Answer:
0.6563 or 65.63% of brook trout caught will be between 12 and 18 inches
Step-by-step explanation:
Mean trout length (μ) = 14 inches
Standard deviation (σ) = 3 inches
The z-score for any given trout length 'X' is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex] e interval
For a length of X =12 inches:
[tex]z=\frac{12-14}{3}\\z=-0.6667[/tex]
According to a z-score table, a score of -0.6667 is equivalent to the 25.25th percentile of the distribution.
For a length of X =18 inches:
[tex]z=\frac{18-14}{3}\\z=1.333[/tex]
According to a z-score table, a score of 1.333 is equivalent to the 90.88th percentile of the distribution.
The proportion of trout caught between 12 and 18 inches, assuming a normal distribution, is the interval between the equivalent percentile of each length:
[tex]P(12\leq X\leq 18) = 90.88\% - 25.25\%\\P(12\leq X\leq 18) = 65.63\%[/tex]