If an insurance company has 1 policies, and each has 0.1 probability of making a claim, what is the standard deviation of the fraction of policies which result in a claim?

0.003

Step-by-step explanation:

Since each of the policy have the same chance of success, then it follows a binomial distribution (only two outcome; success or failure).

α = √(p(1-p)/n) = √ (0.1 ( 1-0.1)/ 10000) = √ 0.000009 = 0.003

where p = 0.1, α = standard deviation.

S.D = +/- 0.003

The standard deviation of the fraction of policies which result in a claim is +/- 0.003

Step-by-step explanation:

Given;

Probability of making a claim p = 0.1

Number of company policies n = 10000

Standard deviation is the measure of how spread out numbers are.

S.D = √(p(1-p)/n)

Where,

p = 0.1

n = 10000

S.D = √(0.1(1-0.1)/10000)

S.D = √(0.1×0.9/10000)

S.D = √(0.000009)

S.D = +/- 0.003

many reason,

step-by-step explanation:

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb