Question

The time needed to complete a job is normally distributed with a mean of 1 hour and standard deviation of 10 minutes. Probability to complete the job in hone hour or more, but less than 70 minutes is:

Answer 1

Answer:

The value is $P(60 < X < 70) = 0.34134$    

Step-by-step explanation:

From the question we are told that

   The mean is  $\mu = 1 \ hour = 60 \ minutes$

    The standard deviation is  $\sigma = 10 \ minutes$

Generally the 1 hour is equivalent to 60 minutes

   Generally the probability to complete the job in one hour or more, but less than 70 minutes is mathematically represented as

    $P(60 < X < 70) = P(\frac{60 - 60 }{10 } < \frac{X - \mu}{\sigma } < \frac{70 - 60 }{10 } )$

=> $P(60 < X < 70) = P(0 < \frac{X - \mu}{\sigma } < 1 )$

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

=> $P(60 < X < 70) = P(0 <Z < 1 )$

=>     $P(60 < X < 70) = P(Z < 1) - P(Z < 0 )$

From the  z  table  (Z <  1)  and  (Z <  0 ) is  

        $P(Z < 1) = 0.84134$

and  

      $P(Z < 0 ) = 0.5$

=>    $P(60 < X < 70) = 0.84134 - 0.5$

=>    $P(60 < X < 70) = 0.34134$