IB Mathematics AA Binomial distribution, its mean and variance Study Notes
IB Mathematics AA Binomial distribution, its mean and variance Study Notes
IB Mathematics AA Binomial distribution, its mean and variance Notes Offer a clear explanation of Binomial distribution, its mean and variance, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Binomial distribution, its mean and variance.
Binomial Distribution
Binomial Distribution
The binomial distribution models the number of successes in a fixed number of independent trials of a binary (two-outcome) experiment.
\( P(X = r) = \binom{n}{r} p^r (1 – p)^{n – r} \)
Where:
- \( X \) = number of successes
- \( n \) = number of trials
- \( p \) = probability of success on a single trial
- \( r \) = number of successes (0, 1, …, n)
Situations where the binomial distribution is appropriate:
The binomial distribution is a suitable model when:
- The experiment consists of a fixed number \( n \) of trials.
- Each trial has exactly two possible outcomes (success or failure).
- The probability of success \( p \) is the same for each trial.
- The trials are independent (the outcome of one does not affect another).
Example:
A fair coin is tossed 5 times. What is the probability of getting exactly 3 heads?
▶️ Answer/Explanation
\( n = 5 \), \( p = 0.5 \), \( r = 3 \)
\( P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 \)
\( = 10 \times 0.125 \times 0.25 \)
\( = 10 \times 0.03125 = 0.3125 \)
Conclusion: The probability of exactly 3 heads is 0.3125.
Example:
In a large city, 30% of residents are left-handed. A random sample of 12 residents is selected. Let \( X \) be the number of left-handed people in the sample. Find the probability that:
- (a) exactly 4 residents are left-handed
- (b) at least 2 residents are left-handed
▶️ Answer/Explanation
\( n = 12 \) (number of trials)
\( p = 0.3 \) (probability of success = left-handed)
\( q = 1 – p = 0.7 \)
(a): Find \( P(X = 4) \)
\( P(X = 4) = \binom{12}{4} (0.3)^4 (0.7)^8 \)
\( = 495 \times (0.0081) \times (0.05764801) \)
\( = 495 \times 0.0004666 \)
\( \approx 0.2309 \)
(b): Find \( P(X \ge 2) \)
\( P(X \ge 2) = 1 – P(X = 0) – P(X = 1) \)
\( P(X = 0) = \binom{12}{0} (0.3)^0 (0.7)^{12} = 1 \times 1 \times 0.01384 = 0.01384 \)
\( P(X = 1) = \binom{12}{1} (0.3)^1 (0.7)^{11} = 12 \times 0.3 \times 0.01977 = 12 \times 0.005931 = 0.07117 \)
So, \( P(X \ge 2) = 1 – 0.01384 – 0.07117 = 1 – 0.08501 = 0.91499 \)
Mean and Variance of the Binomial Distribution
Mean and Variance of the Binomial Distribution
If \( X \sim \operatorname{Bin}(n, p) \) where:
- \( n \) = number of trials
- \( p \) = probability of success on each trial
then:
\( E(X) = np \quad \text{(mean)} \)
\( \operatorname{Var}(X) = np(1 – p) \quad \text{(variance)} \)
This means the expected number of successes is proportional to the number of trials and the chance of success. The variance measures how spread out the distribution is.
Binomial Probabilities Using Technology
For binomial distributions with larger \( n \) or when exact values are needed, probabilities should be found using a calculator or technology (e.g. GDC, TI-84, Casio fx-CG50).
On GDC or calculator:
- Use
binompdf(n, p, r)to find \( P(X = r) \) - Use
binomcdf(n, p, r)to find \( P(X \le r) \)
This ensures accurate values without tedious manual calculations.
Example:
A fair die is rolled 8 times. Let \( X \) be the number of times a 6 appears. Find the mean, variance, and the probability of getting exactly 2 sixes (using technology).
▶️ Answer/Explanation
\( n = 8 \)
\( p = \frac{1}{6} \)
\( q = 1 – p = \frac{5}{6} \)
\( E(X) = 8 \times \frac{1}{6} = \frac{8}{6} \approx 1.33 \)
\( \operatorname{Var}(X) = 8 \times \frac{1}{6} \times \frac{5}{6} = \frac{40}{36} \approx 1.11 \)
Using technology
On GDC / calculator:binompdf(8, 1/6, 2)
Result: \( P(X = 2) \approx 0.296 \)