To find \( P(2) \), the probability of landing on the number 2 when Dean spins the spinner, we can use the formula for probability:
\[ P(2) = \frac{\text{Number of times 2 appears}}{\text{Total number of spins}} \]
Since the spinner is fair and has 4 equal sections (labeled 1, 2, 3, and 4), the probability of landing on any specific number, including 2, in a single spin is:
\[ P(2) = \frac{1}{4} \]
Now, since Dean spun the spinner a total of 60 times, the expected number of times that the spinner would land on 2 is:
\[ \text{Expected number of 2s} = 60 \times P(2) = 60 \times \frac{1}{4} = 15 \]
However, to find the probability \( P(2) \) for the spins already conducted, we still use the probability formula, which remains:
\[ P(2) = \frac{\text{Count of 2s}}{60} \]
If Dean records the actual counts in his experiment, we would use that to specify the count of 2s. But if the recorded count isn't available and we only consider the fundamentally calculated probability in a long-term frequency sense:
Thus,
\[ P(2) = \frac{1}{4} \quad \text{or} \quad 0.25 \]
This means that the probability \( P(2) \) remains \( 0.25 \) regardless of the results of the spins, as long as the spinner is fair and each outcome from the 4 sections is equally likely.