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.