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: 34, 11, 16, 60, 49, 20, 20.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 95.
μ = 100σ = 5x = 95
z =
3.ProbabilityMedium
A bag holds 8 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 3
P =
4.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 3 correct answers?
n = 10p = 0.2k = 3
P =
5.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7939. Find z to two decimals.
A = 0.7939
z =
6.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 15% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm15%
x = cm
7.Confidence intervalsMedium
A sample of 21 items has mean 74 and standard deviation s = 6. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 21x̄ = 74s = 695%t(20) = 2.086
upper limit =
8.Confidence intervalsHard
The diameters of 11 machined shafts have a sample standard deviation of 4.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.
n = 11s = 4.6 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 79.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 79.5s = 15n = 49
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 54. The population standard deviation is 10. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 54σ = 10n = 4990%
L =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. 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 = 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 = 16?
n = 6ŷ = 16
x =
3Solutions
1.Descriptive statistics
Answer: s = 18.41
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 = 210/7 = 30
Squared deviations
Σ(x − x̄)² = 4² + (−19)² + (−14)² + 30² + 19² + (−10)² + (−10)² = 2034
Standard deviation
s = √(2034/6) = 18.41
Answer
s = 18.41
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 = (95 − 100)/5 = −1
Answer
z = −1
3.Probability
Answer: P = 0.5091
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 = 8/11 · 7/10 = 56/110 = 0.5091
Answer
P = 0.5091
4.Probability
Answer: P = 0.2013
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) = 120·0.23·0.87 = 120·0.008·0.2097 = 0.2013
Answer
P = 0.2013
5.Normal distribution
Answer: z = −0.82
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 − 0.7939 = 0.2061
Read the z-table backwards
Find 0.2061 inside the table and read off z.
z = −0.82
Answer
z = −0.82
6.Normal distribution
Answer: x = 185.3 cm
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 = 178 + 1.04·7 = 185.3 cm
Answer
x = 185.3 cm
7.Confidence intervals
Answer: upper limit = 76.73
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.
Critical value
df = n − 1 = 20, t = 2.086
Margin of error
E = 2.086·6/√21 = 2.731
Interval
74 ± 2.731 → [71.27, 76.73]
Answer
upper limit = 76.73
8.Confidence intervals
Answer: σ = 3.214 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 = 10·4.6² = 211.6
Limits for σ²
211.6/20.483 = 10.33
211.6/3.247 = 65.17
Limit for σ
σ = √10.33 = 3.214 µm
Answer
σ = 3.214 µm
95% interval for σ: [3.214, 8.073] µm.
9.Hypothesis tests & confidence intervals
Answer: t = 2.1
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).
Standard error
SE = 15/√49 = 15/7 = 2.143
Test statistic
t = (79.5 − 75)/2.143 = 2.1
Answer
t = 2.1
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: L = 51.65
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 = 1.645·10/√49 = 2.35
Interval limit
L = 54 − 2.35 = 51.65
Answer
L = 51.65
11.Linear regression & correlation
Answer: r² = 0.8411
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 = 33/6 = 5.5
ȳ = 254/6 = 42.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 39
Syy = Σ(y − ȳ)² = 103.3
Coefficient of determination
r² = 39²/(17.5·103.3) = 0.8411
Answer
r² = 0.8411
About 84.11% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 5.639
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 = 118/6 = 19.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30
Regression line
b = −30/17.5 = −1.714
a = 19.67 − (−1.714)·3.5 = 25.67
ŷ = 25.67 − 1.714·x
Solve for x
x = (16 − 25.67)/(−1.714) = 5.639
Answer
x = 5.639
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