Suppose the average automobile insurance claim for the population is $4,560 with standard deviation of $600. What is the percentage of probability that a sample of 100 claim files will yield an average claim of $4,500 or less?

0.15866

Step-by-step explanation:

Given that :

Population mean = μ = 4560

Standard deviation, σ = 600

Sample size, n = 100

Using the relation :

(x - μ) ÷ σ/sqrt(n)

P(x ≤ 4500) :

(4500 - 4560) ÷ 600/sqrt(100)

P(x ≤ 4500) = -60 / (600/10)

P(x ≤ 4500) = - 60 / 60

P(x ≤ 4500) = - 1

P(Z ≤ - 1) :

Using the Z probability calculator ;

P(Z ≤ - 1) = 0.15866

Hence, P(x ≤ 4500) = 0.15866

4/6 = 2/3

so it should take 3/3 hours per man. so 9 hours is the answer 3*3 =9

step-by-step explanation:

1,589.63 in.^3

step-by-step explanation:

i hope this you : d