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; [tex]X \sim U (a,b)[/tex]
where;
a = -0.5 and b = 0.5
Also;
Since The error on each loss is independently and uniformly distributed
Then;
[tex]\sum X _1 \sim N ( n \mu , n \sigma^2)[/tex]
where;
n = 2000
Mean [tex]\mu = \dfrac{a+b}{2}[/tex]
[tex]\mu = \dfrac{-0.5+0.5}{2}[/tex]
[tex]\mu =0[/tex]
[tex]\sigma^2 = \dfrac{(b-a)^2}{12}[/tex]
[tex]\sigma^2 = \dfrac{(0.5-(-0.5))^2}{12}[/tex]
[tex]\sigma^2 = \dfrac{(0.5+0.5)^2}{12}[/tex]
[tex]\sigma^2 = \dfrac{(1.0)^2}{12}[/tex]
[tex]\sigma^2 = \dfrac{1}{12}[/tex]
Recall:
[tex]\sum X _1 \sim N ( n \mu , n \sigma^2)[/tex]
[tex]n\mu = 2000 \times 0 = 0[/tex]
[tex]n \sigma^2 = 2000 \times \dfrac{1}{12} = \dfrac{2000}{12}[/tex]
For 95th percentile or below
[tex]P(\overline X < 95}) = P(\dfrac{\overline X - \mu }{\sqrt{{n \sigma^2}}}< \dfrac{P_{95}- 0 } {\sqrt{\dfrac{2000}{12}}}) =0.95[/tex]
[tex]P(Z< \dfrac{P_{95} } {\sqrt{\dfrac{2000}{12}}}) = 0.95[/tex]
[tex]P(Z< \dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}}) = 0.95[/tex]
[tex]\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1- 0.95[/tex]
[tex]\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} = 0.05[/tex]
From Normal table; Z > Â 1.645 = 0.05
[tex]\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1.645[/tex]
[tex]{P_{95}\sqrt{12} } = 1.645 \times {\sqrt{{2000}}}[/tex]
[tex]{P_{95} = \dfrac{1.645 \times {\sqrt{{2000}}} }{\sqrt{12} } }[/tex]
[tex]\mathbf{P_{95} = 21.236}[/tex]
the 95th percentile for the sum of the rounding errors is 21.236
