The problem asks us to compute the line of best fit (regression line) predicting success (denoted by $Y$) from study times (denoted by $X$). We are given the following summary statistics: Mean of $X$, $\bar{X} = 1.61$ Standard deviation of $X$, $s_x = 1.12$ Mean of $Y$, $\bar{Y} = 2.95$ Standard deviation of $Y$, $s_y = 0.99$ Correlation coefficient, $r = 0.65$

1. Problem Description

The problem asks us to compute the line of best fit (regression line) predicting success (denoted by $Y$) from study times (denoted by $X$). We are given the following summary statistics:

Mean of $X$, $\bar{X} = 1.61$

Standard deviation of $X$, $s_x = 1.12$

Mean of $Y$, $\bar{Y} = 2.95$

Standard deviation of $Y$, $s_y = 0.99$

Correlation coefficient, $r = 0.65$

2. Solution Steps

The equation for the line of best fit is given by:

$Y = a + bX$

where $b$ is the slope and $a$ is the y-intercept.

First, we calculate the slope $b$ using the formula:

$b = r \cdot \frac{s_y}{s_x}$

Plugging in the given values:

$b = 0.65 \cdot \frac{0.99}{1.12} = 0.65 \cdot 0.8839285714 \approx 0.5745535714$

Next, we calculate the y-intercept $a$ using the formula:

$a = \bar{Y} - b \cdot \bar{X}$

Plugging in the given values:

$a = 2.95 - 0.5745535714 \cdot 1.61 = 2.95 - 0.92503125 \approx 2.02496875$

Rounding $a$ and $b$ to two decimal places, we have $a \approx 2.02$ and $b \approx 0.57$.

So, the line of best fit is approximately:

$Y = 2.02 + 0.57X$

3. Final Answer

The equation of the line of best fit is $Y = 2.02 + 0.57X$.