Chapter 7.3, Problem 4Q

To determine

The range and standard deviation of the given data set.

Answer to Problem 4Q

Range for the data for juniors: $11$

Range for the data for seniors: $17$

Standard Deviation for the data for juniors = $3.427$

Standard Deviation for the data for seniors = $5.74$

Explanation of Solution

Given:

The given data set is:

  $\begin{array}{l}Juniors:19,15,20,10,14,21,18,15\\ Seniors:22,19,29,32,15,26,30,19\end{array}$

Calculation:

The range and standard deviation of the given data can be calculated as:

Order the given data set for juniors:

  $10,14,15,15,18,19,20,21$

Range of the given data set is the difference between the largest and smallest data value.

Therefore, range for the data for juniors:

  $21-10=11$

Order the given data set for seniors:

  $15,19,19,22,26,29,30,32$

Therefore, range for the data for seniors:

  $32-15=17$

Now, to find the standard deviation for juniors:

Step 1: Calculate the mean.

Mean = $\frac{10+14+15+15+18+19+20+21}{8}=16.5$

Step 2: Calculate the deviation for each data value.

Step 3: Calculate the square of each deviation.

Step 4: Calculate the mean of squared deviations/variance.

Step 5: Make use of calculator to find the square roots of the variance.

  

Standard Deviation = $\sqrt{\frac{94}{8}}=3.427$

Now, to find the standard deviation for seniors:

Step 1: Calculate the mean.

Mean = $\frac{192}{8}=24$

Step 2: Calculate the deviation for each data value.

Step 3: Calculate the square of each deviation.

Step 4: Calculate the mean of squared deviations/variance.

Step 5: Make use of calculator to find the square roots of the variance.

  

Standard Deviation = $\sqrt{\frac{264}{8}}=5.74$