Answer:
10.26cm
Step-by-step explanation:
With 10,000 shellfish approximately normally distributed with mean length of 10 cm and standard deviation 0.2cm, we can do a reverse normal probability to find out the length that corresponds to the top 10% (or larger than 90%) of the shellfish length:
[tex]\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} = 0.9[/tex]
[tex]e^{-\frac{1}{2}x^2} = \sqrt{2\pi}*0.9 = 2.256[/tex]
[tex]-\frac{1}{2}x^2 = ln(2.256) = 0.8136 [/tex]
[tex]x = 10.26cm[/tex]
