Question

A grocery store receives deliveries of corn from two farms, one in Iowa and the other in Ohio. Both farms produce ears of corn with mean weight 1.26 pounds. The standard deviation of the weights of the ears of corn from the farm in Ohio is 0.01 pound greater than that from the farm in Iowa. A randomly selected ear of corn from the farm in Iowa weighed 1.39 pounds, which has a standardized score of 1.645 for the distribution of weights for the Iowa corn. If an ear of corn from the farm in Ohio weighs 1.39 pounds, how many standard deviations from the mean is the weight with respect to the Ohio distribution

Answer 1

Answer:

The number of standard deviation above the mean is  $z_o = 1.4607$

Step-by-step explanation:

From the question we are told that

   The mean weight of  ears of  corn from each farm is  $\mu=1.26$

   The standard deviation of ears of corn from Iowa is $\sigma_1$

   The standard deviation of ears of corn from Ohio is

       $\sigma_2= 0.01 + \sigma_1$

     

  The weight of a random selected ear of corn from Iowa is  x =  1.39 pound

  The standardized score is  z =1.645

   The weight of a random selected ear of corn from Ohio is  $x_1 = 1.39\ pound$

Generally the standardized score of weight of corn from Iowa is mathematically  represented as

        $z = \frac{ 1.39 - 1.26 }{\sigma_1 }$

=>     $1.645 = \frac{ 1.39 - 1.26 }{\sigma_1 }$  

=>      $\sigma_1 = \frac{ 1.39 - 1.26 }{ 1.64 5}$  

=>      $\sigma_1 = 0.0790$  

Generally the standardized score of weight of corn from Ohio is mathematically  represented as

         $z_o = \frac{ 1.39 - 1.26 }{\sigma_2 }$

=>     $z_o= \frac{ 1.39 - 1.26 }{0.01 + \sigma_1 }$  

=>      $z_o = \frac{ 1.39 - 1.26 }{ 0.089}$  

=>      $z_o = 1.4607$

Because the value is positive it means that this value represent the number of standard deviation above the mean.