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3 years ago

A random sample of house sizes in major city has a sample mean of

Mathematics

1 answer:

Kazeer [188]3 years ago

5 0

Answer:

A random sample of house sizes in major city has a sample mean of ¯x=1204.9 x ¯ = 1204.9 sq ft and sample standard deviation of s=124.6 sq ft. Use the Empirical Rule to determine the approximate percentage of house sizes that lie between 955.7 and 1454.1 sq ft.:

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Answer:yes

Step-by-step explanation:

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3 years ago

#10 i The table shows the admission costs (in dollars) and the average number of daily visitors at an amusement park each the pa

lions [1.4K]

The line of best fit is a straight line that can be used to predict the

average daily attendance for a given admission cost.

Correct responses:

The equation of best fit is; $\underline{ \hat Y = 1,042 - 4.9 \cdot X_i}$

The correlation coefficient is; r ≈<u> -0.969</u>

<h3>Methods by which the line of best fit is found</h3>

The given data is presented in the following tabular format;

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|}Cost, (dollars), x&20&21&22&24&25&27&28&30\\Daily attendance, y&940&935&940&925&920&905&910&890\end{array}\right]$

The equation of the line of best fit is given by the regression line

equation as follows;

$\hat Y = \mathbf{b_0 + b_1 \cdot X_i}$

Where;

$\hat Y$ = Predicted value of the<em> i</em>th observation

b₀ = Estimated regression equation intercept

b₁ = The estimate of the slope regression equation

$X_i$ = The <em>i</em>th observed value

$b_1 = \mathbf{\dfrac{\sum (X - \overline X) \cdot (Y - \overline Y) }{\sum \left(X - \overline X \right)^2}}$

$\overline X$ = 24.625

$\overline Y$ = 960.625

$\mathbf{\sum(X - \overline X) \cdot (Y - \overline Y)} = -433.125$

$\mathbf{\sum(X - \overline X)^2} = 87.875$

Therefore;

$b_1 = \mathbf{\dfrac{-433.125}{87.875}} \approx -4.9289$

Therefore;

The slope given to the nearest tenth is b₁ ≈ -4.9

$b_0 = \mathbf{\dfrac{\left(\sum Y \right) \cdot \left(\sum X^2 \right) - \left(\sum X \right) \cdot \left(\sum X \cdot Y\right)} {n \cdot \left(\sum X^2\right) - \left(\sum X \right)^2}}$

By using MS Excel, we have;

n = 8

∑X = 197

∑Y = 7365

∑X² = 4939

∑Y² = 6782675

∑X·Y = 180930

(∑X)² = 38809

Therefore;

$b_0 = \dfrac{7365 \times 4939-197 \times 180930}{8 \times 4939 - 38809} \approx \mathbf{1041.9986}$

The y-intercept given to the nearest tenth is b₀ ≈ 1,042

The equation of the line of best fit is therefore;

$\underline{\hat Y = 1042 - 4.9 \cdot X_i}$

The correlation coefficient is given by the formula;

$\displaystyle r = \mathbf{\dfrac{\sum \left(X_i - \overline X) \cdot \left(Y - \overline Y \right)}{ \sqrt{\sum \left(X_i - \overline X \right)^2 \cdot \sum \left(Y_i - \overline Y \right)^2} }}$