Q. 6.1 - Elementary Statistics

Question

From the paper "Effects of Chronic Nitrate Exposure on Gonad Growth in Green Sea Urchin Strongylocentrotus droebachiensis" by S. Siikavuopio et al., we found that weights of adult green sea urchins are normally distributed with mean $52g$ and standard deviation $17.2g$.

Part (a): Find the percentage of adult green sea urchins with weights between $50g$and $60g$.

Part (b): Obtain the percentage of adult green sea urchins with weight above $40g$.

Part (c): Determine and interpret the $90th$ percentile for the weights.

Part (d): Find and interpret the $6th$ decile for the weights.

Step-by-Step Solution

Answer

Part (a): The percentage of adult green sea urchins with weights between $50g$ and $60g$ is $22.86\%$.

Part (b):  The percentage of adult green sea urchins with weight above $40g$ is $75.8\%$.

Part (c): The $90th$percentile for the weight is $74.02g$. $90\%$of the urchins have weight below $74.02g$.

Part (d): The $6th$ decile for the weights $56.3g$ $60\%$ of the urchins have weight below $56.3g$.

Part (a) Step 1. Given information.

The given mean is $52g$ and standard deviation is $17.2g$.

Part (a) Step 2. Draw the figure with weights between 50 g and 60 g .

Drawing the figure delimiting the x-values which are $50g$ and $60g$,

The z-scores corresponding to the x-values are given below,

$$\begin{gathered} x=50\to \\ =\frac{50-52}{17.2} \\ =-0.12 \\ x=60\to \\ =\frac{60-52}{17.2} \\ =0.47 \end{gathered}$$

The area to the left of $-0.12$ is $0.4522$ and the area to the left of $0.47$ is $0.6808$. The required area shaded is between $0.6808-0.4522=0.2286$.

On interpreting, we can say $22.86\%$ of adult sea urchin have weight between $50g,60g$.

Part (b) Step 1. Draw the figure with weights above 40 g .

Drawing the figure delimiting the x-values which is $40g$,

The z-scores corresponding to the x-values are given below,

$$\begin{gathered} z=\frac{40-52}{17.2} \\ =-0.7 \end{gathered}$$

The area to the left of $-0.7$ is $0.242$and the area to the right of $-0.7$ is  $0.758$. The required area shaded is $0.758$.

On interpreting, we can say $75.8\%$ of adult sea urchin have weight above $40g$.

Part (c) Step 1. Determine the 90 th percentile for the weights.

The z-score corresponding to $P_{90}$ is the one having an area $0.9$ to its left under the standard normal curve. From standard normal table, that z-score is $1.28$.

The x-value having the z-score of $1.28$, the length that is $1.28$ standard deviation below the mean. It is $52+1.28\left(17.2\right)=74.02$

The $90$th percentile for the weight is $74.02g$.

On interpreting, we can say $90\%$ of the urchins have weight below $74.02g$.

Part (d) Step 1. Determine the 6 t h decile for the weights.

The z-score corresponding to $P_{60}$, sixth decile is the one having an area $0.6$to its left under the standard normal curve. From standard normal table, that z-score is $0.25$.

 

Now, we must find the x-value having the z-score $0.25$, the length is $0.25$ standard deviations below the mean. It is $52+0.25\left(17.2\right)=56.3$

The $6$th decile for the weights $56.3g$.

On interpreting, we can say, $60\%$ of the urchins have weight below $56.3g$.