The number of ships to arrive at a harbor on any given day is a random variable represented by x. The probability distribution of x is as follows. (Give your answers correct to two decimal places.)

Question

x 10 11 12 13 14
P(x) 0.37 0.09 0.05 0.14 0.35

(a) Find the mean of the number of ships that arrive at a harbor on a given day.
12.01 is this answer I got

(b) Find the standard deviation, ó, of the number of ships that arrive at a harbor on a given day. and I got 19.19

Answer 1

First step:
Check to see if it is a discrete probability distribution by summing all probabilities. If it does not equal one, either the given distribution is incomplete, or it is not a discrete distribution.

Second step:
Mean (expected value of X, E(X), μ)
=ΣX*P(X)
=10*0.37+11*0.09....+14*0.35
=12.01 (so your answer is correct)

Third step:

Note that for a discrete distribution, the standard deviation will never be outside the domain of the distribution. 19.19 is outside the domain, so it cannot be right.

Variance σ²= Σ(X-μ)² * P(X)
=(10-12.01)²*0.37+...(14-12.01)²*0.35
=3.1 (approx.)
Standard deviation
=√(σ²)
=1.8 (approx.)

Answer 2

I came up with standard deviation of 2.88 when I worked it out like you have it laid out. But is that right???? I also got 10.58 and that is not right. I have one more chance at this one...

Answer 3

Check your calculations!
Did you get standard deviation or σ² as 2.88?
Don't understand how you got 2.88 and 10.58. Can you post what you did?

For the variance, I have
1.49+0.09+0.00+0.14+1.39=3.1