Chapter 6.6, Problem 31AYU

(a)

To determine

To find: The scatter diagram for the data of one period.

Explanation of Solution

Given:

The given table is shown in table 1

Table 1

Month, $x$	Average Monthly Temperature, $°\text{F}$
January 1	25.5
February 2	29.6
March 3	41.4
April 4	52.4
May 5	62.8
June 6	71.9
July 7	75.4
August 8	73.2
September 9	66.6
October 10	54.7
November 11	43.0
December 12	30.9

Calculation:

The scatter plot of the given data is shown in Figure 1

  

Figure 1

(b)

To determine

To find: The sinusoidal function of the given form.

Answer to Problem 31AYU

The required sinusoidal function is $y=24.95\sin \left(\frac{\pi }{6}x-\frac{2\pi }{3}\right)+50.45$ .

Explanation of Solution

Given:

The given form of the sinusoidal function is,

  $y=A\sin \left(\omega x-\varphi \right)+B$

Calculation:

Consider the given sine function is,

  $y=A\sin \left(\omega x-\varphi \right)+B$

Consider the amplitude $A$ of the sine function is,

  $A=\frac{\text{largest}\text{data}\text{value}-\text{smallest}\text{data}\text{value}}{2}$

Then,

  $\begin{array}{c}A=\frac{75.4-25.5}{2}\\ =24.95\end{array}$

Consider the expression for the vertical shift of the sinusoidal function is,

  $B=\frac{\text{largest}\text{data}\text{value}+\text{smallest}\text{data}\text{value}}{2}$

Then,

  $\begin{array}{c}B=\frac{75.4+25.5}{2}\\ =50.45\end{array}$

Consider the expression for the time period of the data is,

  $\begin{array}{l}T=\frac{2\pi }{\omega }\\ \omega =\frac{2\pi }{T}\end{array}$

As the data repeats after every 12 months, then,

  $\begin{array}{c}\omega =\frac{2\pi }{12}\\ =\frac{\pi }{6}\end{array}$

Consider the horizontal shift is determined by dividing the 12 in four equal parts as,

  $\left[0,3\right],\left[3,6\right],\left[6,9\right],\left[9,12\right]$

Since, the curve increases in the interval of $\left[0,3\right]$ and decreases in the interval of $\left[3,9\right]$ . The local maximum value is at $x=3$ . For the data the maximum value of the function occurs at $x=7$ . Then, shift the graph of the function as,

  $7-3=\text{4}\text{units}$ .

Then, the expression for the sinusoidal function is,

  $\begin{array}{c}y=24.95\sin \left(\frac{\pi }{6}\left(x-4\right)\right)+50.45\\ =24.95\sin \left(\frac{\pi }{6}x-\frac{2\pi }{3}\right)+50.45\end{array}$

(c)

To determine

To find: The scatter plot for the function found in part (b).

Answer to Problem 31AYU

The required plot is shown in Figure 2

Explanation of Solution

Consider the sinusoidal function is,

  $y=24.95\sin \left(\frac{\pi }{6}x-\frac{2\pi }{3}\right)+50.45$

The graph for the sinusoidal model is shown in Figure 2

  

Figure 2

(d)

To determine

To find: The sinusoidal function of the best fit model.

Answer to Problem 31AYU

The best fit model is $y=22.7\sin \left(\frac{\pi }{6}x-\frac{\pi }{2}\right)+58.042$ .

Explanation of Solution

The best fit model with the help of graphing calculator is,

  $y=22.7\sin \left(\frac{\pi }{6}x-\frac{\pi }{2}\right)+58.042$

(e)

To determine

To find: The sinusoidal function of best fit on the scatter diagram of the data.

Answer to Problem 31AYU

The plot for the best fit model is shown in Figure 3

Explanation of Solution

Consider the best fit model is,

  $y=22.7\sin \left(\frac{\pi }{6}x-\frac{\pi }{2}\right)+58.042$

The sinusoidal plot for the best fit model is shown in Figure 3

  

Figure 3