The mean and variance of a Binomial variate are 2.4 and 1.44 respectively. Find the parameters, `P(X=2)` and `P(1 lt X le4)`.

Answer

To solve the problem, we need to find the parameters of a binomial distribution given the mean and variance, and then calculate specific probabilities.

### Step-by-Step Solution:

1. **Identify the Mean and Variance:**
- Given the mean \( \mu = 2.4 \) and variance \( \sigma^2 = 1.44 \).
- For a binomial distribution, the mean is given by:
\[ \mu = n \cdot p \]
- The variance is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \).

2. **Set Up the Equations:**
- From the mean:
\[ n \cdot p = 2.4 \quad \text{(Equation 1)} \]
- From the variance:
\[ n \cdot p \cdot q = 1.44 \quad \text{(Equation 2)} \]

3. **Express \( q \) in terms of \( p \):**
- Since \( q = 1 - p \), we can substitute \( q \) into Equation 2:
\[ n \cdot p \cdot (1 - p) = 1.44 \]

4. **Divide Equation 1 by Equation 2:**
- Dividing Equation 2 by Equation 1:
\[ \frac{n \cdot p \cdot (1 - p)}{n \cdot p} = \frac{1.44}{2.4} \]
- This simplifies to:
\[ 1 - p = \frac{1.44}{2.4} = 0.6 \]
- Thus, we find:
\[ p = 1 - 0.6 = 0.4 \]

5. **Substitute \( p \) back to find \( n \):**
- Substitute \( p \) into Equation 1:
\[ n \cdot 0.4 = 2.4 \]
- Solving for \( n \):
\[ n = \frac{2.4}{0.4} = 6 \]

6. **Calculate \( q \):**
- Now, we can find \( q \):
\[ q = 1 - p = 1 - 0.4 = 0.6 \]

7. **Find \( P(X = 2) \):**
- Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- For \( k = 2 \):
\[ P(X = 2) = \binom{6}{2} (0.4)^2 (0.6)^{4} \]
- Calculate \( \binom{6}{2} = \frac{6!}{2!(6-2)!} = 15 \):
\[ P(X = 2) = 15 \cdot (0.4)^2 \cdot (0.6)^4 \]
- Calculate \( (0.4)^2 = 0.16 \) and \( (0.6)^4 = 0.1296 \):
\[ P(X = 2) = 15 \cdot 0.16 \cdot 0.1296 = 0.31104 \]

8. **Find \( P(1 < X \leq 4) \):**
- This can be calculated as:
\[ P(X = 2) + P(X = 3) + P(X = 4) \]
- We already have \( P(X = 2) = 0.31104 \).
- Calculate \( P(X = 3) \):
\[ P(X = 3) = \binom{6}{3} (0.4)^3 (0.6)^{3} = 20 \cdot (0.4)^3 \cdot (0.6)^3 \]
- Calculate \( (0.4)^3 = 0.064 \) and \( (0.6)^3 = 0.216 \):
\[ P(X = 3) = 20 \cdot 0.064 \cdot 0.216 = 0.27648 \]
- Calculate \( P(X = 4) \):
\[ P(X = 4) = \binom{6}{4} (0.4)^4 (0.6)^{2} = 15 \cdot (0.4)^4 \cdot (0.6)^{2} \]
- Calculate \( (0.4)^4 = 0.0256 \) and \( (0.6)^2 = 0.36 \):
\[ P(X = 4) = 15 \cdot 0.0256 \cdot 0.36 = 0.137781 \]
- Now sum these probabilities:
\[ P(1 < X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.31104 + 0.27648 + 0.137781 = 0.725301 \]

### Final Answers:
- \( P(X = 2) \approx 0.311 \)
- \( P(1 < X \leq 4) \approx 0.725 \)