Question

A probability distribution of the claim sizes for an auto insurance policy appears in the following table. Claims 3 Probability 0.15 0.15 0.05 0.05 0.3 0.3 Find the percentage of claims within one standard deviation of the mean. Preview %

Answer 1

From the given distribution, the following Table iscalculated:

x	p	p x	p x2
1	0.15	0.15	0.15
2	0.15	0.3	0.6
3	0.05	0.15	0.45
4	0.05	0.2	0.8
5	0.3	1.5	7.5
6	0.3	1.8	10.8
Total =		$\sum px=4.1$	$\sum px^{2}=20.3$

So,

Mean = $\mu$ = $\sum px=4.1$

Standard Deviation = $\sigma =\sqrt{\sum px^{2}-(\sum px)^{2}}=\sqrt{20.3-4.1^{2}}=\sqrt{3.49}=1.8682$

So,

$\mu -\sigma =4.1-1.8682=2.2318$ = 2 (Round to integer)

$\mu +\sigma =4.1+1.8682=5.9682$ = 6 (Round to integer)

From the given discrete probability distribution, the probability within $\mu \pm \sigma$ is given by:

P(X = 2) + P(X=3) +P(X=4) + P(X=5) + P(X=6) = 0.15 +0.05 + 0.05 + 0.3 + 0.3 = 0.95 = 95%

So,

Answer is:

95%