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Mrs. Ryan observed the relationshipbetween the number of tutorialsattended and semester grade for eachof the students in her Algebra Iclasses. The table below shows some ofthe resulting data.NumberotTutorialsAttended9SemesterGrade99491006378857761Which of the following shows anappropriate equation for the line ofbest fit?

Mrs. Ryan observed the relationshipbetween the number of tutorialsattended and semester-example-1

1 Answer

6 votes
Answer:

y = 72.35 + 2.95x

Explanations:

Let the number of tutorials attended be represented by x

Let the semester grade be represented by y

x = { 9, 4, 0, 0, 6, 1}

y = {99, 91, 63, 78, 85, 77}

Calculate the mean of x


\begin{gathered} \bar{X}\text{ = }(9+4+0+0+0+6+1)/(5) \\ \bar{X}\text{ =}(20)/(6) \\ \bar{X}\text{ =}3.33 \end{gathered}

Calculate the mean of y


\begin{gathered} \bar{Y}\text{ =}(99+91+63+78+85+77)/(6) \\ \bar{Y}\text{ = }(493)/(6) \\ \bar{Y}\text{ =}82.17 \end{gathered}
m\text{ = }\frac{\sum ^n_(i\mathop=1)(x_i-\bar{X}\text{ )(y}_i-\bar{Y)}\text{ }}{\sum ^n_{i\mathop{=}1}(x_i-\bar{X}\text{ )}^2}
\begin{gathered} m\text{ = }((9-3.33)(99-82.17)+(4-3.33)(91-82.17)+(0-3.33)(63-82.17)+(0-3.33)(78-82.17)+(6-3.33)(85-82.17)+(1-3.33)(77-82.17))/(\mleft(9-3.33\mright)^2+\mleft(4-3.33\mright)^2+(0-3.33)^2+(0-3.33)^2+(6-3.33)^2+(1-3.33)^2) \\ \text{m = }\frac{198.67}{^{}67.33} \\ \text{m = }2.95 \end{gathered}

Calculate the y-intercept using the formula below:


\begin{gathered} b\text{ = }\bar{Y}-m\bar{X}\text{ } \\ \text{b = 82.17-2.95(3.33)} \\ \text{b = 82.17-}9.82 \\ b\text{ = }72.35 \end{gathered}

The equation of a line is:

y = mx + b

Substitute m = 2.95 and b = 72.35 into the equation of a line given above:

y = 2.95x + 72.35

User Satyabrat Sahoo
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