Question

In a certain lake, trout average 12 in. in length with standard deviation 2.75 in. and the bass average 4 lb. in weight with standard deviation 0.8 lb. If Deion caught an 18-in trout and Keri caught a 6-lb bass, which fish was the better catch?

Answer 1

Answer:

The bass fish was the better catch

Step-by-step explanation:

From the question we are told that

     The  population mean for trout is  $\mu_1 = 12 \ in$

     The  standard deviation is  $\sigma_1 = 2.75 \ in$

      The  population mean for  base  is  $\mu _2 = 4 \ lb$

      The standard deviation is  $\sigma_2 = 0.8 \ lb$

      The number of  trout caught   $x_1 = 18$

     The number of  bass caught  $x_2 = 6$

Generally z-value(standardized value ) for the of number  trout caught  is mathematically represented as

        $z_1 = \frac{x_1 - \mu_1}{\sigma_1 }$

substituting value

       $z_1 = \frac{18 - 12}{2.75 }$

       $z_1 = 2.18$

Generally z-value(standardized value ) for the of number  bass caught  is mathematically represented as

        $z_2 = \frac{x_2 - \mu_2}{\sigma_2 }$

substituting value

       $z_2 = \frac{6 - 4}{0.8 }$

       $z_2 = 2.5$

From our calculation we see that  $z_2 > z_1$

The  fish that was the better catch is the bass fish