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The head-to-tail length of a species of fish is normally distributed, with amean of 17.4 cm and a standard deviation of 3.2 cm. One fish is 14.3 cmlong. What is the z-score of the length of this fish? Round your answer to twodecimal places.A. -0.67B. 0.67C. -0.97D. 0.97

The head-to-tail length of a species of fish is normally distributed, with amean of-example-1
User Zap
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Use the next formula to find the z-score:


\begin{gathered} z=(x-\mu)/(\sigma) \\ \\ \mu:mean \\ \sigma:standard\text{ }deviation \end{gathered}
\begin{gathered} z=(14.3cm-17.4cm)/(3.2cm) \\ \\ z=(-3.1)/(3.2)\approx-0.97 \end{gathered}

Then, the z-score for the given situation i s

User Prijupaul
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