = 1287
4.Probability
Answer: P = 0.2816
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) = 45·0.252·0.758 = 45·0.0625·0.1001 = 0.2816
Answer
P = 0.2816
5.Normal distribution
Answer: P = 0.5125
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 = (23 − 32)/6 = −1.5
z₂ = (33.2 − 32)/6 = 0.2
Area between
P = Φ(0.2) − Φ(−1.5) = 0.5793 − 0.06681 = 0.5125
Answer
P = 0.5125
6.Normal distribution
Answer: x = 258.2 g
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) = 0.95
z from the table
z ≈ 1.64
Back to the original units
x = 250 + 1.64·5 = 258.2 g
Answer
x = 258.2 g
7.Confidence intervals
Answer: E = 4.83 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 = 121.5 − 123.5 = −2
Standard error
SE = √(7²/100 + 11²/40) = 1.875
Margin of error
E = 2.576·1.875 = 4.83
Answer
E = 4.83 h
The interval contains 0, so the data do not show a significant difference.
8.Confidence intervals
Answer: upper limit = 5.122 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 = (6·2² + 4·2²)/10 = 4
sp = 2
Margin of error
E = 1.812·2·√(1/7 + 1/5) = 2.122
Interval
3 ± 2.122 → [0.878, 5.122]
Answer
upper limit = 5.122 kg
9.Hypothesis tests & confidence intervals
Answer: t = −3.2
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).
Standard error of the mean difference
SE = 7.7/√16 = 1.925
Test statistic
t = −6.16/1.925 = −3.2
Answer
t = −3.2
With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.
10.Hypothesis tests & confidence intervals
Answer: L = 72.41
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 = 2.576·4/√64 = 1.288
Interval limit
L = 73.7 − 1.288 = 72.41
Answer
L = 72.41
11.Linear regression & correlation
Answer: r² = 0.9879
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 = 119/6 = 19.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 161
Syy = Σ(y − ȳ)² = 374.8
Coefficient of determination
r² = 161²/(70·374.8) = 0.9879
Answer
r² = 0.9879
About 98.79% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 2.75
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 = 95/5 = 19
ȳ = 361/5 = 72.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 186
Regression line
b = 186/40 = 4.65
a = 72.2 − 4.65·19 = −16.15
ŷ = −16.15 + 4.65·x
Predicted value
ŷ = −16.15 + 4.65·16 = 58.25
Residual
e = 61 − 58.25 = 2.75
Answer
e = 2.75
The point lies above the line.
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