Statistics and Probability

1. a) As the number of hours studied increases, the test score also increases. This is a positive association — the data shows a clear upward trend.

b) The data points are fairly tightly clustered along a linear pattern with no obvious outliers. There is a cluster of students in the $2$–$4$ hour range with scores between $58$ and $78$. No points deviate dramatically from the overall trend.

2. a) Substitute $x = 4.5$:

$y = 8(4.5) + 42 = 36 + 42 = 78$

The predicted score is $78$.

b) Set $y = 60$ and solve for $x$:

$60 = 8x + 42$ $18 = 8x$ $x = \frac{18}{8} = 2.25$

The model predicts the student studied approximately $2.25$ hours (about $2$ hours and $15$ minutes).

c) The slope of $8$ means that for each additional hour of studying, the predicted test score increases by $8$ points. It represents the rate of change of test score with respect to study time.

3. a) Total students who prefer soccer: $40$ out of $100$.

$\frac{40}{100} = 40\%$

b) 8th graders who prefer swimming: $15$ out of $35$ total 8th graders.

$\frac{15}{35} = \frac{3}{7} \approx 0.4286 = 42.9\%$

4. Divide each cell by its row total:

| | Basketball | Soccer | Swimming | |---|---|---|---| | 6th Grade | $\frac{15}{30} = 50\%$ | $\frac{10}{30} = 33.3\%$ | $\frac{5}{30} = 16.7\%$ | | 7th Grade | $\frac{12}{35} = 34.3\%$ | $\frac{18}{35} = 51.4\%$ | $\frac{5}{35} = 14.3\%$ | | 8th Grade | $\frac{8}{35} = 22.9\%$ | $\frac{12}{35} = 34.3\%$ | $\frac{15}{35} = 42.9\%$ |

Yes, preferences differ by grade level. Basketball is most popular among 6th graders ($50\%$), soccer is most popular among 7th graders ($51.4\%$), and swimming is most popular among 8th graders ($42.9\%$). There is a shift in preference from basketball to swimming as grade level increases.

5. a) Slope ($-15$): For each \1$increase in price,$15$ fewer units are predicted to be sold. This reflects an inverse relationship between price and demand.

$y$-intercept ($300$): When the price is \0$(free), the model predicts$300$ units would be "sold." This is a theoretical starting value and may not be realistic.

b) Substitute $x = 12$:

$y = -15(12) + 300 = -180 + 300 = 120$

The model predicts $120$ units would be sold at \12$.

c) Set $y = 0$ and solve:

$0 = -15x + 300$ $15x = 300$ $x = 20$

The model predicts zero units sold at a price of \20$.

d) At $x = 25$:

$y = -15(25) + 300 = -375 + 300 = -75$

The model predicts $-75$ units, which is impossible (you cannot sell a negative number of products). The linear model is only valid within a reasonable domain — in this case, roughly \0$to$$20$. Beyond$$20$, the model breaks down and should not be used for predictions. This is an example of extrapolation beyond the range of the data.