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

The bass fish was the better catch

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 = (x_1 - \mu_1)/(\sigma_1 )`

substituting value

`z_1 = (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 = (x_2 - \mu_2)/(\sigma_2 )`

substituting value

`z_2 = (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