You are correct. Let's correct the calculation and recompute \( P(X = 5) \).

Given:
\[ P(X = 5) = \binom{12}{5} (0.45)^5 (0.55)^7 \]

We need to correctly compute \((0.55)^7\):

\[ (0.55)^7 \approx 0.0152 \]

Now, let's recompute \( P(X = 5) \) with the correct value:

\[ \binom{12}{5} = \frac{12!}{5!7!} = 792 \]

\[ (0.45)^5 \approx 0.0186 \]

\[ (0.55)^7 \approx 0.0152 \]

Now, recomputing:

\[ P(X = 5) = 792 \times 0.0186 \times 0.0152 \]

\[ P(X = 5) \approx 792 \times 0.0186 \times 0.0152 \approx 0.224 \]

### (ii) Calculate the probability that this batch contains more plants with white flowers than colored ones:

For more plants with white flowers than colored ones, \( X > 6 \):

\[ P(X > 6) = \sum_{k=7}^{12} P(X = k) \]

Let's calculate each term:

\[ P(X = k) = \binom{12}{k} (0.45)^k (0.55)^{12-k} \]

We need to compute \( P(X = 7), P(X = 8), P(X = 9), P(X = 10), P(X = 11), \) and \( P(X = 12) \).

### Corrected Calculations for \( P(X > 6) \)

Using the binomial formula:

\[ P(X = 7) = \binom{12}{7} (0.45)^7 (0.55)^5 \]
\[ P(X = 8) = \binom{12}{8} (0.45)^8 (0.55)^4 \]
\[ P(X = 9) = \binom{12}{9} (0.45)^9 (0.55)^3 \]
\[ P(X = 10) = \binom{12}{10} (0.45)^{10} (0.55)^2 \]
\[ P(X = 11) = \binom{12}{11} (0.45)^{11} (0.55)^1 \]
\[ P(X = 12) = \binom{12}{12} (0.45)^{12} (0.55)^0 \]

1. \( \binom{12}{7} = 792 \), \( (0.45)^7 \approx 0.00356 \), \( (0.55)^5 \approx 0.0503 \)
\[ P(X = 7) \approx 792 \times 0.00356 \times 0.0503 \approx 0.141 \]

2. \( \binom{12}{8} = 495 \), \( (0.45)^8 \approx 0.00160 \), \( (0.55)^4 \approx 0.0915 \)
\[ P(X = 8) \approx 495 \times 0.00160 \times 0.0915 \approx 0.072 \]

3. \( \binom{12}{9} = 220 \), \( (0.45)^9 \approx 0.00072 \), \( (0.55)^3 \approx 0.1664 \)
\[ P(X = 9) \approx 220 \times 0.00072 \times 0.1664 \approx 0.026 \]

4. \( \binom{12}{10} = 66 \), \( (0.45)^{10} \approx 0.00032 \), \( (0.55)^2 \approx 0.3025 \)
\[ P(X = 10) \approx 66 \times 0.00032 \times 0.3025 \approx 0.0064 \]

5. \( \binom{12}{11} = 12 \), \( (0.45)^{11} \approx 0.00014 \), \( (0.55)^1 \approx 0.55 \)
\[ P(X = 11) \approx 12 \times 0.00014 \times 0.55 \approx 0.0009 \]

6. \( \binom{12}{12} = 1 \), \( (0.45)^{12} \approx 0.00006 \), \( (0.55)^0 = 1 \)
\[ P(X = 12) \approx 1 \times 0.00006 \times 1 \approx 0.00006 \]

Summing these probabilities:

\[ P(X > 6) \approx 0.141 + 0.072 + 0.026 + 0.0064 + 0.0009 + 0.00006 \approx 0.246 \]

### Summary

1. \( P(X = 5) \approx 0.224 \)
2. \( P(X > 6) \approx 0.246 \)
3. Mean \( \mu = 5.4 \)
4. Variance \( \sigma^2 = 2.97 \)

Would you like further details on any part or have any additional questions? 

Feel free to ask questions or choose from the following:

1. Detailed steps for computing \( P(X > 6) \)
2. Explanation of the binomial coefficient calculation
3. Calculation of higher order binomial probabilities
4. Additional examples of binomial distribution problems
5. Application of the binomial theorem in different contexts

**Tip:** When solving binomial distribution problems, always double-check intermediate calculations to ensure accuracy.

Learn how to calculate probabilities using binomial distribution with detailed steps. This problem involves computing specific probabilities and understanding the binomial coefficient. Suitable for advanced high school or college-level students.

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