Question

A company rounds its losses to the nearest dollar. The error on each loss is independently and uniformly distributed on [–0.5, 0.5]. If the company rounds 2000 such claims, find the 95th percentile for the sum of the rounding errors.

Answer 1

Answer:

the 95th percentile for the sum of the rounding errors is 21.236

Step-by-step explanation:

Let consider X to be the rounding errors

Then; $X \sim U (a,b)$

where;

a = -0.5 and b = 0.5

Also;

Since The error on each loss is independently and uniformly distributed

Then;

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

where;

n = 2000

Mean $\mu = \dfrac{a+b}{2}$

$\mu = \dfrac{-0.5+0.5}{2}$

$\mu =0$

$\sigma^2 = \dfrac{(b-a)^2}{12}$

$\sigma^2 = \dfrac{(0.5-(-0.5))^2}{12}$

$\sigma^2 = \dfrac{(0.5+0.5)^2}{12}$

$\sigma^2 = \dfrac{(1.0)^2}{12}$

$\sigma^2 = \dfrac{1}{12}$

Recall:

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

$n\mu = 2000 \times 0 = 0$

$n \sigma^2 = 2000 \times \dfrac{1}{12} = \dfrac{2000}{12}$

For 95th percentile or below

$P(\overline X < 95}) = P(\dfrac{\overline X - \mu }{\sqrt{{n \sigma^2}}}< \dfrac{P_{95}- 0 } {\sqrt{\dfrac{2000}{12}}}) =0.95$

$P(Z< \dfrac{P_{95} } {\sqrt{\dfrac{2000}{12}}}) = 0.95$

$P(Z< \dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}}) = 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1- 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} = 0.05$

From Normal table; Z >   1.645 = 0.05

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1.645$

${P_{95}\sqrt{12} } = 1.645 \times {\sqrt{{2000}}}$

${P_{95} = \dfrac{1.645 \times {\sqrt{{2000}}} }{\sqrt{12} } }$

$\mathbf{P_{95} = 21.236}$

the 95th percentile for the sum of the rounding errors is 21.236