Giving out brainliest me ! 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 4.5 inches

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

Formula to find z score is

Length = 28.2, mean = 30 , standard deivation = 4.5

z = -0.4

So z-score = -0.4

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