An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is --$250 and the standard deviation of the loss is ơ-$1000. the distribution of losses is strongly right-skewed: many policies have $0 loss, but a few have large losses. an auditor examines a random sample of 10,000 of the company's policies. if the company's clairm is correct, what's the probability that the average loss from fire in the sample is no greater than $275?

the probability that the average loss from fire in the sample is no greater than $275 is 0.9938

Step-by-step explanation:

given information:

mean, μ = $250

std deviation, σ = $1000

random sample, n = 10000

x = $275

P( ∠_ 275) = P(z < (z ∠ (x - μ)/(σ/√n))

                    = P (z ∠ (275 - 250)/(1000/√10000)

                    = P(z ∠ 2.5)

                     =  0.9938

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

trapezoid

step-by-step explanation:

first i plotted all my points on a grid graph

i made sure all of them were in the correct quadrant

then i saw it was a trapezoid or a quadrilateral