Binomial Distribution: Mean and Variance
Problem 1: Mean = 3, Variance = 4?
Given: Is it possible a binomial distribution has a mean of 3 and a variance of 4?
Solution:
- Mean: \( \mu = np \)
- Variance: \( \sigma^2 = npq \), where \( q = 1 – p \)
- Given: \( np = 3 \), \( npq = 4 \)
- Dividing variance by mean:
\[
\frac{npq}{np} = \frac{4}{3} \Rightarrow q = \frac{4}{3}
\] - But \( q \) cannot be greater than 1, so this is not possible.
Problem 2: Find distribution if mean + variance = 4.8 for 5 trials
Given: \( n = 5 \), \( np + npq = 4.8 \)
Solution:
- \( 5p + 5p(1 – p) = 4.8 \)
- \( 10p – 5p^2 = 4.8 \Rightarrow 100p – 50p^2 = 48 \Rightarrow 25p^2 – 50p + 24 = 0 \)
- Solving: \( p = \frac{4}{5}, q = \frac{1}{5} \)
Binomial Distribution:
\[
P(X = x) = {5 \choose x} \left( \frac{4}{5} \right)^x \left( \frac{1}{5} \right)^{5 – x}, \quad x = 0, 1, 2, 3, 4, 5
\]
Probabilities:
- \( P(0) = \frac{1}{3125} \), \( P(1) = \frac{20}{3125} \), \( P(2) = \frac{160}{3125} \)
- \( P(3) = \frac{640}{3125} \), \( P(4) = \frac{1280}{3125} \), \( P(5) = \frac{1024}{3125} \)
Problem 3: Mean = 30, SD = 5
Find: (i) \( n, p, q \), (ii) Skewness, (iii) Kurtosis
Part (i):
- Mean: \( np = 30 \)
- SD: \( \sqrt{npq} = 5 \Rightarrow npq = 25 \)
- \( q = \frac{25}{30} = \frac{5}{6} \Rightarrow p = \frac{1}{6}, n = 180 \)
Part (ii): Skewness
\[
\gamma_1 = \frac{q – p}{\sqrt{npq}} = \frac{\frac{5}{6} – \frac{1}{6}}{5} = \frac{2}{15}
\]
Part (iii): Kurtosis
\[
\beta_2 = 3 + \frac{1 – 6pq}{npq} = 3 + \frac{1 – \frac{5}{6}}{25} = 3 + \frac{1}{150}
\Rightarrow \gamma_2 = \frac{1}{150} > 0
\]
The distribution is leptokurtic.
Problem 4: Tossing 7 coins 128 times
| No. of Heads | Frequency |
|---|---|
| 0 | 7 |
| 1 | 6 |
| 2 | 19 |
| 3 | 35 |
| 4 | 30 |
| 5 | 23 |
| 6 | 7 |
| 7 | 1 |
Assumption: Coin is unbiased → \( p = q = \frac{1}{2} \)
\[
P(x) = {7 \choose x} \left( \frac{1}{2} \right)^7, \quad f(x) = 128 \cdot P(x)
\]
Expected Frequencies:
- \( f(0) = 1, f(1) = 7, f(2) = 21, f(3) = 35, f(4) = 35, f(5) = 21, f(6) = 7, f(7) = 1 \)
| X | Observed | Expected |
|---|---|---|
| 0 | 7 | 1 |
| 1 | 6 | 7 |
| 2 | 19 | 21 |
| 3 | 35 | 35 |
| 4 | 30 | 35 |
| 5 | 23 | 21 |
| 6 | 7 | 7 |
| 7 | 1 | 1 |
Problem 5: 800 families, 4 children — Expected with 3 boys, 1 girl
- \( N = 800, n = 4, p = q = \frac{1}{2} \)
- \[
P(X = 3) = {4 \choose 3} \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^1 = \frac{1}{4}
\] - \[
E = 800 \cdot \frac{1}{4} = 200
\] - Expected families = 200
Video
📡 Reach PostNetwork Academy
- 🌐 Website: www.postnetwork.co
- 🎥 YouTube: @postnetworkacademy
- 📘 Facebook: postnetworkacademy
- 💼 LinkedIn: postnetworkacademy
- 👨💻 GitHub: postnetworkacademy