Chapter 1.3, Problem 69E

(a)

To determine

To find: The standard deviation of reflectance of smolts data.

Answer to Problem 69E

Solution: The standard deviation of the provided data is 8.80.

Explanation of Solution

Calculation: Follow the steps given below to obtain standard deviation of the data,

Step 1: Enter the data into Minitab worksheet.

Step 2: Go to Stat, select Basic Statistics and click on Display Descriptive Statistics.

Step 3: In the dialogue box that appears select the name of the column containing the provided data under the field marked as Variables.

Step 4: Next click on Statistics tab and tick mark the option against standard deviation and click on OK twice to obtain the output.

The value of the standard deviation as obtained from Minitab is 8.80.

(b)

To determine

To find: The quartiles of reflectance in data.

Answer to Problem 69E

Solution: The first quartile is $\underset{\_}{Q_1=43.79}$ and third quartile is $\underset{\_}{Q_3=57.02}$

Explanation of Solution

Calculation: Follow the steps given below to obtain the quartiles of the data,

Step 1: Enter the data into Minitab worksheet.

Step 2: Go to Stat, select Basic Statistics and click on Display Descriptive Statistics.

Step 3: In the dialogue box that appears select the name of the column containing the provided data under the field marked as Variables.

Step 4: Next click on Statistics tab and tick mark the option against first and third quartile and click on OK twice to obtain the output.

From Minitab output the first and the third quartiles are obtained as 43.79 and 57.02 respectively.

(c)

To determine

To find: The quartiles for the data.

Answer to Problem 69E

Solution: The first quartile $Q_1$ is 43.5 and third quartile $Q_3$ is 57.02.

Explanation of Solution

Calculation: The observations are arranged in the ascending order as:

33.5,37.8,38.3,38.8,42.1,42.1,42.3,42.3,42.7,42.8,42.8,43.2,43.8,43.8,44.3,44.5,45.6,45.6,46.4,46.6,47.3,47.6,47.8,47.9,49.3,49.7,50.4,50.9,51.4,53,53.9,54.7,54.8,55.5,55.9,56.1,56.1,57,57.6,58.3,58.9,59.6,60.4,63.3,63.4,63.6,64.5,68.1,69,69.2

Here, the total number of observations $n=50$  is even. So, the median $\left(M\right)$ is calculated as,

$\begin{array}{ccc}M& =& \frac{\left(\text{value}\text{of}{\left(\frac{n}{2}\right)}^{th}\text{observation}+\text{value}\text{of}{\left(\frac{n}{2}+1\right)}^{th}\text{observation}\right)}{2}\\ & =& \frac{\left(\text{value}\text{of}{\left(\frac{50}{2}\right)}^{th}\text{observation}+\text{value}\text{of}{\left(\frac{50}{2}+1\right)}^{th}\text{observation}\right)}{2}\\ & =& \frac{\left(\text{value}\text{of}\text{}{25}^{th}\text{observation}+\text{value}\text{of}{26}^{th}\text{observation}\right)}{2}\\ & =& \frac{\left(49.3+49.7\right)}{2}\\ & =& 49.5\end{array}$

The first Quartile $\left(Q_1\right)$ is computed as,

$\begin{array}{ccc}{Q}_1& =& {\left(\frac{n+1}{4}\right)}^{th}\text{observation}\\ & =& {\left(\frac{51}{4}\right)}^{th}\text{observation}\\ & =& {\left(12.75\right)}^{th}\text{observation}\\ & =& {\left(13\right)}^{th}\text{observation}\\ & =& \text{43}\text{.79}\end{array}$

The third Quartile $\left(Q_3\right)$ is computed as,

$\begin{array}{ccc}{Q}_3& =& {\left(3\left(\frac{n+1}{4}\right)\right)}^{th}\text{observation}\\ & =& {\left(3\left(\frac{51}{4}\right)\right)}^{th}\text{observation}\\ & =& {\left(38.25\right)}^{th}\text{observation}\\ & =& {\left(39\right)}^{th}\text{observation}\\ & =& \text{57}\text{.02}\end{array}$

Interpretation: The 25% of the observations lie below the first quartile, which is below 43.5 and 25% of the observations lie above the third quartile, that is, 57.02. The median divides the entire data set into two equal halves.

(d)

To determine

To explain: The better measure to represent the spread of data, quartile or standard deviation.

Answer to Problem 69E

Solution: The standard deviation should be preferred over quartile for the provided data to represent the spread of dataset.

Explanation of Solution

The distribution of provided dataset is normal and its does not have any skewness toward left or right, besides that mean and median is almost same, indicating that data is normal, therefore standard deviation should be preferred over quartile to measure the spread in the data.