Question

The annual gas bill for a town household are considered to be normally distributed with a mean of $ 1130 and a standard deviation of $ 150. If one household is randomly selected, what is the probability that the gas bill will be between $900 and $1100?

Answer 1

Answer:

The probability is  $P(900 < X < 1100) = 0.358102$

Step-by-step explanation:

From the question we are told that

   The  population mean is  $\mu = \ 1130$

    The standard deviation is  $\sigma = \ 150$

   

Generally the probability that the gas bill will be between $900 and  $1100 is mathematically represented as

      $P(900 < X < 1100) = P(\frac{900 - 1130 }{150 } < \frac{X - \mu}{\sigma } < \frac{1100 - 1130 }{150 } )$

=>  $P(900 < X < 1100) = P(-1.533 < \frac{X - \mu}{\sigma } < -0.2 )$

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

So

    $P(900 < X < 1100) = P(-1.533 < Z< -0.2 )$

   $P(900 < X < 1100) = P( Z< -0.2 ) - P(Z < -1.533)$

From the z  table  

     $P(Z < -0.2 ) = 0.42074$

and

     $P(Z < -1.533) = 0.062638$

So

     $P(900 < X < 1100) = 0.42074 - 0.062638$

=> $P(900 < X < 1100) = 0.358102$