please, help me with probability

Hello everyone.
I have a doubt with this exercise:
the probability of a car need to be serviced in one month is 0.2. the company has 900 units
What is the probability that more than 200 cars need to be revised in a month?
if I know that p (x = 900) = 0.2 * 900 = 180 and wish to calculate the probability that p (x> = 200) = p (x = 200) + p (x = 201) ....
I do not think this is so and I'm confused.
I will be grateful to you if you help me.
Thank you. bye

Matthew said:

Hello everyone.
I have a doubt with this exercise:
the probability of a car need to be serviced in one month is 0.2. the company has 900 units
What is the probability that more than 200 cars need to be revised in a month?
if I know that p (x = 900) = 0.2 * 900 = 180 and wish to calculate the probability that p (x> = 200) = p (x = 200) + p (x = 201) ....
I do not think this is so and I'm confused.
I will be grateful to you if you help me.
Thank you. bye

Click to expand...

Lets see if this fits the criteria for a Binomial Distribution:
a) a car either needs service of=r does not, which is a yes/no question
b) there is a fixed probability p=0.2 which does not change
c) there is a fixed number of trials, n=900
OK - it does seem to fit the requirements.

Recall (or look up) the properties of the binomial distribution:
The expectation value (mean) is . . . ?
The standard deviation is . . . ?
The numbers are large enough to allow the approximation of the binomial with a Normal Distribution with that same mean and standard deviation.

How much different from the mean is 200? How many standard deviations is that? Use a table of the normal distribution to find the probability of more than that number of standard deviations.

I think that is a pretty complete road map. If you need more help, show us how far you got.