Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.3 years, and standard deviation of 0.7 years. If you randomly purchase 14 items, what is the probability that their mean life will be longer than 12 years?

Answer 1

Answer:

The probability is  $P(\= X > 12 ) = 0.72688$

Step-by-step explanation:

From the question we are told that

   The  mean is  $\mu = 12.3 \ years$

    The standard deviation is  $\sigma = 0.7 \ years$

    The sample size is  n = 14

Generally the standard error of the mean is mathematically represented as

         $\sigma _{x} = \frac{\sigma}{\sqrt{n} }$

=>       $\sigma _{x} = \frac{0.7}{\sqrt{14} }$

=>       $\sigma _{x} = 0.1871$

Generally the probability that their mean life will be longer than 12 years is mathematically represented as

      $P(\= X > 12 ) = P(\frac{\= X - \mu }{\sigma} > \frac{12 - 12.3}{ 0.1871 } )$

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

 

       $P(\= X > 12 ) = P(Z > -0.6034 )$

From the z table  the area under the normal curve to the left corresponding to    -0.6034  is

=>    $P(Z > -0.6034 ) = 0.72688$

=>   $P(\= X > 12 ) = 0.72688$