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Ivan
3 years ago
5

A car insurance company has high-risk, medium-risk, and low-risk clients, who have, respectively, probabilities .04, .02, and .0

1 of filing claims within a given year. The proportions of the numbers of clients in the three categories are .15, .25, and .60, respectively.(a) What is the probability that a random client doesn't file a claim?(b) What proportion of the claims filed each year come from high-risk clients?(c) What is the probability that a random client who didn't file a claim is low- risk?
Mathematics
1 answer:
Paha777 [63]3 years ago
6 0

Answer:

(a) 0.983

(b) 0.353 or 35.3%

(c) 0.604 or 60.4%

Step-by-step explanation:

a) The probability of a random client does not file a claim is equal to the sum of:

1) the probability of a client being high risk and does not file a claim = P(hr)*(1-P(c_hr))

2) the probability of a client being medium risk and does not file a claim = P(mr)*(1-P(c_mr))

and

3) the probability of a client being low risk and does not file a claim = P(lr)*(1-P(c_lr))

P(not claim) = P(hr)*(1-P(c_hr))+P(mr)*(1-P(c_mr))+P(lr)*(1-P(c_lr))

P(not claim) = 0.15*(1-0.04)+0.25*(1-0.02)+0.6*(1-0.01)

P(not claim) = 0.15*0.96+0.25*0.98+0.6*0.99 = 0.983

(b) To know the proportion of claims that come from high risk clients we need to know the total expected claims in every category:

Claims expected by high risk clients = P(c_hr)*P(hr) = 0.04*0.15 = 0.006 claims/client

Claims expected by medium risk clients = P(c_mr)*P(mr) = 0.02*0.25 = 0.005 claims/client

Claims expected by low risk clients = P(c_lr)*P(lr) = 0.01*0.60 = 0.006 claims/client

The proportion of claims done by high risk clients is

Claims by HR clients / Total claims expected = 0.006 / (0.006+0.005+0.006) =  0.006 / 0.017 = 0.3529 or 35,3%

(c)  The probability of being a client of a particular category and who don't file a claim is:

1) High risk: 0.15*(1-0.04) = 0.144

2) Medium risk: 0.25*(1-0.02) =  0.245

3) Low risk: 0.6*(1-0.01) = 0.594

The probability that a random client who didn't file a claim is low- risk can be calculated as:

Probability of being low risk and don't file a claim / Probability of not filing a claim

P(LR&not claim)/P(not claim) = 0.594 / (0.144+0.245+0.594)

P(LR&not claim)/P(not claim) = 0.594 /  0.983 = 0.604 or 60.4%

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mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

\mu = \frac{3X_{1} + 4X_{2}  }{8}

Now

Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

                                    = \frac{31.5}{8} - 4.5

                                    = 3.9375 - 4.5

                                    = - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

                                                             = \frac{1}{64} [{3(57.76) + 4(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [7(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [404.32]  } + 0.3136

                                                             = 6.3175 + 0.3136

                                                              = 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

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