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

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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