Probability of something happening given the mean and the standard deviation.

[DanielJackins]

I'm doing a practice problem for my upcoming midterm and am stuck on a question.

3. The number of hurricanes that occur in the caribbean during hurricane season is a random variable, where the mean number of hurricanes occurring is 8, or E(X) = 8, with a standard deviation of 2.83.
(a) Find the probability, that in any given hurricane season, the number of hurricanes in the caribbean will be greater than 4 and less than 12.

So I figured this follows a binomial distribution. Given that, I noted that E(X)= np = 8, and that SD(X) = sqrt(Var(X)) = sqrt(np(1-p)) = 2.83. Knowing that information, I solved for p in 2.83 = sqrt(np(1-p)) and subbed in np = 8, and eventually got p = -.0011125. Which must be wrong.(?)

I don't even know if I'm on the right track, or am making some silly mistake.

 

[Stephen Tashi]

Perhaps the problem wants you to assume the number of hurricanes is normally distributed since it didn't tell you anything about the "number of trials" = number of opportunities to made a random draw for a hurricane.

What you did is an interesting way of thinking about the problem and I wouldn't say it was "wrong". It's just that the conclusion it produces shows that the assumption of a binomial distribution doesn't work.

(It's also true that the assumption of a normal distribution doesn't really work since the number of hurricanes is a discrete variable and bounded below. However, the normal distribution is frequently applied as an approximation in situations where it can't really be the exact distribution.)

 

[ImaLooser]

I'm doing a practice problem for my upcoming midterm and am stuck on a question.

3. The number of hurricanes that occur in the caribbean during hurricane season is a random variable, where the mean number of hurricanes occurring is 8, or E(X) = 8, with a standard deviation of 2.83.
(a) Find the probability, that in any given hurricane season, the number of hurricanes in the caribbean will be greater than 4 and less than 12.

So I figured this follows a binomial distribution. Given that, I noted that E(X)= np = 8, and that SD(X) = sqrt(Var(X)) = sqrt(np(1-p)) = 2.83. Knowing that information, I solved for p in 2.83 = sqrt(np(1-p)) and subbed in np = 8, and eventually got p = -.0011125. Which must be wrong.(?)

I don't even know if I'm on the right track, or am making some silly mistake.

I haven't checked your math, but assume it is correct. I would assume that they want you to use the normal approximation to the binomial. If you've just been taught that, then I'm sure that's what it is. If you haven't been taught that, then I dunno.

 

[Pythagorean]

You sure this isn't a Poisson question?

 

[Stephen Tashi]

You sure this isn't a Poisson question?

That's a good idea. It avoids a numerical contradiction and it is for a discrete random variable.