Difficulty: Intermediate | Prerequisites: Basic probability, z-score calculation, standard normal table familiarity
The normal distribution is the single most important probability distribution in statistics. It models continuous random variables where values cluster symmetrically around a mean, with the spread determined by the standard deviation. Nearly every inferential technique you will meet later in the course (confidence intervals, hypothesis tests) relies on normality assumptions. You should already be comfortable computing z-scores and reading a standard normal (z) table before working through this material.
These problems practise converting raw values to z-scores, looking up areas under the standard normal curve, and applying conditional probability within a normal model. If you can standardise a value and read a z-table, you can answer all of them.
Normal distribution (Gaussian distribution)
A continuous probability distribution defined by its mean (μ) and standard deviation (σ), producing the familiar bell-shaped curve. In simple terms, it describes data that clusters around an average, with most values falling close to the centre and fewer appearing as you move further out.
Standard deviation (σ)
A measure of how spread out values are from the mean. Think of it as the average distance a data point sits from the centre of the distribution.
Z-score (standard score)
The number of standard deviations a raw value sits above or below the mean, calculated as z = (X − μ) / σ. In simple terms, it translates any normal distribution into the standard normal so you can use one universal table.
Standard normal distribution
A normal distribution with μ = 0 and σ = 1. Every normal distribution can be converted to this one via z-scores, which is why z-tables work.
Conditional probability
The probability of an event occurring given that another event has already occurred, written P(A | B) = P(A ∩ B) / P(B). Think of it as narrowing the sample space to only the outcomes where B is true, then asking how likely A is within that smaller space.
Empirical rule (68-95-99.7 rule)
Roughly 68% of data falls within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ. This is a quick mental shortcut for estimating areas without a table.
Use the formula: z = (X − μ) / σ
A positive z means the value is above the mean; a negative z means it is below.
Once you have the z-score, look up the cumulative probability in a standard normal table.
Standardise both endpoints to z-scores.
Look up each cumulative probability from the z-table.
Subtract the smaller cumulative probability from the larger one.
Worked example (Q1, lightbulb lifespan):
μ = 1450 hours, σ = 8.5 hours
P(1440 < X < 1465):
z₁ = (1440 − 1450) / 8.5 = −1.18
z₂ = (1465 − 1450) / 8.5 = 1.76
P(z < 1.76) ≈ 0.9608
P(z < −1.18) ≈ 0.1190
P(1440 < X < 1465) = 0.9608 − 0.1190 ≈ 0.8415
Answer: (3) 0.8415
Standardise the value to a z-score.
The probability above that z is 1 minus the cumulative probability from the table.
Worked example (Q2a, test results):
μ = 78, σ = 36
P(X > 72):
z = (72 − 78) / 36 = −0.1667
P(z < −0.17) ≈ 0.4325
P(X > 72) = 1 − 0.4325 ≈ 0.5675
When a condition restricts the range (e.g., "given that the grade is above 72"), the denominator becomes the probability of that condition.
Use P(A | B) = P(A ∩ B) / P(B).
Worked example (Q2b, conditional grade probability):
P(X > 84 | X > 72):
z for 84: (84 − 78) / 36 = 0.1667
P(X > 84) = 1 − P(z < 0.17) ≈ 1 − 0.5675 = 0.4325
P(X > 72) ≈ 0.5675 (from part a)
P(X > 84 | X > 72) = 0.4325 / 0.5675 ≈ 0.7621
Same method as "probability between two values," but express the result as a percentage.
Worked example (Q3, women's heights):
μ = 64 inches, σ = 2.75 inches
P(64 < X < 69.5):
z₁ = (64 − 64) / 2.75 = 0
z₂ = (69.5 − 64) / 2.75 = 2.0
P(z < 2.0) ≈ 0.9772
P(z < 0) = 0.5000
P(64 < X < 69.5) = 0.9772 − 0.5000 = 0.4772 ≈ 48%
Answer: (2) 48
Z-score: z = (X − μ) / σ
Probability between two values: P(a < X < b) = P(z < z_b) − P(z < z_a)
Conditional probability: P(A | B) = P(A ∩ B) / P(B)
Normal distribution models appear everywhere: quality control in manufacturing (lightbulb lifespans, as in Q1), standardised test scoring, biological measurements like height and blood pressure, and financial modelling. Insurance companies, for instance, rely heavily on normal distribution assumptions when pricing policies.
Students often forget to subtract cumulative probabilities when finding the area between two z-scores. You need both endpoints, not just one.
A common error is dividing by the variance (σ²) instead of the standard deviation (σ) when computing z-scores. The formula uses σ, not σ².
For conditional probability problems, students frequently compute P(X > 84) on its own and forget to divide by P(X > 72). The "given that" phrase always signals a restricted denominator.
Some students assume the empirical rule (68-95-99.7) gives exact answers. It is an approximation; for precise probabilities, use the z-table.
⚠️ Z-score calculation is nearly guaranteed on any statistics exam. Be able to do it quickly and without a formula sheet prompt.
⚠️ Conditional probability within normal distributions is a favourite exam question because it tests two concepts at once.
⚠️ "Between two values" problems require two z-lookups and a subtraction. Forgetting the subtraction is the most common mark lost.
⚠️ Watch for problems where σ seems unusually large relative to μ (like Q2, where σ = 36 and μ = 78). This does not change the method, but it can produce z-values close to zero that feel "wrong." Trust the formula.
True or False: A z-score of −2.0 means the value is two standard deviations below the mean.
Fill in the blank: To find P(a < X < b), you compute P(z < z_b) ______ P(z < z_a).
True or False: P(X > 80 | X > 70) equals P(X > 80).
Fill in the blank: The standard normal distribution has a mean of ______ and a standard deviation of ______.
True or False: About 95% of data in a normal distribution falls within one standard deviation of the mean.
Answers: 1. True. 2. minus (subtract). 3. False, you must divide P(X > 80) by P(X > 70). 4. 0, 1. 5. False, 95% falls within two standard deviations; one standard deviation captures about 68%.
Q: A lightbulb's lifespan is normally distributed with μ = 1450 and σ = 8.5. What is P(1440 < X < 1465)?
A: Compute z₁ = (1440 − 1450)/8.5 = −1.18 and z₂ = (1465 − 1450)/8.5 = 1.76. Then P = P(z < 1.76) − P(z < −1.18) ≈ 0.9608 − 0.1190 = 0.8415.
Q: Test scores follow N(78, 36). What is the probability a student scores above 72?
A: z = (72 − 78)/36 = −0.167. P(X > 72) = 1 − P(z < −0.17) ≈ 1 − 0.4325 = 0.5675.
Q: Given that a student scored above 72, what is the probability the score exceeds 84?
A: P(X > 84 | X > 72) = P(X > 84) / P(X > 72). Compute z for 84: (84 − 78)/36 = 0.167. P(X > 84) ≈ 0.4325. So P = 0.4325 / 0.5675 ≈ 0.762.
Q: Heights of women are N(64, 2.75). What percentage have heights between 64 and 69.5 inches?
A: z₁ = 0, z₂ = (69.5 − 64)/2.75 = 2.0. P = P(z < 2.0) − P(z < 0) = 0.9772 − 0.5 = 0.4772, so about 48%.
This material connects directly to confidence intervals and hypothesis testing, which use the same z-score mechanics but in reverse (starting from a desired probability and solving for the critical value). It also ties into the Central Limit Theorem, which explains why sample means are approximately normally distributed regardless of the population shape.
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