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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 ​= ​$20,000.
A. The mean of the damage in any year is ​$__. nothing. The standard deviation of the damage in any year is ​$____.
B. Consider an​ "insurance pool" of 100 people whose homes are sufficiently dispersed so​ that, in any​ year, the damage to different homes can be viewed as independently distributed random variables. Let barY denote the average damage to these 100 homes in a year. The expected value of the average damage barY, is ​$____. The probability that exceeds ​$is____.

Sagot :

ogorwyne

Answer:

a. mean = 1000

standard deviation = 4358.9

b. expected value of average damage bar Y = 1000

probability bar y exceeds 2000 = 0.011

Step-by-step explanation:

we have p1 = 95%, y1 = 0, p2 = 5%, y2 = 20000

Mean = (0.95 * 0) + (0.05 * 20000)

= 1000

var(y) = E(y²) - E(Y)²

= we solve for E(y)²

= 0²*0.95 + 20000²*0.05

= 0 + 20000000

then the variance of y = 20000000 - 1000²

=20000000-1000000

= $19000000

standard  deviation is the square root of variance

= √19000000

= 4358.9

2.

a. Expected value of average is also the mean = 1000

b. we are to find probability that barY exceeds 2000

[tex]z=\frac{2000-1000}{\sqrt{19000000/100} }[/tex]

= 1000/435.889

= 2.29

1-p(z≤2.29)

= 1 - 0.989

= 0.011

so the probability that barY exceeds 2000 is 0.011

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