Question

Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. Find the percentage of buyers who paid:

between $150,000 and $152,400 if the standard deviation is $1200.

Answer 1

Answer:

The percentage is  $P(x_1 < X < x_2) = 47.7 \%$

Step-by-step explanation:

From the question we are told that

   The  population mean is  $\mu = \ 150000$

    The standard deviation is  $\sigma = \ 1200$

    The prices we are considering is $x_1 = \ 150000 \to \ x_2 = \ 152400$

Given that the price is normally distributed , the percentage the percentage of buyers who paid between $150,000 and $152,400 is mathematically represented as

      $P(x_1 < X < x_2) = P(\frac{x_1 - \mu}{\sigma } < \frac{X - \mu}{\sigma } < \frac{x_2 - \mu}{\sigma })$

So   $\frac{X - \mu}{\sigma }$  is equal to z (the standardized value of  X  )

   So

        $P(x_1 < X < x_2) = P(\frac{x_1 - \mu}{\sigma } <Z < \frac{x_2 - \mu}{\sigma })$

substituting values

       $P(x_1 < X < x_2) = P(\frac{150000 - 150000}{1200 } <Z < \frac{152400 - 150000}{1200 })$

      $P(x_1 < X < x_2) = P(0<Z < 2)$

      $P(x_1 < X < x_2) = P( Z < 2) - P( Z < 0 )$

From the standardized normal distribution table $P(Z < 2 ) = 0.97725$ and  

   $P(Z < 0) = 0.5$

So  

     $P(x_1 < X < x_2) = 0.97725 - 0.5$

    $P(x_1 < X < x_2) = 0.47725$

The percentage is  $P(x_1 < X < x_2) = 47.7 \%$