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2 years ago

Can anyone please help me?

Mathematics

1 answer:

d1i1m1o1n [39]2 years ago

6 0

Answer:

f(0)

Step-by-step explanation:

The y intersect and x intercept are both 0 so you get f(0)

Hope this helps

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2 years ago

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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2 years ago

weekly wages at a certain factory are normally distributed with a mean of $400 and a standard deviation of 50. find the probabil

GalinKa [24]

Answer:

47.5%

Step-by-step explanation:

Find the z-scores:

z = (x − μ) / σ

z₁ = (300 − 400) / 50 = -2

z₂ = (400 − 400) / 50 = 0

300 is 2 standard deviations below the mean, and 400 is 0 standard deviations from the mean.

From the empirical rule, 95% of a normal distribution is between -2 and +2 standard deviations. Since normal distributions are symmetrical, that means half of that is between -2 and 0 standard deviations.

95% / 2 = 47.5%

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