What are the solutions to these statistics problems: 62. If b = 2 and a = 5, predict ŷ when x = 4? 63. Class size: Range = 50, K = 5. 64. Sturges: N = 64.

$$\hat{y} = 13$$

The predicted value of $\hat{y}$ is 13.

The correct option is C.
$\boxed{\text{C. 13}}$

63. Class size: Range = 50, K = 5
The formula for class size (or class width) is given by:

$$\text{Class Size} = \frac{\text{Range}}{K}$$

$$\text{Class Size} = 10$$

The class size is 10.

The correct option is B.
$\boxed{\text{B. 10}}$

64. Sturges: N = 64
Sturges' formula for the number of classes (K) is:

$$K = 1 + 3.322 \log_{10}(N)$$

$$K = 1 + 3.322 \times 1.80618$$ $$K = 1 + 5.99999$$ $$K \approx 7.0$$

The number of classes is approximately 7.

The question asks for "Sturges: N=64" and provides options. Option B is the formula itself, which evaluates to approximately 7.
The correct option is B.
$\boxed{\text{B. } 1 + 3.322 \log 64}$

65. CI: $\bar{x} = 50$, $\sigma = 10$, $n = 25$, $Z = 1.96$. Margin of error =
The formula for the margin of error (ME) for a confidence interval for the mean when the population standard deviation ($\sigma$) is known is:

$$ME = Z \cdot \frac{\sigma}{\sqrt{n}}$$

$$ME = 3.92$$

The margin of error is 3.92.

The correct option is B.
$\boxed{\text{B. 3.92}}$

66. CI interval = $50 \pm 3.92$
The confidence interval (CI) is given by $\bar{x} \pm ME$.
Given: $\bar{x} = 50$ and $ME = 3.92$.

$$\text{Upper Bound} = \bar{x} + ME = 50 + 3.92 = 53.92$$

The confidence interval is $(46.08, 53.92)$.

The correct option is A.
$\boxed{\text{A. (46.08, 53.92)}}$