The standardized score, that is, the z-score of the caiman that has mass of 15kg is 1.32
Given in the question,
Mean (µ) = 9.81kg
Standard deviation = 3.93 kg
Value = 15 kg
The standardized score ( z-score) for an individual value in a distribution tells us how many standard deviations from the mean the value falls, and in what direction. To find the standardized score ( z -score), compute
z = (value – mean) / standard deviation (1)
Values larger than the mean have positive z -scores and values smaller than the mean have negative z -scores.
So putting the respective values in equation (1) , we get
z = (15 – 9.81)/ 3.93
z = 5.19/3.93
z = 1.32
Hence caiman’s mass is 1.32 standard deviation greater than the mean caiman mass of 9.81kg
Learn more about standardized score here : https://brainly.com/question/23742231
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