Chapter 0.9, Problem 4E

To determine

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

Answer to Problem 4E

  $\text{Range}=49,\text{}19,\text{Variance}=210.446,42.265,\text{}\text{Standard deviation}=\text{14}\text{.5, 6}\text{.5}$

Explanation of Solution

Given information:

  

Calculation:

It is a sample data because we know exactly it is about two different rides and not all of them.

To find mean, add all observations and divide it by total number of observations.

The mean of ride A is 38.4 and mean of ride B is 40.6

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

  $\begin{array}{l}RideA,60-11=49\\ RideB,50-31=19\end{array}$

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.

  $\begin{array}{l}A={461}^2=212521,\text{ 212521/12=17710}\text{.08}\\ B={487}^2=237169,\text{ 237169/12=19764}\text{.08}\end{array}$

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

  $\begin{array}{l}A={11}^2+{22}^2+\cdots =20025\\ B={31}^2+{32}^2+\cdots =20299\end{array}$

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

  $\begin{array}{l}20025-17710.08=2314.916\\ 20229-1976.08=464.916\end{array}$

Fourth, take integers from third step and divide them with total number of integers in set of minus one. Which means 12 − 1 = 11.

  $\begin{array}{l}\frac{2314.916}{11}=210.446\\ \frac{464.916}{11}=42.265\end{array}$

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

  $\begin{array}{l}\sqrt{210.446}=14.5\\ \sqrt{42.265}=6.5\end{array}$

Thus, $\text{Range}=49,\text{}19,\text{Variance}=210.446,42.265,\text{}\text{Standard deviation}=\text{14}\text{.5, 6}\text{.5}$