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: 59, 34, 34, 35, 52.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is above 66?
μ = 70σ = 8x = 66
P =
3.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 12? The order does not matter.
n = 12k = 5
N =
4.ProbabilityHard
12 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 12p = 0.1
P =
5.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 77.5 and 103.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 77.5x2 = 103.75
P =
6.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 65σ = 10P = 10%
x = points
7.Confidence intervalsMedium
A sample of 11 items has mean 37.5 and standard deviation s = 2. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 11x̄ = 37.5s = 295%t(10) = 2.228
lower limit =
8.Confidence intervalsHard
The population standard deviation is 26. 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 4?
σ = 26E = 495%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 76.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 76.4s = 8n = 16
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 499.7. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 499.7σ = 4n = 1699%
L =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 18) is made. Find its residual e = y − ŷ.
n = 6(11, 18)
e =
3Solutions
1.Descriptive statistics
Answer: s = 11.86
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 = 214/5 = 42.8
Squared deviations
Σ(x − x̄)² = 16.2² + (−8.8)² + (−8.8)² + (−7.8)² + 9.2² = 562.8
Standard deviation
s = √(562.8/4) = 11.86
Answer
s = 11.86
2.Descriptive statistics
Answer: P = 0.6915
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
3.Probability
Answer: N = 792
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(12, 5) = (12·11·10·9·8)/(5·4·3·2·1) = 792
Answer
N = 792
4.Probability
Answer: P = 0.659
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) = 0.912 = 0.2824
P(X = 1) = 12·0.1·0.911 = 0.3766
Add them
P = 0.2824 + 0.3766 = 0.659
Answer
P = 0.659
5.Normal distribution
Answer: P = 0.5319
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 = (77.5 − 100)/15 = −1.5
z₂ = (103.75 − 100)/15 = 0.25
Area between
P = Φ(0.25) − Φ(−1.5) = 0.5987 − 0.06681 = 0.5319
Answer
P = 0.5319
6.Normal distribution
Answer: x = 52.18 points
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) = 0.1 → z = −1.282
Back to x
x = 65 + −1.282·10 = 52.18 points
Answer
x = 52.18 points
7.Confidence intervals
Answer: lower limit = 36.16
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 = 10, t = 2.228
Margin of error
E = 2.228·2/√11 = 1.344
Interval
37.5 ± 1.344 → [36.16, 38.84]
Answer
lower limit = 36.16
8.Confidence intervals
Answer: n = 163
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 = (1.96·26/4)² = 162.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 163
Answer
n = 163
9.Hypothesis tests & confidence intervals
Answer: t = 0.7
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
SE = 8/√16 = 8/4 = 2
Test statistic
t = (76.4 − 75)/2 = 0.7
Answer
t = 0.7
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: L = 497.1
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 = 2.576·4/√16 = 2.576
Interval limit
L = 499.7 − 2.576 = 497.1
Answer
L = 497.1
11.Linear regression & correlation
Answer: r² = 0.954
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 = 110/5 = 22
ȳ = 306/5 = 61.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 164
Syy = Σ(y − ȳ)² = 704.8
Coefficient of determination
r² = 164²/(40·704.8) = 0.954
Answer
r² = 0.954
About 95.4% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = −4.443
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 = 87/6 = 14.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 139
Regression line
b = 139/70 = 1.986
a = 14.5 − 1.986·7 = 0.6
ŷ = 0.6 + 1.986·x
Predicted value
ŷ = 0.6 + 1.986·11 = 22.44
Residual
e = 18 − 22.44 = −4.443
Answer
e = −4.443
The point lies below the line.
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