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GaryK [48]
3 years ago
5

An insurance policy sells for ​$1200. Based on past​ data, an average of 1 in 100 policyholders will file a ​$10 comma 000 ​clai

m, an average of 1 in 250 policyholders will file a ​$40 comma 000 ​claim, and an average of 1 in 400 policyholders will file an ​$80 comma 000 claim. Find the expected value​ (to the​ company) per policy sold. If the company sells 30 comma 000 ​policies, what is the expected profit or​ loss?
Business
1 answer:
Tanzania [10]3 years ago
3 0

Answer:

Expected Value = $740

Expected profit = $22.2m

Explanation:

We can easily calculate the expected value and expected profit/loss in this situation by some minor working

Expected values = Expected Claim - per policy cost

Expected profit/loss = (Expected claim - per policy cost) x number of policies

As you can see per policy cost and no of policies are given in the question data we just need to find expected claim for calculation of expected profit or loss and expected value

Expected Claim = (1/100x$10,000)+(1/250x$40,000)+(1/400x$80,000)

Expected Claim = 100 + 160 + 200

Expected Claim = 460

Now we have a value of expected claim lets put it into Expected profit/loss formula and expected value formula

Expected value = 460-1200

Expected value = -740

-$740 is the value per policy

Expected profit/loss = (460 - $1200 per policy) x 30,000

Expected profit or loss = -22,200,000

Expected loss to the customer = -$22.2 m

Expected profit for the company = $22.2m

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The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almana
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Answer:

a) P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

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And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

c) P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the scores for critical reading of a population, and for this case we know the distribution for X is given by:

X \sim N(502,100)  

Where \mu=502 and \sigma=100

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)

And for this case we want this probability:

P(502-10 < \bar X < 502+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

Part b

Let X the random variable that represent the scores for Math of a population, and for this case we know the distribution for X is given by:

X \sim N(515,100)  

Where \mu=515 and \sigma=100

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)

And for this case we want this probability:

P(515-10 < \bar X < 515+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(505

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

Part c

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X \sim N(494,100)  

Where \mu=494 and \sigma=100

We select a sample of size n=100, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{100}}=10)

And for this case we want this probability:

P(494-10 < \bar X < 494+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

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