Source: HW Assignment 4, Experimental Physics and Engineering Lab
Tags: standard normal distribution, z-score, z-table, normal curve area, inverse z-score, probability, bell curve, Gaussian distribution, PHYS-216
The standard normal distribution has mean μ = 0 and standard deviation σ = 1, and you use a z-table to find the area (probability) under the curve between any two z-values. When working with a non-standard normal distribution, convert raw values to z-scores first, then use the table. Inverse problems work the other way round: start with a probability, find the z-score from the table, then convert back to the raw value.
Standard normal distribution
A normal distribution with mean μ = 0 and standard deviation σ = 1. All other normal distributions can be converted to this one using the z-score formula.
Z-score (standard score)
The number of standard deviations a data point sits from the mean. Tells you where a value falls on the standard normal curve.
Z-table (standard normal table)
A reference table that gives the cumulative area under the standard normal curve to the left of a given z-value. The area represents probability.
Cumulative probability
The probability that a random variable takes a value less than or equal to a given point, i.e. P(Z ≤ z). This is what the z-table directly provides.
Inverse z-score
Working backwards from a known probability to find the corresponding z-value, and then converting to the raw score x.
The z-score transforms any normally distributed variable into standard normal form:
z = (x – μ) / σ
x = the raw value you are interested in
μ = the population mean
σ = the population standard deviation
Once you have z, look it up in the z-table to find cumulative probability P(Z ≤ z).
The z-table gives P(Z ≤ z), the area to the left of z.
Area to the left of z: read directly from the table.
Area to the right of z: calculate as 1 – P(Z ≤ z).
Area between two z-values: subtract the smaller cumulative area from the larger one.
These problems give you μ, σ, and a value x, then ask for a probability.
Problem 2 pattern (P(X ≥ x)): Given μ = 85, σ = 10, find P(X ≥ 70).
Calculate z: z = (70 – 85) / 10 = –1.5
The table gives P(Z ≤ –1.5) = 0.0668
Since we want P(X ≥ 70), take the right tail: 1 – 0.0668 = 0.9332
Answer: 93.32%
Problem 3 pattern (P(X ≥ x)): Given μ = 63, σ = 1.25, find P(X ≥ 64).
Calculate z: z = (64 – 63) / 1.25 = 0.8
P(Z ≤ 0.8) = 0.7881
P(X ≥ 64) = 1 – 0.7881 = 0.2119
Answer: 21.19%
When the distribution is already standard normal (μ = 0, σ = 1), skip the z-score formula and go straight to the table.
(a) Area between z = 0 and z = 2: P(Z ≤ 2) – P(Z ≤ 0) = 0.9772 – 0.5000 = 0.4772
(b) Area to the right of z = 1.26: 1 – P(Z ≤ 1.26) = 1 – 0.8962 = 0.1038
(c) Area to the left of z = –0.80: read directly from table = 0.2119
(d) Area between z = –1.47 and z = 2.39: P(Z ≤ 2.39) – P(Z ≤ –1.47) = 0.9917 – 0.0708 = 0.9209
These problems give you a probability and ask you to find the raw value x.
The reverse formula is:
x = μ + z · σ
Find z from the z-table by locating the probability first, then plug in.
Problem 4 pattern (left-tail probability given): Given μ = 4600, σ = 250, P(X ≤ x) = 0.03.
Look up 0.03 in the body of the z-table to find z ≈ –1.88
x = 4600 + (–1.88 × 250) = 4600 – 470 = 4130 hours
Problem 5 pattern (right-tail probability given): Given μ = 225, σ = 50, P(X > x) = 0.05.
Right-tail probability of 0.05 means left-tail probability of 0.95
Look up 0.95 in the z-table to find z ≈ 1.645
x = 225 + (1.645 × 50) = 225 + 82.25 = 307.25
Z-score (forward): z = (x – μ) / σ
Raw score (inverse): x = μ + z · σ
Right-tail probability: P(X ≥ x) = 1 – P(Z ≤ z)
Area between two z-values: P(a ≤ Z ≤ b) = P(Z ≤ b) – P(Z ≤ a)
⚠️ The z-table gives left-tail area P(Z ≤ z). If the question asks for "greater than" or "at least," you need to subtract from 1. Forgetting this is the most common mistake.
⚠️ For inverse problems, check whether the probability given is left-tail or right-tail before looking up the z-value. A right-tail probability of 0.05 corresponds to a left-tail probability of 0.95.
⚠️ A negative z-score means the value is below the mean. A positive z-score means above. The sign matters when plugging back into x = μ + zσ.
⚠️ Problems 2 and 3 both use the same method (forward z-score, right-tail), just with different context (quiz grades vs. dispensed amounts). Recognise the pattern, not the story.
⚠️ Standard normal problems (Problem 1) skip the z-formula entirely because z = x when μ = 0 and σ = 1.
Q: A population has μ = 100 and σ = 15. What is the probability that a randomly selected value exceeds 130?
A: z = (130 – 100) / 15 = 2.0. P(Z ≤ 2.0) = 0.9772. P(X > 130) = 1 – 0.9772 = 0.0228, or about 2.28%.
Q: For a standard normal distribution, what is the area between z = –1.0 and z = 1.0?
A: P(Z ≤ 1.0) – P(Z ≤ –1.0) = 0.8413 – 0.1587 = 0.6826. About 68.26% of values fall within one standard deviation of the mean.
Q: Given μ = 500 and σ = 100, find x such that P(X ≤ x) = 0.10.
A: Look up 0.10 in the z-table body: z ≈ –1.28. x = 500 + (–1.28 × 100) = 500 – 128 = 372.
Q: Why do you subtract from 1 when the problem says "at least" or "greater than"?
A: Because the z-table gives P(Z ≤ z), the area to the left. "At least x" means the area to the right, which is the complement: 1 – P(Z ≤ z).
Q: A right-tail probability is 0.05. What z-score corresponds to this?
A: Right-tail 0.05 means left-tail 0.95. From the z-table, z ≈ 1.645.
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