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valentina_108 [34]
3 years ago
5

Losses covered by a flood insurance policy are uniformly distributed on the interval (0,2). The insurer pays the amount of the l

oss in excess of a deductible d. The probability that the insurer pays at least 1.20 on a random loss is 0.30. Calculate the probability that the insurer pays at least 1.44 on a random loss.
Mathematics
1 answer:
Cloud [144]3 years ago
5 0

Answer:

The probability that the insurer pays at least 1.44 on a random loss is 0.18.

Step-by-step explanation:

Let the random variable <em>X</em> represent the losses covered by a flood insurance policy.

The random variable <em>X</em> follows a Uniform distribution with parameters <em>a</em> = 0 and <em>b</em> = 2.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

It is provided, the probability that the insurer pays at least 1.20 on a random loss is 0.30.

That is:

P(X\geq 1.2+d)=0.30\\

⇒

P(X\geq 1.2+d)=\int\limits^{2}_{1.2+d}{\frac{1}{2}}\, dx

                0.30=\frac{2-1.2-d}{2}\\\\0.60=0.80-d\\\\d=0.80-0.60\\\\d=0.20

The deductible d is 0.20.

Compute the probability that the insurer pays at least 1.44 on a random loss as follows:

P(X\geq 1.44+d)=P(X\geq 1.64)

                        =\int\limits^{2}_{1.64}{\frac{1}{2}}\, dx\\\\=|\frac{x}{2}|\limits^{2}_{1.64}\\\\=\frac{2-1.64}{2}\\\\=0.18

Thus, the probability that the insurer pays at least 1.44 on a random loss is 0.18.

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\frac{1+\tan x}{\sin x+\cos x}=\sec x is proved

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