Question

question 6
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an insurance company prices its tornado insurance using the following assumptions:

• in any calendar year, there can be at most one tornado.
• in any calendar year, the probability of a tornado is 0.15.
• the number of tornadoes in any calendar year is independent of the number of tornados in any other calendar year.

using the insurance companys assumptions, calculate the probability that there are fewer than 4 tornadoes in a 13 - year period. round your answer to four decimal places.

Explanation:

Step1: Identify the distribution

This is a binomial - distribution problem. Let \(n = 13\) (number of years) and \(p=0.15\) (probability of a tornado in a year). The probability mass function of a binomial distribution is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\). We want to find \(P(X\lt4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\).

Step2: Calculate \(P(X = 0)\)

\[

$$\begin{align*} P(X = 0)&=C(13,0)\times(0.15)^{0}\times(1 - 0.15)^{13-0}\\ &=1\times1\times(0.85)^{13}\\ &\approx0.1209 \end{align*}$$

\]

Step3: Calculate \(P(X = 1)\)

\[

$$\begin{align*} C(13,1)&=\frac{13!}{1!(13 - 1)!}=\frac{13!}{1!12!}=13\\ P(X = 1)&=C(13,1)\times(0.15)^{1}\times(0.85)^{12}\\ &=13\times0.15\times(0.85)^{12}\\ &\approx0.2774 \end{align*}$$

\]

Step4: Calculate \(P(X = 2)\)

\[

$$\begin{align*} C(13,2)&=\frac{13!}{2!(13 - 2)!}=\frac{13\times12}{2\times1}=78\\ P(X = 2)&=C(13,2)\times(0.15)^{2}\times(0.85)^{11}\\ &=78\times0.0225\times(0.85)^{11}\\ &\approx0.2937 \end{align*}$$

\]

Step5: Calculate \(P(X = 3)\)

\[

$$\begin{align*} C(13,3)&=\frac{13!}{3!(13 - 3)!}=\frac{13\times12\times11}{3\times2\times1}=286\\ P(X = 3)&=C(13,3)\times(0.15)^{3}\times(0.85)^{10}\\ &=286\times0.003375\times(0.85)^{10}\\ &\approx0.2078 \end{align*}$$

\]

Step6: Calculate \(P(X\lt4)\)

\[

$$\begin{align*} P(X\lt4)&=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\\ &\approx0.1209 + 0.2774+0.2937+0.2078\\ &\approx0.9008 \end{align*}$$

\]

Answer:

\(0.9008\)