Answer to Question #178789 in Statistics and Probability for Anthony

Question

An insurance company rounds 50000 insurance premiums to the nearest dollar. Assuming that

the fractional parts of the premiums are continuous and uniformly distributed between 0 and 1,

compute the probability that the total amount owing will be altered by more than $60.

You may use R to assist you in answering this question, but you would need to list out all

working details.

Answer 1

Since we are rounding to the nearest dollar here, if the fractional part is more than 0.5, then we would up round it and if the fractional part is less than 0.5, then we would round it downwards to the nearest dollar. Therefore the maximum difference from the real money owed is uniformly distributed here from -0.5 to +0.5

Then, for $U(-0.5,0.5)$ :

mean$=(0.5-0.5)/2=0$

variance$=(0.5+0.5)/12=1/12$

For 50,000 policies, probability that the money altered is more than

$60 is computed here as:

Converting to sample mean, we have:

$P(-60/50000<X<60/50000)=P(-0.0012<X<0.0012)$

Converting to standard normal variable:

$P((-0.0012-0)/(\sqrt{1/12}/\sqrt{50000})<Z<P((0.0012-0)/(\sqrt{1/12}/\sqrt{50000})$

$P(-0.93<Z<0.93)=P(-0.93<Z<0.93)=$

$=0.8238 - 0.1762=0.6476$