Normal Distribution Help

The probability that a person has an unlisted telephone number is 0.15. The district manager of a political action group is phoning people urging them to vote. In a district with 416 households, all households have phones. What is the probability that the district manger will find that:

(a) 50 or fewer households in the district will have unlisted numbers?

I'm having trouble finding out what to do here because there is no standard deviation or means they supply with the problem. This is what I did so far using the normal approximation to the binomial distribution. I'm not sure if I'm using the right formula to calculate this problem.

population means = n*p and o(standard deviation)= square root n*p*q

N = 50 P = .15
U = np = 50(.15) = 7.5
Std dev = sqrt50(.15)(.85) = 2.52
Answer: 50-7.5/2.52 = 16.87
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