Question

A 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 0.11.
• The number of tornadoes in any calendar year is independent of the number of tornados in any other calendar year.

Required:
Using the company's assumptions, calculate the probability that there are fewer than 3 tornadoes in a 14-year period.

Answer 1

Answer:

The  probability is   $P(X < 3) = P(X \le 3-1 = 2 ) = 0.8074$        

Step-by-step explanation:

From the question we are told that

   The  probability of a tornado is $p = 0.11$

    The sample  size is  $n = 14$

     

Since the number of tornadoes in any calendar year is independent of the number of tornadoes in any other calendar year and  there can be  only  outcome so we can evaluate probability using binomial distribution.

The probability of a tornado not occurring is mathematically evaluated

         $q = 1 - p$

=>      $q = 1 - 0.11$

=>      $q = 0.89$

The probability that there are fewer than 3 tornadoes in a 14-year period is mathematically represented as

      $P(X < 3) = P(X \le 3-1 = 2 ) = \left n } \atop {}} \right. C_2 * p ^2 q^{n- 2 } + \left n } \atop {}} \right. C_1 * p ^1 q^{n- 1 } + \left n } \atop {}} \right. C_0 * p ^r q^{n- 0 }$

      $P(X < 3) = P(X \le 3-1 = 2 ) = \left 14 } \atop {}} \right. C_2 * (0.11) ^2 (0.89)^{14- 2 } + \left 14 } \atop {}} \right. C_1 * (0.11) ^1 (0.89)^{14- 1 } + \left 14 } \atop {}} \right. C_0 * (0.11) ^0 (0.89)^{14- 0 }$

       $P(X < 3) = P(X \le 3-1 = 2 ) = 91 * 0.0121*0.247 + 14 * 0.11*0.2198 + 1 * 1 * 0.197$

$P(X < 3) = P(X \le 3-1 = 2 ) = 0.8074$