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: 32, 59, 25, 36, 58, 42, 30.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 612.5?
μ = 500σ = 75x = 612.5
P =
3.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
4.ProbabilityHard
9 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 9p = 0.3
P =
5.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
P =
6.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 60% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points60%
x = points
7.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 143.5 h, known standard deviation 10 h. Model 2: 60 phones, mean 140.5 h, known standard deviation 9 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
n₁ = 60x̄₁ = 143.5 hσ₁ = 10 hn₂ = 60x̄₂ = 140.5 hσ₂ = 9 h95%
lower limit = h
8.Confidence intervalsHard
The population standard deviation is 19. 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 3?
σ = 19E = 395%
n =
9.Hypothesis tests & confidence intervalsMedium
20 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 6.04 and the standard deviation of the differences is 7.3. Find the paired test statistic t for H₀: μd = 0.
n = 20d̄ = 6.04sd = 7.3
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 1.590%
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. 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 = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 63) is made. Find its residual e = y − ŷ.
n = 5(5.5, 63)
e =
3Solutions
1.Descriptive statistics
Answer: s = 13.5
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 = 282/7 = 40.29
Squared deviations
Σ(x − x̄)² = (−8.286)² + 18.71² + (−15.29)² + (−4.286)² + 17.71² + 1.714² + (−10.29)² = 1093
Standard deviation
s = √(1093/6) = 13.5
Answer
s = 13.5
2.Descriptive statistics
Answer: P = 0.9332
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
3.Probability
Answer: P = 0.6651
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)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
4.Probability
Answer: P = 0.196
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.
Zero and one failure
P(X = 0) = 0.79 = 0.04035
P(X = 1) = 9·0.3·0.78 = 0.1556
Add them
P = 0.04035 + 0.1556 = 0.196
Answer
P = 0.196
5.Normal distribution
Answer: P = 0.7881
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (30 − 32)/2.5 = −0.8
Upper tail
P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
6.Normal distribution
Answer: x = 103.8 points
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) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 100 + 0.25·15 = 103.8 points
Answer
x = 103.8 points
7.Confidence intervals
Answer: lower limit = −0.4042 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 = 143.5 − 140.5 = 3
Standard error
SE = √(10²/60 + 9²/60) = 1.737
Margin of error
E = 1.96·1.737 = 3.404
Interval
3 ± 3.404 → [−0.4042, 6.404]
Answer
lower limit = −0.4042 h
The interval contains 0, so the data do not show a significant difference.
8.Confidence intervals
Answer: n = 155
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 = (1.96·19/3)² = 154.1
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 155
Answer
n = 155
9.Hypothesis tests & confidence intervals
Answer: t = 3.7
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 = 7.3/√20 = 1.632
Test statistic
t = 6.04/1.632 = 3.7
Answer
t = 3.7
With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.
10.Hypothesis tests & confidence intervals
Answer: n = 482
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 = (1.645·20/1.5)² = 481.1
Round up
n = 482
Answer
n = 482
11.Linear regression & correlation
Answer: r² = 0.7207
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 = 327/6 = 54.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 37.5
Syy = Σ(y − ȳ)² = 111.5
Coefficient of determination
r² = 37.5²/(17.5·111.5) = 0.7207
Answer
r² = 0.7207
About 72.07% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 5.4
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 = 281/5 = 56.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 28
Regression line
b = 28/10 = 2.8
a = 56.2 − 2.8·5 = 42.2
ŷ = 42.2 + 2.8·x
Predicted value
ŷ = 42.2 + 2.8·5.5 = 57.6
Residual
e = 63 − 57.6 = 5.4
Answer
e = 5.4
The point lies above the line.
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