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The lengths of trout in a lake are normally distributed with a mean of 30 inches and a standard deviation of 4.5 inches.



Enter the z-score of a trout with a length of 28.2 inches.

The lengths of trout in a lake are normally distributed with a mean of 30 inches and a standard deviation of 45 inches Enter the zscore of a trout with a length class=

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MathPhys

Answer:

-0.4

Step-by-step explanation:

z score is:

z = (x - μ) / σ

For x = 28.2, μ = 30, and σ = 4.5:

z = (28.2 - 30) / 4.5

z = -0.4

JeanaShupp

Answer: -0.4

Step-by-step explanation:

Given: Mean : [tex]\mu=30\text{ inches}[/tex]

Standard deviation : [tex]\sigma=4.5\text{ inches}[/tex]

The formula to calculate z-score is given by :-

[tex]z=\dfrac{X-\mu}{\sigma}[/tex]

For X = 28.2 inches, we have

[tex]z=\dfrac{28.2-30}{4.5}\\\\\Rigahtarrow\ z=-0.4[/tex]

Hence, the z-score of a trout with a length of 28.2 inches.= -0.4

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