Question

In any​ year, the weather can inflict storm damage to a home. From year to​ year, the damage is random. Let Y denote the dollar value of damage in any given year. Suppose that in 95% of the years Y= $0 but in 5% of the years Y= $19,152.

A) The mean of the damage in any year is $

B) The standard deviation of the damage in any year is $

C) Consider an "insurance pool" of 100 people whose homes are sufficiently dispersed so that, in any year, the damage to different homes can be viewed as independently distributed random variables. Let Y bar denote the average damage to these 100 homes in a year.

E(Y bar), the expected value of the average damage Y bar is $

D) The probability that Y bar exceeds $2,000 is

please solve A through D formatting the answer to the corresponding letter.

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Answer 1

The mean of the damage is

E(Y) = p1y1 + p2y2

= 0.95*0 + 0.05*19152

= 957.6

b)

variance = E(Y - E(Y))² = p1 (y1 - E(Y))² + p2 (y2 - E(Y))²

Variance = 0.95*(0 - 957.6)² + 0.05*( 19152 - 957.6)²

= 17422957.47

standard deviation = (17422957.47)^0.5

= 4174.08

c) the expected value of the average damage: is also E(Y)= E( ȳ) = 957.6

d) P(  ȳ > 2000 ) = P( Z > [2000 - E( ȳ) ]/ s_ȳ]

s_{ȳ (} standard error of ȳ) = (variance /n)^0.5

n=100

s_ȳ = 4174.08 / 10 = 417.41

P( Z > [2000 - E( ȳ) ]/ s_ȳ] = P( Z > [ 2000 - 957.6] / 417.41 ) = P( Z > 2.497)

= 1 - P( Z < 2.497) =0.0063