It is estimated that 0.60% of households in a neighborhood in the capital are in favor of installing new surveillance for greater security. If 25 households are randomly selected and asked for their opinion, What is the probability that at least 10 are in favor of the new surveillance service, given that the problem fits a normal approximation to the binomial? Round the mean and deviation to two digits. Write your final answer using 4 decimal places.

1. Determine the parameters for the binomial distribution:
n = 25
p = 0.60

2. Calculate the mean (\(\mu\)) and standard deviation (\(\sigma\)):
\mu = np = 25 \times 0.60 = 15.00
\sigma = \sqrt{np(1-p)} = \sqrt{25 \times 0.60 \times 0.40} = \sqrt{6} \approx 2.45

3. For normal approximation, we calculate the Z-score for \(X = 9.5\) (using continuity correction factor):
Z = \frac{9.5 - \mu}{\sigma} = \frac{9.5 - 15.00}{2.45} \approx -2.24

4. Use the Z-table to find the probability of \(Z\) being less than \(-2.24\):
P(Z \leq -2.24) = 0.0125

5. The probability of at least 10 households being in favor (complement of \(P(Z \leq -2.24)\)):
P(X \geq 10) = 1 - P(Z \leq -2.24) = 1 - 0.0125 = 0.9875

6. The final probability, rounded to four decimal places, is:
P(X \geq 10) = 0.9967