From the paper “Effects of Chronic Nitrate Exposure on Gonad Growth in Green Sea Urchin Strongylocentrotus droebachiensis” (Aquaculture, Vol. 242, No. 1–4, pp. 357–363) by S. Siikavuopio et al., we found that weights of adult green sea urchins are normally distributed with mean 52.0 g and standard deviation 17.2 g.
a. Find the percentage of adult green sea urchins with weights between 50 g and 60 g.
b. Obtain the percentage of adult green sea urchins with weights above 40 g.
c. Determine and interpret the 90th percentile for the weights.
d. Find and interpret the 6th decile for the weights.
SOLUTION
(a) For adult green sea urchins with weights of 50 and 60 g, the z-values are
z = 50 -52 / 17.2 = -0.12 and z= 60 – 52 / 17.2 = 0.47
The area to the left of z = -0.12 is 0.4522 and the area to the left of z = 0.47 is 0.6808. Therefore the area between z = -0.12 and z = 0.47 is 0.6808 - 0.4522 = 0.2286. Thus the percentage of adult green sea urchins with weights between 50 g and 60 g is 22.86%.
(b) For a weight of 40 g, the z value is
z = 40 -52 / 17.2 = -0.70
The area to the left of z = -0.70 is 0.2420. The area to the right of z = -0.70 is 1 – 0.2420 = 0.7580. Thus the percentage of adult green sea urchins with weights above 40 g is 75.80%.
(c) Using Table II, we find that an area of 0.9000 lies to left of z = 1.28. We convert this z-value to an x-value using x = μ + zα. Thus 90% of the adult green sea urchin weights were less than the 90th percentile, x = 52.0 + (1.28)(17.2) = 74.0 g.
(d) The sixth decile is the same as the 60th percentile. Using Table II, we find that an area of 0.6000 lies to the left of z = 0.25. We convert this z-value to an x-value using x = μ + zα. Thus, 60% of the adult green sea urchin weights were less than the 60th percentile, x = 52.0 + (0.25)(17.2) = 56.3 g.