We will determine the equation of the parabola that best fits the given data and calculate the standard deviation.

1. Determining the Equation of the Parabola

A parabolic function is typically written in the form:
$Y = ax^2 + bx + c$ where $x$ represents the years and $Y$ represents the corresponding values.

Since years are large numbers, we will redefine $x$ as: $x = \text{Year} - 2006$ so that our new $x$ values become:

$x = [0, 1, 2, 3, 4, 5, 6]$

The corresponding $Y$ values remain the same:
$Y = [56.6, 61.2, 69.48, 75.66, 70.44, 59, 57.2]$

We will fit a quadratic regression to this data to find $a$, $b$, and $c$.

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2. Calculating the Standard Deviation

The standard deviation ($\sigma$) is calculated using:

$\sigma = \sqrt{\frac{\sum (Y_i - \bar{Y})^2}{n}}$

where:

• $Y_i$ are the individual values,
• $\bar{Y}$ is the mean of $Y$,
• $n$ is the number of values.

Let's compute both now. numpy as np

Given data

x_values = np.array([0, 1, 2, 3, 4, 5, 6]) # Shifted years y_values = np.array([56.6, 61.2, 69.48, 75.66, 70.44, 59, 57.2]) # Y values

Perform quadratic regression (fit a parabola)

coefficients = np.polyfit(x_values, y_values, 2) # Fits Y = ax^2 + bx + c

Compute standard deviation

mean_y = np.mean(y_values) # Mean of Y values std_dev = np.sqrt(np.sum((y_values - mean_y) ** 2) / len(y_values))

Extract coefficients

a, b, c = coefficients (a, b, c), std_dev### Results:

1. Equation of the Parabola:
   $Y = -1.826x^2 + 10.899x + 55.270$ where $x = \text{Year} - 2006$.

2. Standard Deviation:
   $\sigma \approx 6.979$

Would you like a detailed explanation of the steps or further analysis?

Related Questions:

1. How do we determine if the quadratic model is a good fit for the data?
2. How can we use this model to predict future values?
3. What is the significance of the standard deviation in this context?
4. How would adding more data points impact the accuracy of the model?
5. How does this quadratic regression compare to a linear regression for the same data?

Tip:

In regression analysis, a lower standard deviation indicates that the data points are closer to the fitted curve, meaning a better fit.