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C.5. At two years of age, sardines inhabiting Japanese waters have a length distribution that is approximately normal with mean
20.20cm and standard deviation
0.65cm. How long are the longest
15% of all these sardines?

C. $5$ . At two years of age, sardines inhabiting Japanese waters have a length distribution that is approximately normal with mean $20.20 \mathrm{~cm}$ and standard deviation $0.65 \mathrm{~cm}$ . How long are the longest $15 \%$ of all these sardines?

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Interpret measures of center and variability

Full solution

Q. C.

5

. At two years of age, sardines inhabiting Japanese waters have a length distribution that is approximately normal with mean

20.20 \mathrm{~cm}

and standard deviation

0.65 \mathrm{~cm}

. How long are the longest

15 \%

of all these sardines?

Identify z-score for $85$ th percentile: Identify the z-score that corresponds to the longest $15\%$ of the distribution. $\newline$ We need to find the z-score for the $85$ th percentile since the longest $15\%$ is the top $85\%$ . $\newline$ Using a z-table or calculator, the z-score for the $85$ th percentile is approximately $1.04$ .
Use z-score formula: Use the z-score formula to find the length. $\newline$ The formula is: $X = \mu + (z \times \sigma)$ , where $X$ is the length we want to find, $\mu$ is the mean, $z$ is the z-score, and $\sigma$ is the standard deviation. $\newline$ $X = 20.20 + (1.04 \times 0.65)$
Calculate the length: Calculate the length. $\newline$ $X = 20.20 + (1.04 \times 0.65)$ $\newline$ $X = 20.20 + 0.676$ $\newline$ $X = 20.876 \, \text{cm}$

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