Chapter 6: Q. 6.40 (page 261)

Green Sea Urchins.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. Let $x$ denote weight of adult green sea urchins.
a. Sketch the distribution of the variable $x$ .
b. Obtain the standardized version, $z,$ of $x$ .
c. Identify and sketch the distribution of $z$ .
d. The percentage of adult green sea urchins with weights between $50$ g and $60$ g is equal to the area under he standard normal curve between and .
e. The percentage of adult green sea urchins with weights above $40$ g is equal to the area under the standard normal curve that lies to the of _ .

Short Answer

Expert verified

Part $(a)$ : The distribution of $x$ is as follows :

Part $(b)$ : The standardized version of $x$ interms of $z$ is, $z = \frac{x - 52.0}{17.2}$ .

Part $(c)$ : The distribution of $z$ is as follows :

Part $(d)$ : The percentage of adult green sea urchins with weights between $50$ g and $60$ g is equal to the area under the standard normal curve between $- 0.12$ and $0.47$ .

Part $(e)$ : The percentage of adult green sea urchins with weights above $40$ g is equal to the area under the standard normal curve that lies to therightof $- 0.70$ .

Step by step solution

Part a Step 1. Given information

Let $x$ denote the weight of adult green sea urchins.

Part a Step 2. The distribution of x is as follows:

Part b Step 1. The standardized version of x interms of z is,

$z = \frac{x - 52.0}{17.2}$ .

Part c Step 1. The distribution of z is as follows:

Part d Step 1. Substitute x=50 in the standardized version of x.

$z = \frac{50 - 52.0}{17.2} = - 0.12$

Substitute $x = 60$ in the standardized version of $x$ .

$z = \frac{60 - 52.0}{17.2} = 0.47$

Therefore, the percentage of adult green sea urchins with weights between $50$ g and $60$ g is equal to the area under the standard normal curve between $- 0.12$ and $0.47$ .

Part e Step 1. Substitute x=40 in the standardized version of x.

$z = \frac{40 - 52.0}{17.2} = - 0.70$

Therefore, the percentage of adult green sea urchins with weights above $40$ g is equal to the area under the standard normal curve that lies to the right of $- 0.70$ .