Chapter 0.9, Problem 6E

To determine

To find: Data is sample or population and then find range, variance and standard deviation.

Answer to Problem 6E

  $\text{Range}=15,\text{}12,\text{Variance}=14.16,10.85,\text{}\text{Standard deviation}=\text{3}\text{.76, 3}\text{.29}$

Explanation of Solution

Given information:

  

Calculation:

It is a population data because we know it is all about students.

The mean of A is 36.66 and mean of B is 43.77

The range is the difference between the highest and lowest value.

  $\begin{array}{l}A,15-0=15\\ B,13-1=12\end{array}$

The variance is the average square deviation from the mean.

First add all numbers from each set and calculate the square then divide it with the number of integers in set.

  $\begin{array}{l}A={243}^2=59049,\\ B={234}^2=54756\\ A\to \frac{59049}{30}=1968.3,B\to \frac{54756}{30}=1825.2\end{array}$

Second we take every integer from each set and calculate square of them and add all together.

  $\begin{array}{l}A=0^2+1^2+\cdots =2379\\ B=1^2+2^2+\cdots =2140\end{array}$

Third, we subtract integers from second step with ones from first step.

  $\begin{array}{l}2379-1968.3=410.7\\ 2140-1825.2=314.79\end{array}$

Fourth, take integers from third step and divide them with total number of integers in set.

  $\begin{array}{l}\frac{410.7}{30}=13.69\\ \frac{314.79}{30}=10.493\end{array}$

The standard deviation is the average deviation from the mean. It is found by taking the square root of the variance.

  $\begin{array}{l}\sqrt{13.69}=3.7\\ \sqrt{10.493}=3.23\end{array}$

Thus, $\text{Range}=15,\text{}12,\text{Variance}=13.69,10.493,\text{}\text{Standard deviation}=\text{3}\text{.7, 3}\text{.239}$

The Class A is more varied from mean which can be estimated by comparing it with Standard deviation. The class A is more varied from mean when compared to class B.