Statistics and Probability

Question

A probability distribution of the claim sizes for an auto insurance policy appears in the following table.

ClaimsProbability
1	0.25
2	0.2
3	0.1
4	0.05
5	0.4

Find the percentage of claims within one standard deviation of the mean.    %

Answer 1

The mean of the distribution, E(X) is the weighted average of size with respect to probability.

E(X) = ∑_i=1,n x_iw_i

E(X) = 1 * .25 + 2 * .2 + 3 * .1 + 4 * .05 + 5 * .4 = 3.15

Then to calculate the standard deviation we need to calculate the (deep breath) square root of the average of the squared deviations from the mean.

σ = √( ∑_{i=1, n} w_i * (x_i-E(X))²) )

(Many computations later) σ = 1.68

Then we see which sizes fit inside a range from E(X) - σ to E(X) + σ, or (3.15 - 1.68, 3.15 + 1.68).

Range: (1.47, 4.8)

Inside this range we have the sizes 2, 3, and 4.

When we sum the probabilities of these sizes together, we get 0.35 or 35% of claims which are less than 1 standard deviation away from the mean.

I hope this helps!