Question 1180787
Here's how to solve this probability problem using the normal approximation to the binomial distribution:

**1. Define the parameters:**

* p = 0.6 (probability of the event in a single trial)
* n = 1000 (number of trials)
* ε = 0.05 (maximum allowed deviation of the relative frequency from the probability)

**2.  Calculate the mean and standard deviation of the number of successes:**

Since we have a large number of independent trials, we can use the normal approximation to the binomial distribution.

* Mean (μ) = n * p = 1000 * 0.6 = 600
* Standard deviation (σ) = sqrt(n * p * (1 - p)) = sqrt(1000 * 0.6 * 0.4) = sqrt(240) ≈ 15.49

**3. Convert the problem to a standard normal distribution problem:**

We want to find the probability that the relative frequency (which is the number of successes divided by the number of trials) deviates from p by less than ε.  This can be written as:

P(|(number of successes / n) - p| < ε)

Or, in terms of the number of successes (x):

P(|x/n - p| < ε)

Multiplying by n:
P(|x - np| < nε)

Substituting our values:
P(|x - 600| < 1000 * 0.05)
P(|x - 600| < 50)

This is equivalent to:

P(550 < x < 650)

Now, we convert to z-scores:

z_lower = (550 - 600) / 15.49 ≈ -3.23
z_upper = (650 - 600) / 15.49 ≈ 3.23

**4. Find the probability using the standard normal distribution table or calculator:**

We want to find P(-3.23 < z < 3.23).  This is equal to:

P(z < 3.23) - P(z < -3.23)

Using a standard normal table or calculator:

P(z < 3.23) ≈ 0.9994
P(z < -3.23) ≈ 0.0006

Therefore, P(-3.23 < z < 3.23) ≈ 0.9994 - 0.0006 ≈ 0.9988

**Answer:**

The probability that the relative frequency of occurrence of the event deviates from its probability by less than 0.05 is approximately 0.9988.