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4 years ago

Consider the first four terms of an arithmetic sequence below.

Mathematics

1 answer:

VMariaS [17]4 years ago

6 0

7, 13, 19 and 25 have a common difference: 6.
6 added to 7 gives us 13; 6 added to 13 gives us 19, and so on.

Explicit formula: a(n) = 7 + 6(n-1), where 7 is the first term and n is the counter (1, 2, 3, ...).

The first term is 7 (given). This corresponds to n=1.

The second term is a(2) = 7 + 6(2-1), or 7 + 6, or 13. This corresponds to n = 2.

and so on.

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The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =

gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy is given as :

$f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0$

Thus, the expected total claim amount $\mu$ = 1000

The variance of the total claim amount $\sigma ^2 = 1000^2$

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

$P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})$

$P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})$

$P(X>1100n) = P(Z> \dfrac{10*100}{1000})$

$P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345$

$\mathbf{P(X>1100n) = 0.158655}$

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

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3 years ago

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saul85 [17]

The answer to this question is A). 39 b). 56

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