Continuous uniform distribution and Normal approximations flashcards

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Continuous uniform (rectangular) distribution U[a, b]
A continuous distribution in which every value between a and b is equally likely: constant pdf on [a, b], zero outside.
Rounding error of a value rounded to the nearest k
True value − recorded value; modelled by U[−k/2, k/2], mean 0.
Continuity correction
Treating each integer k of a discrete variable as the interval k − 0.5 to k + 0.5 when it is approximated by a continuous (normal) variable.
Normal approximation to a discrete distribution
Replacing a binomial or Poisson variable by a normal variable with the same mean and variance, plus a continuity correction.
Cumulative distribution function F(x)
F(x) = P(X ≤ x); for U[a, b] it rises in a straight line from 0 at x = a to 1 at x = b.
pdf of X ~ U[a, b]
f(x) = 1/(b − a) for a ≤ x ≤ b; f(x) = 0 otherwise. (Yes)
Mean and variance of X ~ U[a, b]
E(X) = (a + b)/2; Var(X) = (b − a)²/12. (Yes)
cdf of X ~ U[a, b]
F(x) = 0 for x < a; (x − a)/(b − a) for a ≤ x ≤ b; 1 for x > b. (No — derive it)
E(X²) from the mean and variance
E(X²) = Var(X) + [E(X)]². (Yes)
Normal approximation to B(n, p)
B(n, p) ≈ N(np, np(1 − p)) when n is large and p is close to ½. (No — learn it)
Normal approximation to Po(λ)
Po(λ) ≈ N(λ, λ) when λ is large (more than about 10). (No — learn it)
Standardising a normal variable
Z = (Y − μ)/σ, with σ the square root of the variance. (No — learn it)
Shape of the pdf of U[a, b]
A rectangle of width b − a and height 1/(b − a), total area 1.
Shape of the cdf of U[a, b]
Flat at 0 up to a, a straight line from (a, 0) to (b, 1), then flat at 1.
B(20, 0.1) drawn with N(2, 1.8)
Bars skewed to the right; the normal curve fits badly and even gives area below 0.
Binomial bars drawn with the matching normal curve
Each bar for k has width 1, from k − 0.5 to k + 0.5; the area under the curve over that width approximates P(X = k).
Po(λ) bars as λ increases
The right skew fades and the shape becomes bell-shaped and symmetrical about λ.
Why is P(c < X < d) for U[a, b] just (d − c)/(b − a)?
The pdf is constant, so the area is a rectangle: width (d − c) × height 1/(b − a).
Why is a continuity correction needed?
The discrete variable takes whole numbers only; the normal variable is continuous, so P(Y = k) = 0. Each integer must become a bar of width 1.
When should B(n, p) be approximated by a Poisson rather than a normal distribution?
When n is large and p is small: the distribution is skewed with mean ≈ variance. Use Po(np).
Why does a Poisson approximation to a binomial need no continuity correction?
Both distributions are discrete, so integers map to integers.
Does a uniform distribution have a mode?
No: every value in [a, b] is equally likely. Its mean and median are both (a + b)/2.
Derivation of Var(X) for U[a, b] in outline
E(X²) = (b³ − a³)/(3(b − a)) = (a² + ab + b²)/3; subtract ((a + b)/2)² to get (b − a)²/12.
T ~ U[5, 35]. Find P(T > 26).
(35 − 26)/30 = 0.3.
X ~ U[a, b] with E(X) = 7 and Var(X) = 3. Find a and b.
a + b = 14 and (b − a)² = 36, so b − a = 6: a = 4, b = 10.
Standard deviation of the error when times are rounded to the nearest 0.1 s.
Error ~ U[−0.05, 0.05]; σ = 0.1/√12 = 0.0289 s.
X ~ B(120, 0.35). Use a normal approximation to find P(X ≤ 38).
Y ~ N(42, 27.3); P(Y < 38.5) = P(Z < −0.67) = 1 − 0.7486 = 0.2514.
X ~ Po(30). Use a normal approximation to find P(X > 35).
Y ~ N(30, 30); P(Y > 35.5) = P(Z > 1.00) = 1 − 0.8413 = 0.1587.
X ~ B(100, 0.5). Use a normal approximation to find P(X = 50).
Y ~ N(50, 25); P(49.5 < Y < 50.5) = 2Φ(0.1) − 1 = 0.0796.
400 trials, p = 0.0075. Approximate P(X ≤ 2).
n large, p small: Po(3); P(X ≤ 2) = 0.4232.
R ~ U[2, 4]. Find E(R²).
Var(R) + [E(R)]² = 4/12 + 9 = 28/3.
P(X ≥ 56) for binomial X, normal approximation
56 is included, so P(Y > 55.5) — not 56.5.
P(X < 20) for a discrete X, normal approximation
Fewer than 20 means up to 19: P(Y < 19.5).
Writing N(μ, σ²) for an approximation
The second number is the variance: N(42, 27.3), then divide by √27.3 when standardising.
Giving a uniform pdf
Always include the range and “0 otherwise”: f(x) = 1/30, 5 ≤ x ≤ 35; 0 otherwise.
Finding the largest m with P(X ≥ m) ≥ 0.9 by a normal approximation
Solve (m − 0.5 − μ)/σ ≤ −1.2816, then round down; check the integers either side.
Rounding z for the normal table
Round z to 2 d.p. and give the probability from the table to 4 d.p.; calculator values are also acceptable.

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