Binimial Distribution vs. Normal Approximation* to the Binomial
Harold fails to study for his statistics nal. The nal has 100 multiple-choice questions,
each with 5 choices. Harold has no choice but to guess randomly at all 100 questions.
What is the probability that Harold will get at least 30% on the test?
(Note: The probability of successfully guessing for each question is 20%.)
n = 100
p = 1/5 = 0.2
Harold needs at least 30% (i.e., at least 30 correct answers)
→ P(X ≥ 30)
Rule of thumb:
#1 Binomial distribution when calculators are available:
- gives the exact probabiltiy
#2 Normal approximation when calculators are not available:
- works well if np ≥ 10 and n(1-p) ≥ 10
1. Binomial distribution:
P( X ≥ 30)
= 1 - P(X ≤ 29)
=1≈ 0.0112
[84: BinomCDF: n=100, p=0.2, x=29]
[nS: BinomCDF: n=100, p=0.2, lower=0, upper=29]
2. Normal approximation to the binomial
X~Bin(n = 100, p = 0.2)
p̂ = x/n
p̂ ~N(p, √(p(1-p)/n)) = N(0.2, √0.2*0.8/100) = N(0.2, 0.04)
µₚ̂ = 0.2
σₚ̂ = 0.04
P(p̂ ≥ 0.30)
Apply continuity correction factor:
Z* = (p̂ - µₚ̂)/σₚ̂ = (0.295 - 0.2)/0.04
P(p̂ ≥ 29.5/100) = P(Z ≥ Z*) = P(Z ≥ 2.375) = 0.0088
fi
fi
***Without continuity correction factor (WRONG but acceptable in AP):
Z* = (p̂ - µₚ̂)/σₚ̂ = (0.3 - 0.2)/0.04
P(p̂ ≥ 30/100) = P(Z ≥ Z*) = P(Z ≥ 2.5) = 0.0062