A kayak-rental company needs to replace many of its kayaks, and it wants to ensure it has enough kayaks to meet the demand from customers during the summer season. On average, it rents out 42 kayaks each day with a standard deviation of 4. The company has 45 kayaks. Use a calculator or spreadsheet program to find the probability that the company will have enough kayaks on any given day. Round the answer to the nearest tenth.(1 point) The probability that there are enough kayaks on any given day is

1 answer

To determine the probability that the kayak rental company has enough kayaks on any given day, we can model the number of kayaks rented out each day as a normal distribution.

Given:

  • Mean (μ) = 42 kayaks per day
  • Standard deviation (σ) = 4 kayaks
  • Number of kayaks available = 45

We want to find the probability that the number of kayaks rented out (X) is less than or equal to 45:

\[ P(X \leq 45) \]

To find this probability, we first need to standardize the value using the z-score formula:

\[ z = \frac{X - \mu}{\sigma} \]

Substituting the values we have:

\[ z = \frac{45 - 42}{4} = \frac{3}{4} = 0.75 \]

Next, we look up the z-score of 0.75 in the standard normal distribution table or use a calculator or spreadsheet to find the corresponding probability.

Using a standard normal distribution calculator, we find:

\[ P(Z \leq 0.75) \approx 0.7734 \]

This means that the probability of renting out 45 or fewer kayaks on any given day is approximately 0.7734.

Rounding this to the nearest tenth, we have:

\[ \text{Probability that there are enough kayaks available} \approx 0.8 \]

Thus, the probability that the company will have enough kayaks on any given day is 0.8 (or 80%).