Question

Population of 10,000 shellfish are approximately normally distributed with mean length of 10 cm and standard deviation 0.2cm. What length corresponds to the top 10% of shellfish lengths?

Answer 1

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:

$\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} = 0.9$

$e^{-\frac{1}{2}x^2} = \sqrt{2\pi}*0.9 = 2.256$

$-\frac{1}{2}x^2 = ln(2.256) = 0.8136$

$x = 10.26cm$