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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?

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6 votes

Answer:

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

User Aimiliano
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