Chapter 4.6, Problem 3AGP

To determine

To calculate: The total attendance at a community pool for the temperature ${90}^{\circ }C$ .

Answer to Problem 3AGP

The attendance is 439.45

Explanation of Solution

Given Information: A table consisting of the values of attendance and corresponding temperatures.

Calculation:

An equation of the best fit line is given by,

  $y=a+bx$ …… (1)

• Here, a is the intercept and bis the slope.

Find $xy\text{ and }{x}^2$ ,

$x$	$y$	$xy$	$x^2$
92	395	36340	8464
102	309	31518	10404
97	332	32204	9409
104	307	31928	10816
108	251	27108	11664
110	178	19580	12100
111	164	18204	12321
111	149	16539	12321
107	273	29211	11449
105	299	31395	11025

Sum of every column is,

  $\begin{array}{c}{\displaystyle \sum x}=1047\\ {\displaystyle \sum y}=2657\\ {\displaystyle \sum x}y=274027\\ {\displaystyle \sum x^2}=109973\end{array}$

Calculate a as,

  $\begin{array}{c}a=\frac{{\displaystyle \sum y\cdot {{\displaystyle \sum x}}^2}-{\displaystyle \sum x\cdot {\displaystyle \sum x}y}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{2657\times 109973-1047\times 274027}{10\times 109973-{1047}^2}\\ =1502.98\end{array}$

Calculate b as,

  $\begin{array}{c}b=\frac{n\cdot {\displaystyle \sum xy}-{\displaystyle \sum x\cdot {\displaystyle \sum y}}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{10\times 274027-1047\times 2657}{10\times 109973-{1047}^2}\\ =-11.817\end{array}$

Substitute the values of a and b in equation (1),

  $\begin{array}{c}y=a+bx\\ y=-11.817x+1502.98\end{array}$

Hence, the equation of best line is,

  $y=-11.817x+1502.98$

Now, when the temperature is 90 degrees,

The value of x will be 90. The value of y from equation (1) is,

  $\begin{array}{c}y=-11.817x+1502.98\\ =-11.817\left(90\right)+1502.98\\ =439.45\end{array}$