To find the probability that the kayak rental company has enough kayaks on any given day, we need to model the daily demand for kayaks, which is given as normally distributed with a mean (\(\mu\)) of 42 kayaks and a standard deviation (\(\sigma\)) of 4 kayaks.
The company has 45 kayaks, so we want to find the probability that the demand (X) is less than or equal to 45 kayaks. This can be represented as:
\[ P(X \leq 45) \]
To do this, we first need to calculate the z-score for 45 kayaks. The z-score is calculated using the following formula:
\[ z = \frac{X - \mu}{\sigma} \]
Substituting the values into the formula:
\[ z = \frac{45 - 42}{4} = \frac{3}{4} = 0.75 \]
Next, we will look up this z-score in the standard normal distribution table or use a calculator to find the probability corresponding to \(z = 0.75\).
Using a standard normal distribution table or calculator, we find:
\[ P(Z \leq 0.75) \approx 0.7734 \]
This means that the probability that the demand for kayaks does not exceed 45 is approximately 0.7734, or 77.34%.
Rounding this to the nearest tenth:
\[ P(X \leq 45) \approx 77.3% \]
Thus, the probability that the company will have enough kayaks on any given day is 77.3%.