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: 49, 17, 55, 38, 14, 59, 34.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 180.
μ = 170σ = 10x = 180
z =
3.ProbabilityMedium
A bag holds 4 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 4
P =
4.ProbabilityHard
7 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?
n = 7p = 0.5
P =
5.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0475. Find z to two decimals.
A = 0.0475
z =
6.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points70%
x = points
7.Confidence intervalsMedium
A sample of 25 items has mean 74 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 25x̄ = 74s = 7.595%t(24) = 2.064
upper limit =
8.Confidence intervalsHard
The population standard deviation is 37. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?
σ = 37E = 1.590%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 208.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 208.7s = 12n = 16
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 121.7. The population standard deviation is 5. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 121.7σ = 5n = 2590%
L =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
12.Linear regression & correlationHard
The table gives 5 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 = 20?
n = 5ŷ = 20
x =
3Solutions
1.Descriptive statistics
Answer: s = 17.72
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 = 266/7 = 38
Squared deviations
Σ(x − x̄)² = 11² + (−21)² + 17² + 0² + (−24)² + 21² + (−4)² = 1884
Standard deviation
s = √(1884/6) = 17.72
Answer
s = 17.72
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 = (180 − 170)/10 = 1
Answer
z = 1
3.Probability
Answer: P = 0.2143
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 = 4/8 · 3/7 = 12/56 = 0.2143
Answer
P = 0.2143
4.Probability
Answer: P = 0.0625
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) = 0.57 = 0.007813
P(X = 1) = 7·0.5·0.56 = 0.05469
Add them
P = 0.007813 + 0.05469 = 0.0625
Answer
P = 0.0625
5.Normal distribution
Answer: z = 1.67
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 − 0.0475 = 0.9525
Read the z-table backwards
Find 0.9525 inside the table and read off z.
z = 1.67
Answer
z = 1.67
6.Normal distribution
Answer: x = 59.8 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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 65 − 0.52·10 = 59.8 points
Answer
x = 59.8 points
7.Confidence intervals
Answer: upper limit = 77.1
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 = 24, t = 2.064
Margin of error
E = 2.064·7.5/√25 = 3.096
Interval
74 ± 3.096 → [70.9, 77.1]
Answer
upper limit = 77.1
8.Confidence intervals
Answer: n = 1647
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 = (1.645·37/1.5)² = 1646
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1647
Answer
n = 1647
9.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 12/√16 = 12/4 = 3
Test statistic
t = (208.7 − 200)/3 = 2.9
Answer
t = 2.9
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: L = 120.1
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).
Margin of error
E = 1.645·5/√25 = 1.645
Interval limit
L = 121.7 − 1.645 = 120.1
Answer
L = 120.1
11.Linear regression & correlation
Answer: r = 0.9132
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 = 120/6 = 20
ȳ = 336/6 = 56
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 184
Syy = Σ(y − ȳ)² = 580
Correlation coefficient
r = 184/√(70·580) = 0.9132
Answer
r = 0.9132
12.Linear regression & correlation
Answer: x = 4.364
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 = 15/5 = 3
ȳ = 115/5 = 23
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22
Regression line
b = −22/10 = −2.2
a = 23 − (−2.2)·3 = 29.6
ŷ = 29.6 − 2.2·x
Solve for x
x = (20 − 29.6)/(−2.2) = 4.364
Answer
x = 4.364
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