The Normal distribution flashcards

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Normal distribution N(μ, σ²)
A continuous distribution with a symmetrical bell-shaped curve, mean μ and variance σ² (standard deviation σ).
Standard Normal distribution
Z ~ N(0, 1²): the Normal distribution with mean 0 and standard deviation 1; the one with printed tables.
Φ(z)
The cumulative probability P(Z < z) for Z ~ N(0, 1²): the area under the standard Normal curve to the left of z. (Yes (table))
Percentage point of Z for probability p
The value z with P(Z > z) = p, i.e. probability p above it; e.g. p = 0.05 gives z = 1.6449. (Yes (table))
z-value (standardised value) of x
z = (x − μ)/σ: the number of standard deviations x lies above the mean (negative if below).
Standardising a Normal variable
Z = (X − μ)/σ, where Z ~ N(0, 1²) (No — learn it)
Symmetry of the standard Normal table
Φ(−z) = 1 − Φ(z) (No — derive it)
Upper-tail probability for Z
P(Z > z) = 1 − Φ(z) (No — derive it)
Probability between two values of X ~ N(μ, σ²)
P(a < X < b) = Φ((b − μ)/σ) − Φ((a − μ)/σ) (No — derive it)
Value with a given z-value
x = μ + zσ (No — derive it)
Effect on the Normal curve of increasing μ (σ fixed)
The curve slides to the right; its shape does not change.
Effect on the Normal curve of doubling σ (μ fixed)
Same centre; the curve becomes twice as wide and half as tall, so the area stays 1.
Proportions within 1, 2 and 3 standard deviations of the mean
About 68.3%, 95.4% and 99.7%.
Sketch for P(Z < −a), a > 0
Shade the left tail beyond −a; by symmetry it equals the right tail beyond a, so the area is 1 − Φ(a).
Sketch for finding μ and σ from two tails
Mark both given values on the curve with their tail areas; the lower value gets a negative z, the upper a positive z.
Why is P(X ≤ a) the same as P(X < a) for a Normal variable?
X is continuous, so P(X = a) = 0.
Properties that follow from symmetry about μ
Mean = median = mode = μ, and P(X < μ) = P(X > μ) = 0.5.
How many probability statements are needed to find unknown μ and σ?
One per unknown: one for μ or σ alone, two (giving simultaneous equations) for both.
When is a Normal model unsuitable for a quantity that cannot be negative?
When 0 is not far below the mean, so the model gives a noticeable probability to negative values (e.g. mean 14, sd 10: P(X < 0) = 0.0808).
Percentage points table or Φ table in reverse?
Use the percentage points table if the probability is listed (0.1, 0.05, 0.025, 0.01, …); otherwise find the nearest value inside the Φ table. No interpolation.
X ~ N(30, 2.5²). Find P(27 < X < 34).
z = −1.2 and 1.6: 0.9452 − (1 − 0.8849) = 0.8301 = 0.830 (3 s.f.).
M ~ N(62, 4²). Find P(M > 68 | M > 60).
P(M > 68) = 1 − Φ(1.5) = 0.0668; P(M > 60) = Φ(0.5) = 0.6915; 0.0668/0.6915 = 0.0966.
T ~ N(240, 15²). Find the value exceeded by 5% of T.
z = 1.6449, t = 240 + 1.6449 × 15 = 264.7.
T ~ N(240, 15²). Find the value below which 20% of T lies.
z = −0.8416 (lower tail), t = 240 − 0.8416 × 15 = 227.4.
X ~ N(80, σ²) and P(X > 92) = 0.0228. Find σ.
Φ(z) = 0.9772 gives z = 2.00; (92 − 80)/σ = 2, so σ = 6.
V ~ N(μ, 0.8²) and P(V < 20) = 0.1. Find μ.
(20 − μ)/0.8 = −1.2816, so μ = 20 + 1.2816 × 0.8 = 21.03.
30% of L is below 17 and 10% is above 25. Find μ and σ.
μ − 0.5244σ = 17 and μ + 1.2816σ = 25; 1.806σ = 8, σ = 4.43, μ = 19.3.
P(T < 36) = 0.0668 and P(T > 52) = 0.0062. Find μ and σ.
z = −1.5 and 2.5: μ − 1.5σ = 36, μ + 2.5σ = 52, so 4σ = 16, σ = 4, μ = 42.
X ~ N(50, 16): what do you divide by when standardising?
σ = √16 = 4, not 16: the second parameter is the variance.
Sign of z for a value in the lower tail
Negative. If less than half the area is below the value, it is below the mean, so z < 0.
Middle 90% of a Normal distribution: which z?
5% in each tail, so z = ±1.6449 (not 1.2816).
What to write when using the tables
The standardising line with numbers, then z to 2 d.p. and the table value to 4 d.p.; final probability to 3 s.f.
Sense-check for a found μ or σ
σ must be positive, and μ must lie on the correct side of each given value (above a value with a small lower tail).

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