Question

g The average lifespan for a certain type of vehicle is 8 years and follows an exponential distribution. A lot contains 200 of these vehicles, brand new. What is the probability that 50 or more of them fail in the first 2 years

Answer 1

Answer:

The probability is $P(A \ge 50 ) = 0.14321$

Step-by-step explanation:

The average life span is  $\pi = 8$

The sample size is n =  200

Generally we are told that average lifespan for a certain type of vehicle follows an exponential distribution

i.e  

             $X \~ \ \ \ exp(8)$

Gnerally the probability distribution function for exponential distribution is

       $f(x) = \left \{ {{\lambda * e^{-\lambda * x }\ \ \ x\ge 0 } \atop {0 \ \ \ \ \ \ \ \ \ \ x < 0}} \right.$

Now $\lambda$ is the  constant which is evaluated as

      $\lambda = \frac{1}{\pi}$

=>   $\lambda = \frac{1}{8}$

=>   $\lambda = 0.125$

Generally the that the vehicles will fail in less than 2 years is  mathematically represented as

         $P(X \le 2 )= p= \int\limits^2_0 {\lambda e^{- \lambda * x} } \, dx$

        $P(X \le 2 )= p= [ {0.125 * - \frac{1}{0.125 } e^{- 0.125 * x} } ]|\left 2 } \atop {0}} \right.$

       $P(X \le 2 )= p= [ {0.125 * - \frac{1}{0.125 } e^{- 0.125 * 2} } ] - [ {0.125 * - \frac{1}{0.125 } e^{- 0.125 * 0} } ]$

     $P(X \le 2 )= p= -0.77880 + 1$

     $P(X \le 2 )= p= 0.2212$

Generally the probability that the vehicle will fail in the first 2 years follows a binomial  distribution  with the mean evaluated as

         $\mu = n * p$

=>      $\mu = 200 * 0.2212$

=>      $\mu = 44.24$

The standard deviation is

       $\sigma = \sqrt{np(1 - p)}$    

      $\sigma = \sqrt{200 * 0.2212 (1 - 0.2212)}$  

      $\sigma = 5.86$

Using normal approximation of binomial  distribution the  probability that 50 or more of them fail in the first 2 years is mathematically represented as

   $P(A \ge 50 ) = 1 - P(A < 50 )$

Applying continuity correction

    $P(A \ge 50.5 ) = 1 - P(A < 50.5 )$

Here

    $P(A < 50.5 ) = P(\frac{A - \mu }{\sigma} < \frac{50 - 44.24}{5.87} )$

Generally  

       $\frac{A - \mu }{\sigma} = Z (The \ standardized \ value \ of A )$

=> $P(A < 50.5 ) = P( Z < 1.0664 )$

From the z-table the probability value  of  ( Z  <  1.0664   ) is  

     $P( Z < 1.0664 ) = 0.85679$

So  

   $P(A \ge 50 ) = 1 - 0.85679$

=> $P(A \ge 50 ) = 0.14321$