Source: Midterm Make Up Assignment S26-4-1-1
Tags: normal distribution, standard deviation, variance, confidence interval, z-score, z-table, central tendency, mean, median, mode, percentile, population parameter, sample statistic
This set of topics covers how normal distributions scale when data is stretched or compressed, how to build and reverse-engineer confidence intervals using z-scores, how to find percentile cutoffs from a z-table, and what "central tendency" means. These are the statistical backbone of the midterm.
Normal distribution
A symmetric, bell-shaped probability distribution fully described by its mean (μ) and standard deviation (σ). Written as X ~ N(μ, σ²).
Variance (σ²)
The average squared deviation from the mean. The square of the standard deviation.
Standard deviation (σ)
The square root of the variance. Measures spread in the same units as the data.
Z-score
The number of standard deviations a value sits from the mean: z = (x − μ) / σ. Converts any normal distribution to the standard normal N(0, 1).
Confidence interval (CI)
A range of values, computed from sample data, that is expected to contain the true population parameter with a given probability (the confidence level).
Standard error (SE)
The standard deviation of the sampling distribution of the mean: SE = σ / √n. Measures how much the sample mean is expected to vary from sample to sample.
Central tendency
A single value that represents the "centre" of a dataset. The three main measures are the mean, the median, and the mode.
If every data point's distance from the mean is multiplied by a constant c, the standard deviation is also multiplied by c, and the variance is multiplied by c².
Formally: if (Xᵢ − μ)_exp = c × (Xᵢ − μ)_common, then:
σ_exp = c × σ_common
σ²_exp = c² × σ²_common
Given: common population has mean 58 cm and variance 78 cm². Experimental population has the same mean, but each stem's distance from the mean is 3 times the common distance.
σ_common = √78 ≈ 8.83 cm
σ_exp = 3 × 8.83 = 26.50 cm
That is all there is to it. The scaling factor applies directly to the standard deviation.
A confidence interval for a population mean (when σ is known) takes the form:
x̄ ± z_(α/2) × (σ / √n)
where α = 1 − confidence level.
To find just the lower bound:
Lower bound = x̄ − z_(α/2) × (σ / √n)
Given: n = 176, x̄ = 140 mmHg, σ = 34 mmHg, confidence level = 85%.
Step 1: Find α and z.
α = 0.15, so α/2 = 0.075.
From the z-table, P(Z < z) = 0.925 gives z ≈ 1.44.
Step 2: Compute the standard error.
SE = 34 / √176 = 34 / 13.27 ≈ 2.56
Step 3: Lower bound.
Lower = 140 − 1.44 × 2.56 = 140 − 3.69 = 136.31
Sometimes the question gives you the lower bound and asks for the sample mean. Rearrange:
x̄ = lower bound + z_(α/2) × (σ / √n)
Given: n = 26, σ = 4.4, confidence level = 92%, lower bound = 50.
Step 1: α = 0.08, α/2 = 0.04. From the z-table, P(Z < z) = 0.96 gives z ≈ 1.75.
Step 2: SE = 4.4 / √26 ≈ 0.863.
Step 3: x̄ = 50 + 1.75 × 0.863 = 50 + 1.51 ≈ 51.51, which rounds to 52.
When a problem gives you a percentage and asks for the boundary value, you are looking up a percentile on the standard normal, then converting back to the original scale:
x = μ + z × σ
The trick is figuring out which tail the percentage refers to.
Given: air filter catches 90% of particles. Mean diameter = 0.5 μm, σ = 0.2 μm. Find the largest diameter that passes through.
The filter catches the largest 90%, so the 10% that pass through are the smallest. The cutoff is the 10th percentile.
Step 1: P(X ≤ x) = 0.10, so z = −1.28 (from z-table, since P(Z < −1.28) ≈ 0.10).
Step 2: x = 0.5 + (−1.28)(0.2) = 0.5 − 0.256 = 0.244 μm.
These come up repeatedly and are worth memorising:
Confidence level | α/2 | z |
|---|---|---|
80% | 0.10 | 1.28 |
85% | 0.075 | 1.44 |
90% | 0.05 | 1.645 |
92% | 0.04 | 1.75 |
95% | 0.025 | 1.96 |
99% | 0.005 | 2.576 |
The three standard measures are:
Mean: the arithmetic average of all values.
Median: the middle value when data is sorted.
Mode: the most frequently occurring value.
All three describe the "centre" of a dataset, though they can differ substantially for skewed distributions.
The exam statement "means, modes, and medians are all descriptors of central tendency" is True.
Confidence interval (known σ):
x̄ ± z_(α/2) × (σ / √n)
Standard error:
SE = σ / √n
Z-score:
z = (x − μ) / σ
Percentile cutoff:
x = μ + z × σ
Variance scaling:
If Y = c(X − μ) + μ, then σ_Y = |c| × σ_X
⚠️ When scaling distances from the mean by a factor c, the standard deviation scales by c and the variance scales by c². Do not accidentally scale the variance by c instead of c².
⚠️ For confidence intervals, always check whether the question asks for the lower bound, upper bound, or both. Computing the wrong one costs full marks.
⚠️ The z-value depends on the confidence level. Double-check α/2 before looking up the table. A 92% CI uses z ≈ 1.75, not 1.96 (which is for 95%).
⚠️ For the air filter problem, think about which tail the "passing through" particles occupy. The filter catches the big ones, so the ones that escape are in the lower tail (small diameters).
⚠️ "Population standard deviation" vs. "sample standard deviation" matters. If σ is given as the population value, use z. If you only have a sample standard deviation s, you would use t (though these midterm problems all give σ).
⚠️ Rounding instructions vary by question. Some ask for 2 decimal places, some for 0, some for 3. Read each question's instructions.
Q: A population has variance 50. If each observation's distance from the mean is doubled, what is the new variance?
A: New variance = 2² × 50 = 200.
Q: For a 90% confidence interval, what z-value do you use?
A: z = 1.645 (since α/2 = 0.05, and P(Z < 1.645) = 0.95).
Q: A sample of 100 has mean 75, population σ = 10. What is the upper bound of a 95% CI?
A: Upper = 75 + 1.96 × (10/√100) = 75 + 1.96 × 1 = 76.96.
Q: Scores are normally distributed with mean 500 and σ = 100. What score marks the 95th percentile?
A: z = 1.645 for the 95th percentile. x = 500 + 1.645 × 100 = 664.5.
Q: Is the mode always a measure of central tendency?
A: Yes. Mean, median, and mode are all classified as measures of central tendency, though the mode may not be near the centre in heavily skewed or multimodal distributions.
Q: A 88% confidence interval has a lower bound of 30. If n = 36 and σ = 6, what is the sample mean?
A: α = 0.12, α/2 = 0.06, z ≈ 1.555. SE = 6/√36 = 1. x̄ = 30 + 1.555 × 1 = 31.555 ≈ 31.56.
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