Given the data above:

(1) The range is the highest value minus the lowest value. In this case the range is 32.6-16.5=16.1

(2) The arithmetic mean of the temperatures is approximately 25.34 while the median temperature is 26.

(3) You could model the data given with a sinusoidal function. In...

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Given the data above:

(1) The range is the highest value minus the lowest value. In this case the range is 32.6-16.5=16.1

(2) The arithmetic mean of the temperatures is approximately 25.34 while the median temperature is 26.

(3) You could model the data given with a sinusoidal function. In real life, temperatures will on average vary with the season, increasing in the summer and decreasing in the winter (at least at latitudes away from the equator.)

The plot of the given data shows one period of the sinusoid. (A sinusoid is a function that can be represented as a sine or cosine function.)

(4) You will need to plot the points.

(5)(6) T(t)=asin[b(t-c)]+d:

T(t) is the temperature at time t. Here T is measured in degrees and is the average temperature for a month. t is time measured in months.

a: represents the amplitude. This is the amount the function goes above or below the midline. (If a<0 then the graph of the base function is reflected across the horizontal axis.) Here a represents the amount the temperature varies from the mean.

b: is used to find the period and vice versa. `p=(2pi)/b` . This represents the number of oscillations in a period of time. Here b sets the period at 12 months.

c: is a horizontal translation of the base graph; usually called a phase shift in this type of application. If you use a cosine function, then there will be no phase shift as the function begins at its minimum. (Just have a<0) If you use a sine function, you will need to translate 3 units left or right depending on the sign of a.

d: is a vertical translation of the base graph. This sets the midline for the graph. Here d is the average yearly temperature.

(7) a: `a=(32.6-16.5)/2=8.05`

b: since the period should be 12 months, `b=(2pi)/p=(2pi)/12=pi/6`

c: Again if we use cosine there is no need for a horizontal translation, c=0.

d: The midline is the average yearly temperature so `d~~25.34`

`T(t)=-8.05cos(pi/6 t)+25.34` is a possible model for the data.

If you use sine then you must use a phase shift of 3 units. If a>0 then shift 3 units left so c=3

`T(t)=8.05sin[pi/6(t-3)]+25.34`

**My calculator gives a sine regression : `y~~8.555sin(.4773x-1.315)+24.5599`

The graph of `T(t)=-8.05cos(pi/6t)+25.34` :