Question

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 $52.0\mathrm{g}$ and standard deviation $17.2\mathrm{g}$. 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 $50\mathrm{g}$ and $60\mathrm{g}$ is equal to the area under the standard normal curve between and

e. The percentage of adult green sea urchins with weights above $40\mathrm{g}$ is equal to the area under the standard normal curve that lies to the of

Short answer

a). The distribution of the variable,

b). The standardized version,

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

c). The distribution of $z$,

d). The percentage of adult green sea urchins having a weight between $50\mathrm{g}$and $60\mathrm{g}$is equal to the area under the standard curve between the points $-0.12$and $0.47$.

e). The percentage of adult green sea urchins having weight above $40\mathrm{g}$is equal to the area under the standard curve that lies between $-0.70$and infinity.

Step-by-step solution

Part (a) Step 1: Given Information

Given data:

Mean$=52.0\mathrm{g}$.

Standard deviation $=17.2\mathrm{g}$.

Part (a) Step 2: Explanation

The histogram of the distribution is roughly bell-shaped when the variable is roughly distributed regularly.

The distribution can be obtained as,

Part (b) Step 1: Given Information

Given data:

Mean $=52.0\mathrm{g}$.

Standard deviation $=17.2\mathrm{g}$.

Part (b) Step 2: Explanation

The histogram of the distribution is roughly bell-shaped when the variable is roughly distributed regularly.

The standardized version will be equal to $z=\frac{x-\text{mean}}{\text{standard deviation}}$

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

Part (c) Step 1: Given Information

Given data:

Mean $=52.0\mathrm{g}$.

Standard deviation $=17.2\mathrm{g}$.

Part (c) Step 2: Explanation

The histogram of the distribution is roughly bell-shaped when the variable is roughly distributed regularly.

The distribution of $z$will be

Part (d) Step 1: Given Information

Given data:

Mean $=52.0\mathrm{g}$.

Standard deviation $=17.2\mathrm{g}$.

Part (d) Step 2: Explanation

The standardized version will be equal to

$z=\frac{x-\text{mean}}{\text{standard deviation}}$

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

For $x=50$,

$z=\frac{50-52}{17.2}$

$z=-0.12$

For $x=60$,

$z=\frac{60-52}{17.2}$

$z=0.47$

Part (e) Step 1: Given Information

Given data:

Mean $=52.0\mathrm{g}$.

Standard deviation $=17.2\mathrm{g}$.

Part (e) Step 2: Explanation

The Standardized version will be equal to

$z=\frac{x-\text{mean}}{\text{standard deviation}}$$z=\frac{x-52.0}{17.2}$

When $x=40$,

$z=\frac{40-52}{17.2}$

$z=-0.70$