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Salsk061 [2.6K]
2 years ago
13

Question not on textbook: An insurance company has 25,000 automobile policy holders. If the yearly claim of a policy holder is a

random variable with mean 320 and standard deviation 540, use CLT to approximate the probability that the total yearly claim exceeds 8.3 million
Mathematics
1 answer:
Lynna [10]2 years ago
3 0

Answer:

There is 0.00022 probability that the yearly income of the insurance claims will exceed 8.3 millions.

Step-by-step explanation:

Given data:

Sample Size = n = 25,000

Mean of yearly claim = Mu = 320

Standard Deviation of yearly claim = SD = 540

What we have to find out?

Probability of total yearly claims exceeding 8.3 million?

As the total no. of policy holders is 25,000 therefore the targeted value of claim will be calculated as follows:

Targeted value of policy claim = x = Total Yearly claim / No. of Policy holders

x = (8.3 * 1.000.000) / 25,000

x = 332

Central Limit Theorem (CLT):

From Central theorem we know that: z(x) = (x - Mu)/(SD/√n) ; Equation 1

By putting values in equation 1 we have:

z(332) = (332-320)/(540 / √25,000)

z(332) = 12 / (540/158.114)

z(332) = 3.51

By using the normal distribution tables we get:

P(x <= 332) = Phi(z) = Phi(3.51)

P(x <= 332) = 0.99978

By using the probability unity method we will get:

P(x > 332) = 1 - P(x <= 332)

P(x > 332) = 1 - 0.99978

P( x > 332) = 0.00022

Thus we get the probability of total yearly claim exceeding 8.3 million equals to 0.00022.

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Step-by-step explanation:

The <em>probaility</em> of an event is defined as the number of favorable outcomes divided by the number of total possible events.

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1. <u>As </u><u>first step</u><u>, you may draw a table to find the </u><u>sample space</u><u> (set of all possible outcomes)</u>.

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The results of the rolling two dice are summarized in this table:

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First roll

  1                                       (1,1)  (1,2)    (1,3)   (1,4)   (1,5)  (1,6)

  2                                     (2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

  3                                     (3,1)  (3,2)  <u>(3,3)</u>   (3,4)  (3,5)  (3,6)

  4                                     (4,1)   (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

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<u></u>

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c) Number of outcomes for the event rolling two 3s (3,3): 1

3) <u>Next, you can calculate the probabilities:</u>

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2 years ago
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<h2>Answer:</h2>

The correct options are:

        Choice A

       Choice C

and  Choice E

<h2>Step-by-step explanation:</h2>

We are asked to find the value of:

                   60% of 94

We know that: it is represented as:

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Choice A)

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We know that:

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0.06\times 940

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Choice B)

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This option is incorrect.

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Choice C)

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This option is correct.

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60\%\ of 94\ is:\\\\\\=\dfrac{60}{100}\cdot 94\\\\\\=\dfrac{3}{5}\cdot 94

Choice D)

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         0.6\cdot 94

Choice E)

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